NCERT Solutions For Class 12 Maths Chapter 8

NCERT Solutions Class 12 Maths Application of Integrals

Students finding difficulties in their maths exam preparations can refer these NCERT solutions. We at BYJU’S provide free NCERT solutions for class 12 Maths chapter 8. These solutions are available in PDF format which can be easily downloaded and used by the students in their preparations.

NCERT Solutions for class 12 maths chapter 8 Application of Integrals includes a complete set of solved questions from the NCERT maths textbook. Students can practice more questions from these materials as it includes various solved MCQ, and both short and long type questions. NCERT Solutions for class 12 maths chapter 8 pdf covers solutions for various questions, including finding the area between two curves, determining the average value of a function, etc.

NCERT Solutions Class 12 Maths Chapter 8 Exercises

Area between two curves

CASE – 1:

1

For finding the Area bounded by the curve y = f(x), x-axis and the lines x = a, x = b let us consider a very thin vertical strip of length y and width dx. Therefore, Area of the strip (dA) = y dx [Since, y = f(x)]

Hence the total Area enclosed by the curve y = f(x), x- axis and the lines x = a, x = b:

\(\boldsymbol{A =\int_{a}^{b} dA=\int_{a}^{b}y\;dx}\)

Therefore, A = \(\int_{a}^{b}f(x)\;dx\)

CASE – 2:

2

Similarly the Area bounded by the curve x = g(y), y-axis and the lines y = c, y = d is given by:

\(A=\int_{c}^{d} dA=\int_{c}^{d} x\;dy\)

Therefore, A = \(\int_{c}^{d}g(y)\;dy\)

CASE – 3:

3

If the curve lies below x-axis, then the Area bounded by the curve y = f(x), x-axis and the lines x = a, x = b will come negative.

Since, the area cannot be negative therefore we will neglect the negative sign and considering its absolute value only.

\(A = \left |\int_{a}^{b}f(x)\;dx \right|\)

CASE – 4:

1

As shown above, some portion of the curve lies above the x-axis and some portion of the curve lies below the x-axis.

Here in this case, Area-1 > 0 and Area-2 < 0

Therefore, total Area bounded by the curve y= f(x), x-axis and the lines x=a, x=b is given by: A = |A|+B

NOTE:

\(\boldsymbol{\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}}\)

Example 1: Find the area enclosed by equation: x2 + y2 = 9

Sol:

Equation x2 + y2 = 32 represents the circle with centre (0,0) and radius 3 units.

Therefore, y = \(\sqrt{3^{2}-x^{2}}\)

5

From the figure, the Area enclosed by the circle = 4 × (Area enclosed by the curve ABOA)

Now, Area enclosed by the curve ABOA:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx = \int_{0}^{3} \sqrt{9-x^{2}}\;dx\)

Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin^{-1}\frac{x}{3} \right | _{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin^{-1}(1) – \frac{0}{2} \sqrt{9-0}-\frac{0}{2} \sin^{-1}\frac{0}{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9}{2}\times \frac{\pi }{2}=\frac{9\pi }{4}}\) unit2

Therefore, the Area enclosed by the circle = 4 × (area enclosed by curve ABOA)

\(\boldsymbol{\Rightarrow }\) \(4\times \frac{9\pi }{4}=9\pi\)

Hence the Area enclosed by the circle = 9π unit2

 

 

Example 2: Find the area enclosed by the curve \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)

Sol:

Equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\) represents an ellipse with major axis = 3 units and minor axis = 2 units

6Since, \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)

\(\boldsymbol{\Rightarrow }\) 4x2 + 9y2 = 36

\(\boldsymbol{\Rightarrow }\) 9y2 = 62 – (2x)2

Therefore, \(y=\sqrt{\frac{6^{2}-(2x)^{2}}{9}}\)

Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)

Now, Area enclosed by the curve ABOA = \(\int_{0}^{3} y\;dx\)

Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\times \int_{0}^{3}\sqrt{6^{2}-(2x)^{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\int_{0}^{3}\;\frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\left | \frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6} \right |_{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}[\frac{6}{2}\sqrt{36-36}+18\sin^{-1}(1)]=3\pi }\)

Therefore, the Area enclosed by the curve ABOA = 3π unit2

Hence, the total Area enclosed by an ellipse ABA’B’ = 4×3π = 12π unit2.

Area of region bounded by the curve and line:

Example 3: Find the area of the region enclosed by the curve y = x2 and the line y = 3.

Sol:

7

y = x2 represents a parabola, symmetrical about y-axis as shown in the above figure. By substituting y =9 in equation of parabola y = x2 we will get coordinates of point M.

i.e. x2 = 9

Therefore, x = +3 or -3

Hence the coordinates of point M = (3, 9)

The area of the region bounded by the curve y = x2 and the line y = 9 is the area enclosed by curve POMP.

Now, Area of region POMP = 2(area of region AOMA)

Since, x2 = y

Therefore, x = \(\sqrt{y}\)

Thus, the Area of region bounded by the curve AOMA:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{9}x\;dy\) = \(\int_{0}^{9}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{0}^{9}=\frac{2}{3}\times (9^{\frac{3}{2}})}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times 27 = 18}\)unit2

Therefore, the Area of region bounded by the curve AOMA = 18 unit2

Hence, the Area of region bounded by the curve POMP = 2 × (area of region bounded by curve AOMA) = 36 unit2

Example 4: Find the area enclosed by the curve x2 + y2 = 50, x-axis and the line y = x in the 1st quadrant.

Sol:

Equation x2 + y2 = 50 represents a circle with radius \(\sqrt{50}\) units.

Since y = x

Therefore x2 + x2 = 50 [for points of intersection of both the curves]

Hence, x = +5 and -5

Similarly, y = +5 and -5

Thus the coordinates of point B are (5, 5)

8

Form the figure, the area of region bounded by the curve x2 + y2 = 50, x = y and x – axis in the 1st quadrant is the Area enclosed by the curve OMABO.

i.e. Area of triangle OMB + Area under the curve MBAM.

Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;\;=\;\frac{1}{2}\times 5\times 5 =\frac{25}{2}\) unit2.

Now, the Area under the curve MBAM = \(\int_{5}^{5\sqrt{2}}y\;dx\)

Since, x2 +y2 = 50. Therefore, y2 = 50 – x2

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{50-x^{2}}}\)

Therefore, the Area under curve MBAM [x2 + y2 = 50]:

Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{5\sqrt{2}}\sqrt{50-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{50-x^{2}}+\frac{50}{2}\;\sin^{-1}\frac{x}{5\sqrt{2}} \right |_{5}^{5\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{5\sqrt{2}}{2}\times\ \sqrt{50-50}]+[\frac{50}{2}\;\sin^{-1}(1)]-[\frac{5}{2}\times \sqrt{50-25}]-[\frac{50}{2}\times \sin^{-1}\frac{1}{\sqrt{2}}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{0+\frac{25\pi }{2}-\frac{25}{2}-\frac{25\pi }{4}=\frac{25}{4}(\pi -2)}\)unit2

Therefore, the Area under the curve MBAM = \(\boldsymbol{\frac{25}{4}(\pi -2)}\) unit2

Now, the total Area under the shaded region = Area of triangle OMB + Area under the curve MBAM

\(\boldsymbol{\Rightarrow }\) \(\frac{25}{2}+\frac{25}{4}(\pi -2)=\frac{25}{2}+\frac{25\pi }{4}-\frac{25}{2}\boldsymbol{=\frac{25\pi }{4}}\)unit2

Hence, the Area of shaded region OMABO = \(\boldsymbol{\frac{25\pi }{4}}\)unit2

Exercise 8.1

Q.1: Find the area enclosed by the curve y = x2 and the lines y = 2, y = 4 and the y-axis.

Sol:

Equation y = x2 represents a parabola symmetrical about y-axis.

9

The Area of the region bounded by the curve y = x2, y = 2, and y = 4, is the Area enclosed by the curve AA’B’BA.

Now, the Area of region AA’B’BA = 2 (Area of region ABNMA)

Since, x2 = y

Therefore, x = \(\sqrt{y}\)

Thus, the Area of region bounded by the curve ABNMA [y = x2]:

\(\boldsymbol{\Rightarrow }\) \(\int_{2}^{4}x\;dy\) = \(\int_{2}^{4}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{2}^{4}=\frac{2}{3}\times [(4^{\frac{3}{2}})-(2)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times (8-2\sqrt{2}) = \frac{4}{3}(4-\sqrt{2})}\) unit2

Therefore, the Area of region bounded by the curve ABNMA = \(\boldsymbol{\frac{4}{3}(4-\sqrt{2})}\)unit2

Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)= \(\boldsymbol{\frac{8}{3}(4-\sqrt{2})}\)unit2

 

Q.2: Find the area enclosed by the curve y2 = 4x and lines x = 1, x = 3 and the x- axis in the first quadrant.

