# Class 12 RD Sharma Solutions – Chapter 21 Areas of Bounded Regions – Exercise 21.2

**Question 1. Find the area in the first quadrant bounded by the parabola y = 4x**^{2} and the lines x = 0, y = 1,** and y = 4.**

^{2}and the lines x = 0, y = 1

**Solution:**

From the question it is given that,

Lines, x = 0, y = 1, y = 4

Parabola y = 4x

^{2}… [equation (i)]So, equation (i) represents a parabola with vertex (0, 0) and axis as y – axis. x = 0 is y – axis and y = 1, y = 4 are line parallel to x – axis passing through (0, 1) and (0, 4) respectively, as shown in the rough sketch below,

Now, we have to find the area of ABCDA,

Then, the area can be found by taking a small slice in each region of width Δy,

And length = x

The area of sliced part will be as it is a rectangle = x Δy

So, this rectangle can move horizontal from y = 1 to x = 4

The required area of the region bounded between the lines = Region ABCDA

Given, y = 4x

^{2}x =

On integration, we get,

Now, applying limits we get,

Therefore, the required area is square units.

**Question 2. Find the area of the region bounded by x**^{2} = 16y, y = 1, y = 4,** and the y-axis in the quadrant.**

^{2}= 16y, y = 1, y = 4

**Solution:**

From the question it is given that,

Region in first quadrant bounded by y = 1, y = 4

Parabola x

^{2}= 16y … [equation (i)]So, equation (i) represents a parabola with vertex (0, 0) and axis as y-axis, as shown in the rough sketch below,

Now, we have to find the area of ABCDA,

Then, the area can be found by taking a small slice in each region of width Δy,

And length = x

The area of sliced part will be as it is a rectangle = x Δy

So, this rectangle can move horizontal from y = 1 to x = 4

The required area of the region bounded between the lines = Region ABCDA

Given, x

^{2}= 16y

On integrating we get,

Given, x

^{2}= 16yOn integrating we get,

Now, applying limits we get,

Therefore, the required area is square units.

**Question 3. Find the area of the region bounded by x**^{2} = 4ay and its latus rectum**. **

^{2}= 4ay and its

**Solution:**

We have to find the area of the region bounded by x

^{2}= 4ayThen,

Area of the region =

On integrating we get,

=

Now applying limits,

Therefore, the area of the region is square units.

**Question 4. Find the area of the region bounded by x**^{2} + 16y = 0 and its latus rectum.

^{2}+ 16y = 0 and its latus rectum.

**Solution:**

We have to find the area of the region bounded by x

^{2}+ 16y = 0Then,

Area of the region =

On integrating we get,

=

Now applying limits,

Therefore, the area of the region is square units.

**Question 5. Find the area of the region bounded by the curve ay**^{2} = x^{3}, the y-axis,** and the lines y = a and y = 2a.**

^{2}= x

^{3}, the y-axis

**Solution:**

We have to find the area of the region bounded by curve ay

^{2}= x^{3}, and lines y = a and y = 2a.Then,

Area of the region =

On integrating we get, =

Now applying limits we get, =