RD Sharma Class 12 Ex 21.2 Solutions Chapter 21 Areas of Bounded Regions

Here we provide RD Sharma Class 12 Ex 21.2 Solutions Chapter 21 Areas of Bounded Regions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 21.2 Solutions Chapter 21 Areas of Bounded Regions book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter21
Exercise21.2
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 21.2 Solutions Chapter 21 Areas of Bounded Regions

Question 1. Find the area in the first quadrant bounded by the parabola y = 4x2 and the lines x = 0, y = 1, and y = 4.

Solution:

From the question it is given that,

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

Parabola y = 4x2 … [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

\displaystyle=\int_1^4xdy

Given, y = 4x2

x = \displaystyle\sqrt{\frac{y}{4}}\\ \int_1^4\sqrt{\frac{y}{4}}\ dy\\ \frac{1}{2}\int_1^4\sqrt y\ dy

On integration, we get,

\displaystyle=\frac{1}{2}\left[\frac{2}{3}y\sqrt y\right]_1^4

Now, applying limits we get,

= \frac{1}{2} \left[\left(\frac{2}{3} × 4 × \sqrt4\right) - \left(\frac{2}{3} × 1 × \sqrt1\right)\right]\\ = \frac{1}{2} \left[\frac{16}{3} - \frac{2}{3}\right]\\ = \frac{1}{2} \left[\frac{(16 - 2)}{3}\right]\\ = \frac{1}{2} \left[\frac{14}{3}\right]\\ = \frac{7}{3}

Therefore, the required area is \frac{7}{3}  square units.

Question 2. Find the area of the region bounded by x2 = 16y, y = 1, y = 4, and the y-axis in the quadrant.

Solution:

From the question it is given that,

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

Parabola x2 = 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

\displaystyle\int_1^4x\ dy

Given, x2 = 16y

x=\sqrt{16y}\\ x=4\sqrt{y}\\ =\int_1^44\sqrt4\ dy

On integrating we get,

=4\left[\frac{2}{3}y\sqrt y\right]^4_1\\ =\int_1^4x\ dy

Given, x2 = 16y

x=\sqrt{16y}\\ x=4\sqrt{y}\\ =\int_1^44\sqrt4\ dy

On integrating we get,

=4\left[\frac{2}{3}y\sqrt y\right]^4_1\\ =\int_1^4x\ dy

Now, applying limits we get,

= 4 \left[\left(\frac{2}{3} × 4 × \sqrt4\right) - \left(\frac{2}{3}× 1 × \sqrt1\right)\right]\\ = 4 \left[\frac{16}{3} - \frac{2}{3}\right]\\ = 4 \left[\frac{(16 - 2)}{3}\right]\\ = 4 \left[\frac{14}{3}\right]\\ = \frac{56}{3}

Therefore, the required area is \frac{56}{3}  square units.

Question 3. Find the area of the region bounded by x2 = 4ay and its latus rectum

Solution:

We have to find the area of the region bounded by x2 = 4ay

Then,

Area of the region = \displaystyle2\times\int_0^{2a}\left(a-\frac{x^2}{4a}\right)dx

On integrating we get,

2\times\left[ax-\frac{x^3}{12a}\right]_0^{2a}

Now applying limits,

= 2 × \left[(a (2a - 0)) - \frac{((2a)^3 - 0^3)}{12a}\right]\\ = 2 × \left[(2a^2) - \frac{8a^3}{12a}\right]\\ = 2 × \left[\frac{(24a^3 - 8a^3)}{12a}\right]\\ = 2 × \left[\frac{16a^3}{12a}\right]\\ = 2 × \left[\frac{4a^2}{3}\right]\\ = \frac{8a^2}{3}\\

Therefore, the area of the region is \frac{8a^2}{3}  square units.

Question 4. Find the area of the region bounded by x2 + 16y = 0 and its latus rectum. 

Solution:

We have to find the area of the region bounded by x2 + 16y = 0

Then, 

Area of the region = \displaystyle2\times\int_0^{8}\left[-\frac{x^2}{16}-(-4)\right]dx

On integrating we get,

2\times\left[4x-\frac{x^3}{48}\right]_0^{8}

Now applying limits,

= 2 × \left[(4 (8 - 0)) - \frac{((8)^3 - 0^3)}{48}\right]\\ = 2 × \left[(32) - \frac{512}{48}\right]\\ = 2 × \left[(32)-\frac{32}{3}\right]\\ = 2 × \left[\frac{96-32}{3}\right]\\ = 2 × \left[\frac{64}{3}\right]\\ = \frac{128}{3}\\

Therefore, the area of the region is  \frac{128}{3}  square units.

Question 5. Find the area of the region bounded by the curve ay2 = x3, the y-axis, and the lines y = a and y = 2a.

Solution:

We have to find the area of the region bounded by curve ay2 = x3, and lines y = a and y = 2a.

Then,

Area of the region = \displaystyle\int_a^{2a}(ay^2)^{\frac{1}{3}}\ dy\\ =a^{\frac{1}{3}}\int_a^{2a}y^{\frac{2}{3}}dy

On integrating we get, = a^{\frac{1}{3}}\left[\frac{3}{5}y^{\frac{5}{3}}\right]_0^{2a}

Now applying limits we get, = \frac{3}{5}\left(2^{\frac{5}{3}}-1\right)a^2\ sq\ units

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