Sol:

10

Equation y2 = 4x represents a parabola, symmetrical about the x-axis as shown in the above figure.

The area of the region bounded by the curve y2 = 4x, x = 1, x = 3 and the x-axis is the area enclosed by the curve ABCDA.

Now, the Area of region bounded by the curve ABCDA:

Since, y2 = 4x

Therefore, y = \(2\sqrt{x}\)

Hence, the area of region bounded by the curve ABCDA [y2 = 4x]:

\(\boldsymbol{\Rightarrow }\) \(\int_{1}^{3}y\;dx\) = \(\int_{1}^{3}2\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=\frac{4}{3}\times [(3^{\frac{3}{2}})-(1)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4}{3}\times (3\sqrt{3}-1) = \frac{-4\;+\;12\sqrt{3}}{3}}\) unit2

Therefore, the area of region bounded by the curve ABCDA \(\boldsymbol{= \frac{-4\;+\;12\sqrt{3}}{3}}\)unit2

Q.3: Find the value of k if the line x = k divides the area enclosed by the curve y2 = 9x and the line x = 4 in to two equal parts.

Sol:

11

Equation y2 = 9x represents a parabola, symmetrical about the x-axis as shown in the above figure.

Since, the line x = k divides the Area OCBB’O in to two equal halves and the curve is symmetrical to x-axis. Therefore, Area of the region OADO = Area of the region ABCDA.

The Area of the region bounded by the curve y2 = 9x and the line x = k is the Area of region enclosed by the curve OADO.

Since, y2 = 9x

Therefore, y = \(3\sqrt{x}\)

Hence, the Area of region bounded by the curve OADO:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{k}y\;dx\) = \(\int_{0}^{k}3\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=2\times [(k^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2k^{\frac{3}{2}}}\)unit2

Therefore, the Area of region bounded by the curve OADO = \(\boldsymbol{2k^{\frac{3}{2}}}\)unit2

The Area of the region bounded by the curve y2 = 9x and the lines x = k and x = 4 is the Area under the curve ABCDA

Now, the Area of region bounded by the curve ABCDA [y2 = 9x]:

\(\boldsymbol{\Rightarrow }\) \(\int_{k}^{4}y\;dx\) = \(\int_{k}^{4}3\sqrt{x}\;dx\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{k}^{4}=2\times [(4^{\frac{3}{2}})-(k)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})}\) unit2

Therefore, the Area of region bounded by the curve ABCDA = \(\boldsymbol{\Rightarrow (16-2k^{\frac{3}{2}})}\) unit2

Since, the Area of region OADO = Area of region ABCDA [GIVEN]

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})=2k^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k^{\frac{3}{2}}=4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k = 4^{\frac{2}{3}}}\)

Therefore, the value of k = \(\boldsymbol{4^{\frac{2}{3}}}\)

Q.4: Find the area enclosed by the curve y2 = 16x, y-axis and the line y = 2.

Sol:

12

Equation y2 = 16x represents a parabola, symmetrical about the x-axis as shown in the above figure.

The Area of the region bounded by the curve y2 = 16x, y = 2 and the y-axis is the Area of region enclosed by the curve OABO.

Now, the Area of region enclosed by the curve OABO:

Since, y2 = 16x

Therefore, x = \(\frac{y^{2}}{16}\)

Thus, the Area of region bounded by the curve OABO:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{2}x\;dy\) = \(\int_{0}^{2}\frac{y^{2}}{16}\;dy\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{1}{16}\times \frac{y^{3}}{3}\right |_{0}^{2}=\frac{1}{16}\times [\frac{8}{3}-0}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}}\) unit2

Therefore, Area of region bounded by the curve OABO \(\boldsymbol{=\frac{1}{6}}\)unit2

Q.5: Find the area enclosed by the curve y2 = 25x and the line x = 3

Sol:

13

Equation y2 = 25x represents a parabola, symmetrical about x-axis as shown in the above figure.

The area of the region bounded by the curve y2 = 25x and x = 3 is the Area enclosed by the curve BOCAB.

Now, the Area of region BOCAB = 2(Area of region OABO)

Since, y2 = 25x

Therefore, y = \(5\sqrt{x}\)

Hence, the Area of region bounded by the curve OABO:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx\) = \(\int_{0}^{3}5\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 5\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{0}^{3}=\frac{10}{3}\times [(3^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{10}{3}\times (3\sqrt{3})=10\sqrt{3}}\)unit2

Therefore, the Area of region bounded by the curve OABO\(\boldsymbol{=10\sqrt{3}}\) unit2

Now, the Area of region BOCAB = 2 × (Area of region OABO) \(\boldsymbol{=20\sqrt{3}}\) unit2

Q.6: Find the area enclosed by the curve x2 = 8y, y = 1, y = 9 and the y-axis in the first quadrant.

Sol:

14

Equation x2 = 8y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The area of the region bounded by the curve x2 = 8y, y = 1, y = 9 and the first quadrant is the area enclosed by the curve ABDCA.

Now, the Area of the region ABDCA:

Since, x2 = 8y

Therefore, x = \(2\sqrt{2y}\)

Hence, the Area of region bounded by the curve ABDCA:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{1}^{9}x\;dy\) = \(\int_{1}^{9}2\sqrt{2y}\;dy\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\sqrt{2}\times \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{9}=\frac{4\sqrt{2}}{3}\times [(9^{\frac{3}{2}})-(1)}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4\sqrt{2}}{3}\times (27-1)=104\times \frac{\sqrt{2}}{3}}\) unit2

Therefore, Area of region bounded by the curve ABDCA \(\boldsymbol{=104\times \frac{\sqrt{2}}{3}}\)unit2

Q.7: Find the area bounded by the curve whose equation is x2 = 9y and the line x = 6y – 3.

Sol:

15

Equation x2 = 9y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola x2 = 9y and the line x =6y – 3 is the Area enclosed under the curve ABC0A.

Since, the parabola x2 = 9y and the line x = 6y – 3 intersect each other at points A and C, hence the coordinates of points A and C are given by:

(6y-3)2 = 9y

\(\boldsymbol{\Rightarrow }\) 36y2 – 45y + 9 = 0

By Hit and Trial method solutions of this quadratic equation are:

y = 1 and y = \(\frac{1}{4}\)

Hence, the co-ordinates of point A and point C are (3,1) and \((\frac{-3}{2},\frac{1}{4})\) respectively

Since, x2 = 9y

Therefore, x = \(3\sqrt{y}\)

The Area of region bounded by the curve ABCOA = [Area of region OBCbOArea of region OCbO] + [Area of region ABOaAArea of region OAaO]

The Area enclosed by the curve OBCbO:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3} y\;dx = \boldsymbol{\int_{0}^{3}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(\frac{9}{2}+9)=\frac{9}{4}}\) unit2

The Area enclosed by the curve ABOaA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx} = \boldsymbol{\int_{\frac{-3}{2}}^{0}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(0-\frac{9}{8}+\frac{9}{2})=\frac{9}{16}}\) unit2

The Area enclosed by the curve OAaO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx\; =\;\int_{\frac{-3}{2}}^{0} \frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{27}(0-\frac{-27}{8})=\frac{1}{8}}\)unit2

The Area enclosed by the curve OCbO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3} y\;dx\; = \;\int_{0}^{3}\frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}(\frac{27}{3}-0)=1}\) unit2

Since, the Area of region bounded by the curve ABCOA = [Area of region OBCbO – Area of region OCbO] + [Area of region ABOaA – Area of region OAaO]

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{9}{4}-1]+[\frac{9}{16}-\frac{1}{8}]=\frac{27}{16}}\)

Therefore, the Area of region bounded by the curve ABCOA \(=\frac{27}{16}\) unit2

 

 

Q.8: Find the area enclosed by the curve 4y = x2 and y = |x|

Sol:

16

Equation x2 = 4y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The area of the region bounded by the curve x2 = 4y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)

Now, Area of region OAEO = OABO – OEABO

Since, x2 = 4y

\(\boldsymbol{\Rightarrow }\) \(y=\frac{x^{2}}{4}\)

Now, the Area of region bounded by the curve OEABO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;=\;\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{4}\left | \frac{x^{3}}{3} \right |_{0}^{4}=\frac{16}{3}}\) unit2

Therefore, the area of region bounded by the curve OEABO = \(=\frac{16}{3}\) unit2

Now, the Area of region bounded by the curve OABO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;\Rightarrow \;\int_{0}^{4}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{4}=8}\) unit2

Therefore, the Area of the region bounded by the curve OABO = 8 unit2

Now, Area of region OAEO = Area of region (OABO – OEABO)

\(\boldsymbol{\Rightarrow }\) \(8-\frac{16}{3}\) = \(\frac{8}{3}\) unit2

Therefore, the total Area of shaded region = 2×\(\frac{8}{3}\) = \(\frac{16}{3}\)unit2

Q.9: Find the area enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)

Sol:

17

Equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) represents an ellipse.

Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)

Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)

Now, the Area enclosed by the curve ABOA:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2} y\;dx\;=\; \frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\)

Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) unit2

Therefore, the Area enclosed by the curve ABOA \(\boldsymbol{=\frac{3\pi}{2}}\) unit2

Hence, the total Area enclosed by the ellipse ABA’B’ = 4 × \(\boldsymbol{\frac{3\pi}{2}}\) unit2

= 6π unit2

Q.10: Find the area enclosed by the curve x2 + y2 = 9, line x = \(2\sqrt{2}y\) and the first quadrant.

Sol:

18

Equation x2 + y2 = 9 represents a circle with radius equal to 3units.

Since, x = \(2\sqrt{2}y\)

Therefore (\(2\sqrt{2}y\))2 + y2 = 9 (for points of intersection of both the curves)

Hence, the coordinates of point B = (\(2\sqrt{2}y\),1)

Now, the area of region bounded by the curve x2 + y2 = 9, x = \(2\sqrt{2}y\), and the first quadrant is the area enclosed by the curve OMABO.

i.e. Area of triangle OMB + Area under the curve MBAM.

Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;=\;\frac{1}{2}\times2\sqrt{2}y\times 1 =\sqrt{2}\;y\;\)unit2

Now, the Area under the curve MBAM:

\(\boldsymbol{\Rightarrow }\) \(\int_{2\sqrt{2}}^{3}\;y\;dx\)

Since, x2 +y2 =9

Therefore, y2 = 9 – x2

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{9-x^{2}}}\)

Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2\sqrt{2}}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\;\sin^{-1}\frac{x}{3} \right |_{2\sqrt{2}}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\;\boldsymbol{[\frac{3}{2}\times\ \sqrt{9-9}]+[\frac{9}{2}\;\sin^{-1}(1)]-[\frac{2\sqrt{2}}{2}\times \sqrt{9-8}]-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\) unit2

Therefore, the Area under the curve MBAM:

= \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit2

Now, total Area under the shaded region = Area under the curve MBAM + Area of the triangle OMB

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]+\sqrt{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\) unit2

Hence, the required Area is given by the region OMABO:

\(\boldsymbol{=\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit2

Q.11: Find the area of larger part of circle x2 + y2 = 16 cut off by the line x = \(\frac{4}{\sqrt{2}}\) in the first quadrant.

Sol:

19

Equation x2 + y2 = 16 represents a circle with radius = 4 units.

For coordinates of point B:

\(\boldsymbol{\Rightarrow }\) \((\frac{4}{\sqrt{2}})^{2}+y^{2}=16\)

\(\boldsymbol{\Rightarrow }\) 8 + y2 = 16

\(\boldsymbol{\Rightarrow }\) y = \(2\sqrt{2}\)

Hence, the coordinates of point B are: \(\boldsymbol{(\frac{4}{\sqrt{2}},2\sqrt{2})}\)

The required Area is given by the curve OMBCO:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2\sqrt{2}}y\;dx\;=\;\int_{0}^{2\sqrt{2}}\sqrt{16-x^{2}}\;dx}\)

Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1} \frac{x}{4} \right |_{0}^{2\sqrt{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8\sin^{-1}\frac{1}{\sqrt{2}}]-0}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2}\times 2\sqrt{2}+8\times \frac{\pi }{4}=[4+2\pi] }\) unit2

Therefore, the Area of shaded region OMBCO = [4 + 2π]unit2

 

 

20

Let us assume two curves represented by the equation y = f(x) and y = g(x) in [a, b] as shown in the above figure. In this case the height of an elementary strip will be [f(x) – g(x)] and its width will be dx.

Now, Area of the elementary strip (dA) = [f(x) – g(x)] dx

Hence, the total Area of shaded region (A) = \(\int_{a}^{b} [f(x)-g(x)]\;dx\)

 

 

Example – 1: Find the area enclosed between two curves whose equations are: x2 = 4y and y2 = 4x.

Sol:

21

The Equation x2 = 4y represents a parabola symmetrical about y-axis and the equation y2 = 4x represents a parabola symmetrical about x-axis.

On solving both the equations:

\(\boldsymbol{\Rightarrow }\) \(\left ( \frac{y^{2}}{4} \right )^{2}=4y\)

\(\boldsymbol{\Rightarrow }\) y3 = 64 i.e. y = 4

Which gives x = 4. Hence, the coordinates of point N are (4, 4).

Now, The Area of region enclosed by the curve NAOBN = Area of region enclosed by the curve OANMO – Area of region enclosed by the curve OBNMO.

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}2\sqrt{x}\;dx-\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\times \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{3}}{12} \right ]_{0}^{4}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4}{3}\times (4)^{\frac{3}{2}}-0 \right ]-\left [ \frac{64}{12}-0 \right ]= \frac{16}{3}}\) unit2

Therefore, the Area of shaded region = \(\frac{16}{3}\) unit2

 

 

EXAMPLE – 2: Find the area enclosed by the sides of a triangle whose vertices have coordinates (1, 0) (3, 5) and (5, 4).

Sol:

23

Form the above figure:

Let, A (1, 0), B (3, 5) and C (5, 4) be the vertices of triangle ABC.

Now, the equation of line AB:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{5\;-\;0}{3\;-\;1}\right]\)

\(\boldsymbol{\Rightarrow }\) 2y = 5x – 5

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{5x\;-\;5}{2}}\)

The Equation of line BC:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-5)=(x-3)\times \left[\frac{4-5}{5-3}\right]\)

\(\boldsymbol{\Rightarrow }\) 2y – 10 = 3 – x

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{13\;-\;x}{2}}\)

The Equation of line AC:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{4-0}{5-1}\right]\)

\(\boldsymbol{\Rightarrow }\) 4y = 4x – 4

\(\boldsymbol{\Rightarrow }\) y = x – 1

Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.

The Area under the curve ABMA [2y = 5x – 5]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}y\;dx\;=\;\int_{1}^{3}\frac{5x-5}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{4}-\frac{5x}{2} \right ]_{1}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}=5}\) unit2

Therefore, Area under the curve ABMA = 5 unit2

The Area under the curve MBCN [2y = 13 – x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}\frac{13-x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{13x}{2}-\frac{x^{2}}{4} \right ]_{3}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{65}{2}-\frac{25}{4}-\frac{39}{2}+\frac{9}{4}=9}\) unit2

Therefore, Area under curve MBCN = 9 unit2

The Area under the curve ACNA [y = x – 1]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{5}y\;dx\;=\;\int_{1}^{5}(x-1)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{2}-x \right ]_{1}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{2}-5-\frac{1}{2}+1=8}\) unit2

Therefore, Area under the curve ACNA = 8 unit2

Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCNArea under curve ACNA.

Therefore, the Area of triangle ABC = 5 + 9 – 8 = 6 unit2

Exercise – 8.2

Q.1: Find the area lying above the x-axis enclosed between two curves whose equations are given as: x2 + y2 = 6x and y2 = 3x.

Sol:

24

The Equation y2 = 3x represents a parabola symmetrical about x-axis.

The Equation x2 + y2 = 6x i.e. (x – 3)2 + y2 = 9 represents a circle with centre (3, 0) and radius 3 units.

