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:
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:
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:
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:
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:
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:
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 ODCAO – Area 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:
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
