RD Sharma Solutions for Class 12 Maths Exercise 21.3 Chapter 21 Areas of Bounded Regions are available here. We, at BYJU’S, ensure that you get the best RD Sharma Solutions for Class 12 Maths Chapter 21 study material that can help you excel in your studies with a well-organized way of learning. Your difficulties are addressed rightly, empowering you to achieve excellent marks in your class and exams.
RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 3
Access RD Sharma Solutions for Class 12 Maths Chapter 21 Exercise 3
EXERCISE 21.3
Question. 1
Solution:
From the question, it is given that the area of the region common to the parabolas y2 = 6x and x2 = 6y,
Given, y2 = 6x
y = √(6x)
and x2 = 6y, then y = x2/6
Now applying limits, we get,
= [√6 ((6)3/2/(3/2) – (63/18) – 0]
By simplification, we get,
= 12 square units
Question. 2
Solution:
From the question, it is given that the area of the region common to the parabolas 4y2 = 9x and 3x2 = 16y,
Given, 4y2 = 9x
y2 = 9x/4
y = √(9x/4)
y = (3/2) √x
and 3x2 = 16y
y = 3x2/16
Now applying limits, we get,
= [(4)3/2 – (43/16)]
By simplification, we get,
= [(8) – (64/16)]
= [8 – 4]
= 4 square units
Question. 3
Solution:
From the question, it is given that the area of the region bounded by,
y = √x
y2 = x … [equation (i)]
y = x … [equation (ii)]
So, equation (i) represents a parabola with vertex at (0, 0) and x-axis as its axis and equation (ii) represents a line passing through the origin and intersecting parabola at (0, 0) and (1, 1), is as shown in the rough sketch below,
Now, we have to find the area of AOBPA,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y1 – y2)
The area of the sliced part will be as it is a rectangle = (y1 – y2) Δx
So, this rectangle can move horizontal from x = 0 to x = 1
The required area of the region bounded between the lines = Region AOBPA
Now, applying limits, we get,
= [((2/3) × 1 × √1) – ((1)2/2)] – [0]
= [(2/3) – ½]
= [(4 – 3)/6]
= 1/6
Therefore, the required area = 1/6 square units.
Question. 4
Solution:
From the question, it is given that the area bounded by the curve y = 4 – x2
And line y = 0, y = 3
So, y = 4 – x2
x2 = – (y – 4) … [equation (i)]
y = 0 … [equation (ii)]
y = 3 … [equation (iii)]
So, equation (i) represents a parabola with vertex at (0, 4) and passes through (0, 2), (0, -2) on the x-axis and equation (ii) is the x-axis and cutting the parabola at (2, 0) and (-2, 0). Equation (iii) represents a line parallel to the x-axis passing through (0, 3), as shown in the rough sketch below,
Now, we have to find the area of AOBPA,
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = (y1 – y2) Δx
So, this rectangle can move horizontal from x = 0 to x = 2
The required area of the region bounded between the lines = Region ABDEA
Now, applying the limits, we get,
= 2 [(8 – (8/3)) – 0]
= 2 [(24 – 8)/3]
= 2 [16/3]
= 32/3
Therefore, the required area 32/3 square units.
Question. 5
Solution:
From the question, it is given that,
(x2/a2) + (y2/b2) = 1 … [equation (i)]
(x/a) + (y/b) = 1 … [equation (ii)]
So, equation (i) represents an ellipse with centre at origin and passing through (±a, 0), (0, ±b), and equation (ii) represents a line passing through (a, 0) and (0, b), as shown in the rough sketch below,
Then, from the figure, the region shaded is the required region and substituting (0, 0) in x2/a2 + y2/b2 ≤ 1 gives a true statement and substituting (0, 0) in 1 ≤ x/a + y/b gives a false statement.
Then, the area can be found by taking a small slice in each region of width Δx,
And length = (y – 0) = y
The area of the sliced part will be as it is a rectangle = (y1 – y2) Δx
So, this rectangle can move horizontal from x = 0 to x = a,
Comments