Substituting the equation of parabola in equation of circle:

(x – 3)2 + 3x = 9 \(\boldsymbol{\Rightarrow}\) x2 + 9 – 6x + 3x = 9

\(\boldsymbol{\Rightarrow}\) x2 – 3x = 0 i.e. x = 0 or x = 3 which gives y = 0 and y = ± 3

Therefore, the coordinates of point A above the x-axis are (3, 3)

Now, the Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA

The Area of region bounded by the curve OQAMO [y2 = 3x]:

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{0}^{3}y\;dx=\int_{0}^{3}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times\left [ \frac{2x^{\frac{3}{2}}}{3} \right ]_{0}^{3}=\sqrt{3}\times \frac{2}{3}\times 3^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times \frac{2}{3}\times 3\sqrt{3}=6}\) unit2

Therefore, the Area of region bounded by the curve OQAMO = 6 unit2

Now, the Area of region bounded by the curve ABMA [(x – 3)2 + y2 = 9)]:

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{3}^{6}y\;dx=\int_{3}^{6}\sqrt{9-(x-3)^{2}}\;dx}\\\)

Since, \(\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\\\)

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [ \frac{x-3}{2}\sqrt{\left (3 \right )^{2}-\left ( x-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{x-3}{3} \right ]_{3}^{6}}\\\)

\(\boldsymbol{\Rightarrow \;\;\;\left [\frac {6-3}{2}\sqrt{{9}-\left (6-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{6-3}{3}\right]-\left [ \frac{x-3}{2}\;\sqrt{9-\left (3-3 \right )^{2}}\;+\frac{9}{2}\;\sin^{-1}\frac{3-3}{3}\;\right]}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [0 +\frac{9}{2}\times \sin^{-1}1-0\right ]=\frac{9}{2}\times \frac{\pi }{2}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\frac{9\pi }{4}}\) unit2

Therefore, the Area of region bounded by the curve ABMA \(=\frac{9\pi }{4}\) unit2

The Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA

\(\boldsymbol{\Rightarrow}\) \(6+\frac{9\pi }{4}=\frac{24+9\pi }{4}\) unit2

Therefore, the Area of shaded region (OQABO)\(=\frac{24+9\pi }{4}\) unit2

Q.2: Find the area enclosed between two curves: 9x2 + 9y2 = 4 and (x – \(\frac{2}{3}\))2 + y2 = \(\frac{4}{9}\)

Sol:

25

Equation 9x2 + 9y2 = 4 . . . . . . . (1), represents a circle with centre (0, 0) and radius 3 units.

Equation (x – \(\frac{2}{3}\) )2 + y2 = \(\frac{4}{9}\) . . . . . . . . . . .(2), represents a circle with centre (3, 0) and radius 3 units.

On solving equation (1) and equation (2), we will get:

\(\\\boldsymbol{\Rightarrow }\) \((x-\frac{2}{3})^{2}+\frac{4-9x^{2}}{9}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(x^{2}+\frac{4}{9}-\frac{4x}{3}+\frac{4}{9}-x^{2}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{4}{9}=\frac{4}{3}\;x\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{x=\frac{1}{3}}\) which gives \(\boldsymbol{y=\pm \frac{1}{\sqrt{3}}}\)

Therefore, the coordinates of points M and N are: \((x=\frac{1}{3})\;\;(y=\pm \frac{1}{\sqrt{3}})\)

Now, the Area enclosed by region BMONB = 2 × [Area enclosed by the curve BMOB].

And the Area enclosed by the curve BMOB = Area of region enclosed by the curve (MOPM+MPBM).

Now, the Area of region enclosed by the curve MPBM [9x2 + 9y2 = 4]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{1}{3}}^{\frac{2}{3}}y\;dx=\int_{\frac{1}{3}}^{\frac{2}{3}}\sqrt{\frac{4}{9}-x^{2}}\;dx}\)

Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-x^{2}}+\frac{4}{2\times 9}\sin^{-1}\frac{3x}{2} \right ]_{\frac{1}{3}}^{\frac{2}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac{1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\sin^{-1}(1) \right ] -\left [ \frac{1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\sin^{-1}\frac{1}{2}\right ] }\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{2}{9}\times \frac{\pi }{2} \right ]-\left [ \frac{1}{6}\times \frac{1}{\sqrt{3}}+\frac{2}{9}\times \frac{\pi }{6} \right ]=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]}\) unit2

Therefore, area of region enclosed by the curve MPBM \(=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\) unit2

Now, the Area of region enclosed by the curve MOPM \(\left [ \left ( x-\frac{2}{3} \right )^{2}+y^{2}=\frac{4}{9} \right ]\) :

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{3}}y\;dx=\int_{0}^{\frac{1}{3}}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}\;dx}\\\)

Since, \(\\\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x-\frac{2}{3}}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{x-\frac{2}{3}}{\frac{2}{3}} \right ]_{0}^{\frac{1}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{\frac{1}{3}-\frac{2}{3}}{2}\sqrt{\frac{4}{9}-\left ( \frac{1}{3}-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{\frac{1}{3}-\frac{2}{3}}{\frac{2}{3}} \right]- \left [ \frac{0-\frac{2}{3}}{2}\;\sqrt{\frac{4}{9}-\left ( 0-\frac{2}{3} \right )^{2}}\;+\frac{2}{9}\;\sin^{-1}\frac{0-\frac{2}{3}}{\frac{2}{3}}\; \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\times \sin^{-1}\frac{-1}{2} \right ]- \left [ \frac{-1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\times \sin^{-1}(-1) \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6\sqrt{3}}+\frac{2}{9}\times \frac{-\pi }{6} \right ]-\left [ \frac{2}{9}\times \frac{-\pi }{2} \right ]= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]}\) unit2

Therefore, the Area of region enclosed by the curve MOPM \(= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]\) unit2

Now, the Area enclosed by the curve BMOB = Area of region enclosed by the curve (MOPM+MPBM)

\(\boldsymbol{\Rightarrow }\) \(\left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ] +\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]}\) unit2

Now, the Area enclosed by the curve BMONB = 2 [Area enclosed by the curve BMOB] \(=2\times \left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]\) unit2

Therefore, the Area of shaded region = \(\left [ \frac{8\pi }{27}-\frac{2}{3\sqrt{3}} \right ]\) unit2

Q.3: Find the area lying above the x-axis enclosed between two curves whose equations are given as: 4x2 + 4y2 = 9 and x2 = 4y.

Sol:

26

The Equation x2 = 4y represents a parabola symmetrical about y-axis.

The Equation 4x2 + 4y2 = 9 i.e. x2 + y2 = \(\frac{3}{2}\) represents a circle with centre (0, 0) and radius \(\frac{3}{2}\) units.

Now, on substituting the equation of parabola in the equation of circle we will get:

4(4y) + 4y2 =9 i.e. 4y2 + 16y – 9 = 0

From the above quadratic equation: a = 4, b = 16 and c = -9

Substituting the values of a, b and c in quadratic formula:

\(\\\boldsymbol{\Rightarrow }\) \(\\y=\frac{-16+\sqrt{(16)^{2}-4(4\times-9)}}{2\times 4}\;and\;y=\frac{-16-\sqrt{(16)^{2}-4(4\times -9)}}{2\times 4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{256+144}}{8}\;and\;y=\frac{-16-\sqrt{256+144}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{400}}{8}\;and\;y= \frac{-16-\sqrt{400}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+20}{8}\;and\;y=\frac{-16-20}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(y = \frac{1}{2}\;\;and\;\;y = \frac{-9}{2}\)

Which gives x = \(\pm \;\sqrt{2}\) [Neglecting y = \(\frac{-9}{2}\) as it gives absurd results]

Therefore, the coordinates of point B are \(\left (\sqrt{2},\;\frac{1}{2} \right )\)

Now, Area of region bounded by the curve ODCBO = 2 × (Area of region bounded by the curve OBCO)

Now, Area of region bounded by curve OBCO = (Area of region bounded by the curve OCBMO + Area of region bounded by the curve OBMO)

Area of region bounded by the curve OCBMO [4x 2 + 4y2 = 9]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}\;\;dx}\)

Since, \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\frac{9}{4}- x^{2}}+\frac{9}{2\times 4}\sin^{-1}\frac{2\times x}{3} \right ]_{0}^{\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac {\sqrt{2}}{4}\times \sqrt{{\frac{9}{4}-2}}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right ]\\}\) unit2

Therefore, Area of region bounded by the curve ABMA = \(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]\) unit2

Area of region bounded by the curve OBMO [x2 = 4y]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{12} \right ]_{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{12}-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3\sqrt{2}}}\) unit2

Therefore, the area of region bounded by the curve OBMO = \(\frac{1}{3\sqrt{2}}\) unit2

Now, the Area of region bounded by curve OBCO = [Area of region bounded by the curve OCBMO – Area of region bounded by the curve OBMO]

\(\\\boldsymbol{\Rightarrow }\) \(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-\frac{1}{3\sqrt{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit2

Therefore, Area of region bounded by the curve OBCO = \(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit2

Now, the Area of region bounded by the curve ODCBO = 2 × [Area of region bounded by the curve OBCO]

\(\\\boldsymbol{\Rightarrow }\) \(2\times \left [\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\)

Therefore, the Area of shaded region = \(\left [ \frac{1}{2\sqrt{2}}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit2

Q.4: Find the area enclosed by the sides of a triangle whose vertices have coordinates (-2, 0) (3, 4) and (5, 2).

Sol:

27

Form the above figure:

Let, A (1, 0), B (3, 5) and C (5, 4) be the vertices of triangle ABC

Now, the equation of line AB:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)\;=\;(x+2)\times \left[\frac{4\;-\;0}{3\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) 5y = 4x + 8

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{4x\;+\;8}{5}}\)

The Equation of line BC:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-4)=(x-3)\times \left[\frac{2\;-\;4}{5\;-\;3}\right]\)

\(\\\boldsymbol{\Rightarrow }\) 2y – 8 = 6 – 2x

\(\\\boldsymbol{\Rightarrow }\) y = 7 – x

The Equation of line AC:

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)=(x+2)\times \left[\frac{2\;-\;0}{5\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) 7y = 2x + 4

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{ y= \frac{2\;x+4}{7}}\)

Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.

The Area under the curve ABMA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{3}y\;dx\;=\;\int_{-2}^{3}\frac{4x\;+\;8}{5}\;dx}\\\) = \(\boldsymbol{\left [ \frac{4x^{2}}{10}+\frac{8x}{5} \right ]_{-2}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{36}{10}+\frac{24}{5}-\frac{16}{10}-\frac{-16}{5}=10}\) unit2

Therefore, the Area under the curve ABMA = 10 unit2

The Area under the curve MBCN:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}=(7-x)\;dx}\\\) = \(\boldsymbol{\left [ 7x-\frac{x^{2}}{2} \right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{35-\frac{25}{2}- 21+\frac{9}{2}=6}\) unit2

Therefore, the Area under the curve MBCN = 6 unit2

The Area under the curve ACNA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{5}y\;dx\;=\;\int_{-2}^{5}\frac{2x\;+\;4}{7}\;dx}\\\) = \(\boldsymbol{\left [ \frac{2x^{2}}{14}+\frac{4x}{7}\right ]_{-2}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{7}+\frac{20}{7}-\frac{4}{7}-\frac{-8}{7}=7}\) unit2

Therefore, the Area under the curve ACNA = 7 unit2

Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA.

Therefore, the Area of triangle ABC = 10 + 6 – 7 = 9 unit2

Q.5: Find the area enclosed by the sides of a triangle whose equations are: y = 4x + 2, y = 3x + 2 and x = 5.

Sol:

28

From the above figure,

The Equation of side AC: y = 4x + 2 . . . . . . (1)

The Equation of side BC: y =3x + 2 . . . . . . (2)

And, x = 5 . . . . . . . . . . (3)

From equation (1) and equation (3):

y = 4(5) + 2 = 22 [Since, x = 5]

Therefore, the coordinates of point A are (5, 22).

From equation (2) and equation (3):

y = 3(5) + 2 = 17 [Since, x = 5]

Therefore, the coordinates of point B are (5, 17).

Now, substituting equation (1) in equation (2):

3x + 2 = 4x + 2 i.e. x = 0 and y = 2

Therefore, the coordinates of point C are (0, 2).

Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC.

The Area under the curve ACOMA [y = 4x + 2]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(4x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{50+10-0=60}\) unit2

Therefore, the Area under the curve ABMA = 60 unit2.

The Area under the curve MBCN [y = 3x + 2]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(3x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{75}{2}+10-0=47.5}\) unit2

Therefore, the Area under the curve MBCN = 47.5 unit2.

Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC

\(\\\boldsymbol{\Rightarrow }\) 60 – 47.5 = 12.5 unit2

Therefore, the Area of triangle ABC = 12.5 unit2

Q.6: Find the area enclosed by the curves y = x2 + 3, y = 2x, x = 2 and x =0.

Sol:

29

The Equation y = x2 + 3 represents a parabola symmetrical about y-axis.

On substituting equation of line x = 2 in the equation of parabola we will get the coordinates of point C i.e. (2, 7).

On substituting x = 2 in the equation of line y = 2x we will get the coordinates of point B i.e. (2, 4).

From the above figure,

The Area of region enclosed by the curve ODCBO = Area of region enclosed by the curve ODCAOArea of region enclosed by the curve OBAO

Now, the Area of region enclosed by the curve ODCAO [y = x2 + 3]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(x^{2}+3)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}+3x \right ]_{0}^{2}=\frac{8}{3}+6=\frac{26}{3}}\)unit2

Therefore, the Area of region enclosed by the curve ODCAO = \(\frac{26}{3}\) unit2

Now, the Area of region enclosed by the curve OBAO [y = 2x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [x^{2} \right ]_{0}^{2}=4}\) unit2

Therefore, the Area of region enclosed by the curve OBAO = 4 unit2

Now, the Area of region enclosed by the curve ODCBO = Area of region enclosed by the curve ODCAO – Area of region enclosed by the curve OBAO

\(\Rightarrow \frac{26}{3}-4=\frac{2}{3}\) unit2

Therefore, The Area of shaded region (ODCBO) = \(\frac{2}{3}\)unit2

Q.7: Find the area enclosed between the curve y2 = 6x and line y = 3x.

Sol:

30

Equation y2 = 6x represents a parabola, symmetrical about x-axis.

Now, substituting the Equation of line y = 3x in the equation of parabola:

(3x)2 = 6x \(\Rightarrow x = \frac{2}{3}\) which gives y = 2

Hence the coordinates of point A are \((\frac{2}{3},2)\)

The area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

Now, the Area enclosed by the curve OMABO [y2 = 6x]:

Since, y2 = 6x

Therefore, y = \(\sqrt{6x}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx = \int_{0}^{\frac{2}{3}}\sqrt{6x}\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{6}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{2}{3}}=\sqrt{6}\times \frac{2}{3}\times \left ( \frac{2}{3} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2} \times \sqrt{3}\times \frac{2}{3}\times 2\sqrt{2}\times \frac{1}{3\sqrt{3}}=\frac{8}{9}}\\\) unit2

Therefore, the Area enclosed by the curve OMABO = \(\frac{8}{9}\) unit2

Now, the Area enclosed by the curve OAMO [y = 3x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx\Rightarrow \int_{0}^{\frac{2}{3}}3x\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\left [\frac{x^{2}}{2}\right ]_{0}^{\frac{2}{3}}=\frac{3}{2}\times \left ( \frac{2}{3} \right )^2}\) = \(\boldsymbol{\frac{2}{3}}\) unit2

Therefore, the Area enclosed by the curve OAMO = \(\frac{2}{3}\) unit2

Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8}{9}-\frac{2}{3} = \frac{2}{9}\) unit2

Therefore, the Area enclosed by the curve OABO = \(\frac{2}{9}\) unit2

Miscellaneous Exercise

 

Q.1: Find the area enclosed by the curve whose equations are: y = 2x2, x = 2, x = 3 and x-axis.

Sol:

31

The equation y = 2x2 represents a parabola symmetrical about y-axis.

Now, the Area of region enclosed by the curve ABCDA [y = 2x2]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{3}y\;dx=\int_{2}^{3}2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2x^{3}}{3} \right ]_{2}^{3}=\left [ 18-\frac{16}{3} \right ]=\frac{38}{3}}\) unit2

Therefore, the Area of shaded region \(=\frac{38}{3}\) unit2

Q.2: Find the area enclosed by the curve whose equations are: y = 5x4, x = 3, x = 7 and x-axis.

Sol:

32

The equation y = 2x2 represents a quartic parabola symmetrical about y-axis.

Now, the Area of region enclosed by the curve ABCDA [y = 5x4]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx=\int_{3}^{7}5x^{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{5}}{5} \right ]_{3}^{7}=\left [ 16807-243 \right ]=16564}\) unit2

Therefore, the Area of shaded region = 16807 unit2

Q.3: Find the area enclosed between the curve y2 = 3x and line y = 6x.

Sol:

33

Equation y2 = 3x represents a parabola, symmetrical about x-axis.

Now, substituting the Equation of line y = 6x in the equation of parabola:

(6x)2 = 3x \(\Rightarrow x = \frac{1}{12}\) which gives y = \(\frac{1}{2}\)

Hence the coordinates of point A are \((\frac{1}{12},\frac{1}{2})\)

The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

Now, the Area enclosed by the curve, OMABO [y2 = 3x]:

Since, y2 = 3x

Therefore, y = \(\sqrt{3x}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{12}} y\;dx = \int_{0}^{\frac{1}{12}}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{1}{12}}=\sqrt{3}\times \frac{2}{3}\times \left ( \frac{1}{12} \right )^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3} \times \frac{2}{3}\times \frac{1}{24\sqrt{3}}=\frac{1}{36}}\) unit2

Therefore, the Area enclosed by the curve OMABO = \(\frac{1}{36}\)unit2

Now, the Area enclosed by the curve OAMO [y = 6x]:

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{48}}\)unit2

Therefore, the Area enclosed by the curve OAMO = \(\frac{1}{48}\)unit2

Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{36}-\frac{1}{48} = \frac{1}{144}\)unit2

Therefore, the Area enclosed by the curve OABO = \(\frac{1}{144}\)unit2

Q.4 Find the area enclosed by the curve y = 2x2 and the lines y = 1, y = 3 and the y-axis.

Sol:

34

Equation y = 2x2 represents a parabola symmetrical about y-axis.

The Area of the region bounded by the curve y = 2x2, y = 1, and y = 3, is the Area enclosed by the curve AA’B’BA.

Now, the Area of region AA’B’BA = 2 (Area of region ABNMA)

Since, 2x2 = y

Therefore, x = \(\sqrt{\frac{y}{2}}\)

Thus, the Area of region bounded by the curve ABNMA [y = 2x2]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}x\;dy=\int_{1}^{3}\sqrt{\frac{y}{2}}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{\sqrt{2}}\times {\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{3}=\frac{\sqrt{2}}{3}\times [(3^{\frac{3}{2}})-(1)^{\frac{3}{2}}}}]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{\sqrt{2}}{3}\times (3\sqrt{3}-1) = \frac{\sqrt{2}}{3}(3\sqrt{3}-1)}\)unit2

Therefore, the Area of region bounded by the curve ABNMA = \(\frac{\sqrt{2}}{3}(3\sqrt{3}-1)\)unit2

Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)\(= \frac{2\sqrt{2}}{3}(3\sqrt{3}-1)\)unit2

 

 

Q.5: Find the area enclosed between the curve y2 = 9ax and line y = mx.

Sol:

35

Equation y2 = 9ax represents a parabola, symmetrical about x-axis.

Now, substituting the Equation of line y = mx in the equation of parabola:

9ax = (mx)2 i.e x = \(\frac{9a}{m^{2}}\) which gives y = \(\frac{9a}{m}\)

Hence the co-ordinates of point A are \(\left ( \frac{9a}{m^{2}},\frac{9a}{m}\right )\)

The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

Now, the Area enclosed by the curve OMABO [y2 = 9ax]:

Since, y2 = 9ax

Therefore, y = \(3\sqrt{ax}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{9a}{m^{2}}} y\;dx = \int_{0}^{\frac{9a}{m^{2}}}3\sqrt{ax}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\sqrt{a}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{9\;a}{m^{2}}}=3\sqrt{a}\times \frac{2}{3}\times \left ( \frac{9\;a}{m^{2}} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2\sqrt{a} \times 27\times a\sqrt{a}\times \frac{1}{m^{3}}=\frac{54\;a^{2}}{m^{3}}}\) unit2

Therefore, the Area enclosed by the curve OMABO \(=\frac{54\;a^{2}}{m^{3}}\) unit2

Now, the Area enclosed by the curve OAMO [y = mx]:

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{9\;a}{m^{2}}\times \frac{9\;a}{m}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{81\;a^{2}}{2\;m^{3}}}\) unit2

Therefore, the Area enclosed by the curve OAMO =\(\frac{81\;a^{2}}{2\;m^{3}}\) unit2

Now, the Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

i.e. \(\frac{54\;a^{2}}{m^{3}}-\frac{81\;a^{2}}{2\;m^{3}} = \frac{27\;a^{2}}{2\;m^{3}}\) unit2

Therefore, the Area enclosed by the curve OABO = \(\frac{27\;a^{2}}{2\;m^{3}}\)unit2

 

Q.6: Find the area bounded by the curve whose equation is 3x2 = 4y and the line 2y – 12 = 3x.

Sol:

36

Equation 3x2 = 4y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola 3x2 = 4y and the line 2y -12 = 3x is the Area enclosed under the curve ABC0A.

Since, the parabola 3x2 = 4y and the line 3x = 2y – 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:

Since, \(\;x=\frac{2y-12}{3}\)

\(\\\boldsymbol{\Rightarrow }\) \(3\left ( \frac{2y-12}{3} \right )^{2}=4y\;\;\;i.e. \;\;\;(2y-12)^{2}=12y\)

\(\\\boldsymbol{\Rightarrow }\) 4y2 +144 – 48y – 12 y = 0

\(\\\boldsymbol{\Rightarrow }\) y2 – 15y + 36 = 0

By splitting the middle term Method solutions of this quadratic equation are:

y2 – (12+3)y + 36 = 0 \(\Rightarrow\) y(y – 12) –3(y – 12) = 0

\(\Rightarrow\) (y – 3) (y – 12) = 0

Therefore, y = 12 and y = 3 which gives x = 4 and x = -2 respectively.

Hence, the co-ordinates of point A and point C are (-2, 3) and (4, 12) respectively.

Since, 3x2 = 4y

Therefore, y = \(\frac{3x^{2}}{4}\)

The Area of region bounded by the curve ABCOA = [Area of region aACba] – [Area of region OCbO + Area of region OAaO ]

The Area enclosed by the curve aACBa [2y – 12 = 3x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{4} y\;dx \Rightarrow \int_{-2}^{4}\frac{3x+12}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left | \frac{3x^{2}}{2}+ 12x \right |_{-2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left [ 24+48-6-(-24) \right ]=45}\) unit2

The Area enclosed by the curve OAaO [3x2 = 4y]:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{-2}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(0-\frac{-8}{3}) \right |=2}\) unit2

The Area enclosed by the curve OCbO [3x2 = 4y]:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}\;y\;dx\;\Rightarrow \int_{0}^{4} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{0}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(\frac{64}{3}-0) \right |=16}\) unit2

Now, the Area of region bounded by the curve ABCOA = [ Area of region aACba ] – [ Area of region OCbO + Area of region OAaO ]

\(\\\boldsymbol{\Rightarrow }\) 45 – [2 + 16] = 27 unit2

Therefore, the Area of shaded region ABCOA = 27 unit2

Q.7: Find the area enclosed by the curves {(x , y) : 6y x2 and y = |x|}

Sol:

37

Equation x2 = 6y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by the curve x2 = 6y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)

Now, Area of region OAEO = OABO – OEABO

Since, x2 = 6y

Therefore, \(y=\frac{x^{2}}{6}\)

Now, the Area of region bounded by the curve OEABO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}\frac{x^{2}}{6}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{3}}{3} \right |_{0}^{6}=12}\) unit2

Therefore, the Area of region bounded by the curve OEABO = 12 unit2

Now, the Area of region bounded by the curve OABO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{6}=18}\) unit2

Therefore, the Area of the region bounded by the curve OABO = 18 unit2

Now, Area of region OAEO = Area of region (OABO – OEABO)

\(\\\boldsymbol{\Rightarrow }\) 18 – 12 = 6 unit2

Therefore, the total Area of shaded region = 2 × 6 = 12 unit2

Q.8: Find the area enclosed by the sides of a triangle whose vertices have coordinates (3, 0) (5, 8) and (7, 5).

Sol:

38

Form the above figure:

Let, A (3, 0), B (5, 8) and C (7, 5) be the vertices of triangle ABC.

Now, the equation of line AB:

Since, \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{8-0}{5-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y = 8x – 24

\(\boldsymbol{\Rightarrow }\) y=4x-12

The Equation of line BC:

Since, \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-8)=(x-5)\times \left(\frac{5-8}{7-5}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y – 16 = -3x + 15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{31-3x}{2}}\)

The Equation of line AC:

Since, \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{5-0}{7-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 4y=5x-15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{5x-15}{4}}\)

Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.

The Area under the curve ABMA [y = 4x – 12]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}(4x-12)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}-12x\right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[50-60]-[18-36]=8}\)unit2

Therefore, Area under the curve ABMA = 8 unit2

The Area under the curve MBCN [2y + 3x = 31]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{7}y\;dx\;=\;\int_{5}^{7}\frac{31-3x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times \left [ 31x-\frac{3x^{2}}{2}\right ]_{5}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{217}{2}-\frac{147}{4}]-[\frac{155}{2}-\frac{75}{4}]=13}\) unit2

Therefore, Area under the curve MBCN = 13 unit2

The Area under the curve ACNA [4y = 5x – 15]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx\;=\;\int_{3}^{7}\left ( \frac{5x-15}{4} \right )dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{8}-\frac{15x}{4} \right ]_{3}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{245}{8}-\frac{105}{4} \right ]-\left [ \frac{45}{8}-\frac{45}{4} \right ]=10}\) unit2

Therefore, Area under the curve ACNA = 10 unit2

Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA.

Therefore, the Area of triangle ABC = 8 + 13 – 10 = 11 unit2

Q.9: Find the area enclosed by the sides of a triangle whose equations are: 2x – 4 = y, – 2y = -3x + 6 and x – 3y = -5.

Sol:

39

From the above figure:

The Equation of line AB: 3y = x + 5 . . . . . . (1)

The Equation of line BC: y = 4 – 2x . . . . . . (2)

The Equation of line AC: 2y = 3x – 6 . . . . . . (3)

From equation (1) and equation (2):

3(4 – 2x) = x + 5 i.e. x = 1, which gives y = 2

Therefore, the coordinates of point B are (1, 2)

From equation (2) and equation (3):

2(4 – 2x) = 3x – 6 i.e. x = 2 which gives y = 0

Therefore, the coordinates of point C are (2, 0).

From equation (1) and equation (3):

2y = 3(3y – 5) – 6 i.e. y = 3 which gives x = 4

Therefore, the coordinates of point A are (4, 3).

Now, the Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA

The Area under the curve ABMNA [3y = x + 5]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{4}y\;dx\;=\;\int_{1}^{4}\frac{x+5}{3}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{6}+\frac{5x}{3} \right ]_{1}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{16}{6}+\frac{20}{3}-\frac{1}{6}-\frac{5}{3}=\frac{15}{2}}\) unit2

Therefore, the Area under the curve ABMNA \(=\frac{15}{2}\) unit2

The Area under the curve BMCB [y = 4 – 2x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{2}y\;dx\;=\;\int_{1}^{2}(4-2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 4x-x^{2} \right ]_{1}^{2}}\)

\(\\\boldsymbol{\Rightarrow }\) [8 – 4] – [4 – 1] =1 unit2

Therefore, the Area under the curve MBCN = 1 unit2

The Area under the curve ACNA [2y = 3x – 6]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{4}y\;dx\;=\;\int_{2}^{4}\frac{3x-6}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{4}-\frac{6x}{2} \right ]_{2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3\times 16}{4}-\frac{6\times 4}{2} \right ]- \left [ \frac{3\times 4}{4}-\frac{6\times 2}{2} \right ]=3}\) unit2

Therefore, the Area under the curve ACNA = 3 unit2

The Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA

\(\\\boldsymbol{\Rightarrow }\) \(\frac{15}{2}-1-3 = 3.5\) unit2

Therefore, the Area of triangle ABC = 3.5 unit2

Q.10: Find the area enclosed by the curve 2x2 = y and the line y = 2x + 12 and x – axis.

Sol:

40

Equation 2x2 = y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola 2x2 = y and the line y = 2x + 12 and x-axis is the Area enclosed under the curve ABCOA.

Since, the parabola 2x2 = y and the line y = 2x + 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:

Since, y = 2x + 12

\(\\\boldsymbol{\Rightarrow }\) 2x2 = (2x+12)

\(\\\boldsymbol{\Rightarrow }\) x2 – x – 6 = 0

By splitting the middle term Method solutions of this quadratic equation are:

x2 – (3 – 2)x – 6 = 0 \(\Rightarrow\) x(x – 3) +2(x – 3) = 0

\(\Rightarrow\) (x – 3) (x + 2) = 0

Therefore, x = 3 and x = -2 which gives y = 18 and y = 8 respectively.

Hence, the co-ordinates of point E and point A are (3, 18) and (-2, 8) respectively.

Since, 2x2 = y

Therefore, y = 2x2

The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]

The Area enclosed by the curve ACOA [ 2x2 = y ]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} 2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{2\;x^{3}}{3}\right |_{-2}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left |(0-\frac{-16}{3}) \right |=\frac{16}{3}}\) unit2

The Area enclosed by the curve ABC [ y = 2x + 12 ] :

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area\;of\;\Delta ABC=\frac{1}{2}\times Base\times Altitude}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times |BC|\times |AC|=\frac{1}{2}\times \left | 4 \right |\times |8|=16}\) unit2

The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]

\(\Rightarrow \frac{16}{3}+16=\frac{64}{3}\) unit2

Therefore, the Area of shaded region ABCOA \(=\frac{64}{3}\)unit2

Q.11: Plot the curve y = |x + 4| and hence evaluate \(\int_{-9}^{0}|x+4|\;dx\)

Sol:

From the given equation the corresponding values of x and y are given in the following table.

X -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3
Y 5 4 3 2 1 0 1 2 3 4 5 6 7

Now, on using these values of x and y, we will plot the graph of y = |x + 4|

41

From the above graph, the required Area = the Area enclosed by the curve ABCA + the Area enclosed by the curve CDOC

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-9}^{0}|x+4|\;dx+\int_{-9}^{-4}(x+4)\;dx+\int_{-4}^{0}(x+4)\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-9}^{-4}+\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-4}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | 8-16-\frac{81}{2}+36 \right |+\left | 0-(8-16) \right |=\left | \frac{-25}{2} \right |+8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{41}{2}}\) unit2

Therefore, the area of shaded region = \(\frac{41}{2}\) unit2

 

 

Q.12: Find the area enclosed by the curve y = sin x between 0 x ≤ 2π

Sol:

42

From the above figure, the required Area is represented by the curve OABCD.

Now, the Area bounded by the curve OABCD = the Area bounded by the curve OABO + the Area bounded by the curve BCDB

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\pi } sin(x)\;dx+\left | \int_{\pi }^{2\pi } sin(x)\;dx\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x)\right ]_{0}^{\pi }+\left | \left [ -cos (x) \right ] _{\pi }^{2\pi }\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[-cos(\pi )+cos(0)]+\left | [-cos(2\pi )+cos(\pi )] \right |}\\\)

\(\boldsymbol{\Rightarrow }\) [1+1] + [ |– 1 – 1| ] unit2

Therefore, the area of shaded region = 4 unit2

Q.13: Find the area of smaller region enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) and the line \(\frac{x}{2}+\frac{y}{3}=1\)

Sol:

The Equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) represents an ellipse.

The Equation \(\frac{x}{2}+\frac{y}{3}=1\) represents a line with x and y intercepts as 2 and 3 respectively.

43

Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\)

\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)

Therefore, the Area of smaller region enclosed by the Ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) and the line \(\frac{x}{2}+\frac{y}{3}=1\) is represented by curve ACBA

Now, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – Area enclosed by the curve ABOA

Now, the Area enclosed by the curve ACBOA:

\(\boldsymbol{\int_{0}^{2} y\;dx\Rightarrow\frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\\\)

Since, \(\ \int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) unit2

Therefore, the Area enclosed by the curve ACBOA = \(\boldsymbol{\frac{3\pi}{2}}\)unit2

Now, the Area enclosed by the curve ABOA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area \;of\;\Delta ABO = \frac{1}{2}\times AO\times BO}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times 2\times 3=3}\) unit2

Therefore, the Area enclosed by the curve ABOA = 3 unit2

Since, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – the Area enclosed by the curve ABOA

\(\\\boldsymbol{\Rightarrow }\) \(\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)\)unit2

Therefore, the Area of shaded region \(=\frac{3}{2}(\pi -2)\) unit2

Q.14: Find the area enclosed by the curve |x| + |y| = 2, by using the method of integration.

Sol:

44

Equation |x| + |y| = 2 represent a region bounded by the lines:

x + y = 2 . . . . . . (1)

x – y = 2 . . . . . . (2)

-x + y = 2 . . . . . . (3)

-x – y = 2 . . . . . . (4)

From equations (1), (2), (3) and (4) we conclude that the curve intersects x-axis and y-axis axis at points A (0, 2), B (2, 0), C (0, -2) and D (-2, 0) respectively.

From the above figure:

Since, the curve is symmetrical to x-axis and y-axis. Therefore, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA

Now, the Area of region bounded by the curve ABOA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=4\int_{0}^{2}(2-x)dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2x-\frac{x^{2}}{2} \right ]_{0}^{2}=(4-2)=2}\\\) unit2

Therefore, the Area of region bounded by the curve ABOA = 2unit2

Since, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA

Therefore, the Area of region bounded by the curve ABCDA = (2 × 4) = 8 unit2

Therefore, the Area of shaded region = 8 unit2

Q.15: Find the area which is exterior to curve x2 = 2y and interior to curve x2 + y2 = 15.

Sol:

45

The Equation x2 = 2y represents a parabola symmetrical about y-axis.

The Equation x2 + y2 = 15 represents a circle with centre (0, 0) and radius units.

Now, on substituting the equation of parabola in the equation of circle we will get:

(2y) + y2 = 15 i.e. y2 + 2y – 15 = 0

Now, by splitting the middle term method solutions of this quadratic equation are:

y2 + (5 – 3)y – 15 = 0 \(\Rightarrow\) y(y + 5) – 3(y +5) = 0

\(\Rightarrow\) (y – 3) (y + 5) = 0

Neglecting y = -5 [gives absurd values of x]

Therefore, y = 3 which gives x = \(\pm \sqrt{6}\)

Hence, the coordinates of point B and point D are (\(\sqrt{6}\), 3) (\(-\sqrt{6}\), 3) respectively.

Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB+ Area of region bounded by the curve OAC’O)

Area of region bounded by the curve BAMB [ x 2 + y2 = 15 ]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\sqrt{6}}^{\sqrt{15}}y\;dx=\int_{\sqrt{6}}^{\sqrt{15}}\sqrt{\left ( \sqrt{15} \right )^{2}-x^{2}}\;\;dx}\\\)

Since, \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{15- x^{2}}+\frac{15}{2}\sin^{-1}\frac{x}{\sqrt{15}} \right ]_{\sqrt{6}}^{\sqrt{15}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac {\sqrt{15}}{2}\times \sqrt{{15-15}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{15}}{\sqrt{15}} \right ]-}\\\) \(\\\boldsymbol{\left [\frac {\sqrt{6}}{2}\times \sqrt{{15-6}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{6}}{\sqrt{15}}\right]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{15\pi }{4} \right ]-\left [\frac{3\sqrt{6}}{2}+\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}\right]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{15\pi }{4}=\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5} \right ]}\)unit2

Area of region bounded by the curve OBMO [ x2 = 2y ]:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{6}}y\;dx=\int_{0}^{\sqrt{6}}\frac{x^{2}}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{6} \right ]_{0}^{\sqrt{6}}=\frac{6\sqrt{6}}{6}-0}\\\) = \(\boldsymbol{\sqrt{6}}\) unit2

Area of region bounded by the curve OAC’O:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( \sqrt{15} \right )^{2} }{4}}\\\) = \(\boldsymbol{\frac{15\pi }{4}}\)unit2

Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB + Area of region bounded by the curve OAC’O)

Therefore, the Area of region bounded by the curve BAC’A’DOB:

\(\boldsymbol{\Rightarrow }\) \(2\left [ \frac{15\pi }{4}-\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}+\sqrt{6} +\frac{15\pi }{4}\right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\\\)unit2

Therefore, the Area of shaded region: \(\boldsymbol{ \left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\)unit2

Q.16: Find the area enclosed by the curve y = x3, x-axis and the lines x = -2 and x = 2.

Sol.

46

The Equation y = x3 represents a cubic parabola which intersects the line x = 2 and x = -2 at points A and D respectively

From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}x^{3}\;dx+\left | \int_{-2}^{0}x^{3}\;dx \right |}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{4}}{4} \right ]_{0}^{2}+\left | \left [ \frac{x^{4}}{4} \right ]_{-2}^{0} \right |}\)

\(\\\boldsymbol{\Rightarrow }\) 4 – 0 + 0 + 4 =16 unit2

Therefore, the Area of shaded region = 16 unit2

Q.17: Find the area enclosed by the curve y = x|x|, y – axis and the lines y = -1 and y = 3.

Sol:

Now, y = x|x| is equal to [y = x2] when x > o

And y = x|x| is equal to [y = -x2] when x < o

47

From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}x^{2}\;dx+\left | \int_{-1}^{0}-x^{2}\;dx \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}\right ]_{0}^{3}+\left | \left [ \frac{x^{3}}{3} \right ]_{-1}^{0} \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{9+\frac{1}{3}=\frac{28}{3}}\)unit2

Therefore, the area of shaded region \(=\frac{28}{3}\)unit2

Q.18: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and y – axis, when [0 ≤ x ≤ \(\frac{\pi }{2}\)].

Sol:

y = Cos(x) . . . . . . . . (1)

y = Sin(x) . . . . . . . . . (2)

Now, from equation (1) and equation (2):

Cos (x) = Sin (x) \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\)

\(\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\) \(\Rightarrow\) x = \(\frac{\pi }{4}\)

Therefore, the coordinates of point A are: \(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)

48

Now, from the above figure:

The required Area = Area enclosed by the curve ADMA + Area enclosed by the curve AMOA

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\\\)

Since, \(\int \sin^{-1}y\;dy= y\sin^{-1}y+\sqrt{1-y^{2}}\\\)

And, \(\\\int \cos^{-1}y\;dy= y \cos^{-1}y-\sqrt{1-y^{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ y\sin^{-1}y+\sqrt{1-y^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}+\left [ y\cos^{-1}y-\sqrt{1-y^{2}} \right ]_{\frac{1}{\sqrt{2}}}^{1}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{1}{\sqrt{2}}\times \frac{\pi }{4}+\frac{1}{\sqrt{2}}-(0+1) \right ]+\left [ 0-\left ( \frac{1}{\sqrt{2}}\times \frac{\pi }{4}-\frac{1}{\sqrt{2}} \right ) \right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}-1 \right ]}\)unit2

Therefore, the Area of shaded region = \(\left [\sqrt{2}-1 \right ]\)unit2

Q.20: Find the area which is exterior to curve y2 = 6x and interior to curve x2 + y2 = 16.

Sol:

49

The Equation y2 = 6x represents a parabola symmetrical about x – axis.

The Equation x2 + y2 = 16 represents a circle with centre (0, 0) and radius 4 units.

Now, on substituting the equation of parabola in the equation of circle we will get:

x2 + (6x) = 16 i.e. x2 + 6x – 16 = 0

Now, by splitting of middle term method solutions of this quadratic equation are:

x2 + (8 – 2)x – 16 = 0 \(\Rightarrow\) x(x + 8) – 2(x + 8) = 0

\(\Rightarrow\) (x – 2) (x + 8) = 0

Neglecting x = -8 [gives absurd values of y]

Therefore, x = 2 which gives y = \(\pm 2\sqrt{3}\)

Hence, the coordinates of point B and point D are (2, \(2\sqrt{3}\)) (2, \(-2\sqrt{3}\)) respectively.

Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)

Area of region bounded by the curve BPMOB [x 2 + y2 = 16]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{\left ( 4\right )^{2}-x^{2}}\;\;dx}\\\)

Since, \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{16- x^{2}}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2}{2}\times \sqrt{{16-4}}+8\times \sin^{-1}\frac{1}{2} \right ]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\sqrt{3}+\frac{8\times \pi }{6} \right ]=\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]}\\\)unit2

Therefore, Area of region bounded by the curve BPMOB = \(\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]\)unit2

Area of region bounded by the curve POMP [y2 = 6x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{6x}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\left [ \frac{\sqrt{6}\times 2}{3}\times x^\frac{3}{2} \right ]_{0}^{2}=\frac{2\sqrt{6}}{3}\times 2\sqrt{2}-0\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8\sqrt{3}}{3}\)unit2

Therefore, Area of region bounded by the curve POMP : \(\frac{8\sqrt{3}}{3}\)unit2

Area of region bounded by the curve BOA’B:

\(\\\boldsymbol{\Rightarrow \frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( 4 \right )^{2} }{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) 4π unit2

Therefore, Area of region bounded by the curve BOA’B = 4π unit2

Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)

Therefore, the Area of region bounded by the curve PBA’B’NOP:

\(\\\boldsymbol{\Rightarrow }\) \(2\left [2\sqrt{3}+\frac{4\pi }{3}-\frac{8\sqrt{3}}{3}+4\pi \right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{32\pi }{3}-\frac{4\sqrt{3}}{3} \right ]=\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]}\\\)unit2

Therefore, the Area of shaded region = \(\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]\)unit2

Q.21: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and x – axis, when [0 ≤ x ≤ ].

Sol:

y = Cos(x) . . . . . . . . (1)

y = Sin(x) . . . . . . . . . (2)

Now, from equation (1) and equation (2):

Cos (x) = Sin (x) \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\\\)

\(\\\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\\\) \(\\\Rightarrow\) x = \(\frac{\pi }{4}\)

Therefore, the coordinates of point A are: \(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)

50

Now, from the above figure:

The required Area = Area enclosed by the curve AONA + Area enclosed by the curve ANBA

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x) \right ]_{0}^{\frac{\pi }{4}}+[sin (x)]_{\frac{\pi }{4}}^{\frac{\pi }{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos\frac{\pi }{4}+cos(0)+sin\frac{\pi }{2}-sin\frac{\pi }{4}\right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}} \right ]=[2-\sqrt{2}\;]}\)unit2

Therefore, the Area of shaded region \( = [2-\sqrt{2}\;]\)unit2