NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 - Application of Integrals

Question:1 Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Answer:

The area bounded by the ellipse : $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Area will be 4 times the area of EAB.

Therefore, Area of EAB = $\int_0^4 y\,dx$

$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$

$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$

$= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}$

$= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]$

$= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]$

$= \frac{3}{4}\left [ 4\pi \right ] =3\pi$

Therefore, the area bounded by the ellipse will be $= 4 \times 3\pi = 12\pi$ units.

Question 2: Find the area of the region bounded by the ellipse $\small \frac{x^2}{4}+\frac{y^2}{9}=1$

Answer:

The area bounded by the ellipse : $\small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore, Area of EAB $= \int_0^2 y\ dx$

$= \int^2_{0}3\sqrt{1-\frac{x^2}{4}} dx$

$= \frac{3}{2}\int^2_{0}\sqrt{4-x^2} dx$

$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$

$= \frac{3}{2}\left [ \frac{2\pi}{2} \right ]$

$= \frac{3\pi}{2}$

Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi$ units.

Question 3: Choose the correct answer in the following

Area lying in the first quadrant and bounded by the circle $\small x^2+y^2=4$ and the lines $\small x=0$ and $\small x=2$ is

$\small (A)\hspace{1mm}\pi$

$\small (B)\hspace{1mm}\frac{\pi}{2}$

$\small (C)\hspace{1mm}\frac{\pi}{3}$

$\small (D)\hspace{1mm}\frac{\pi}{4}$

Answer:
The area bounded by circle C(0,0,4) and the line x=2 is

The required area = area of OAB

$\int_0^2 y\,dx = \int_0^2 \sqrt{4 - x^2}\,dx$

$= \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1} \frac{x}{2} \right]_0^2$

$= 2 \cdot \frac{\pi}{2}$

$= \pi$

Hence, The correct answer is $\pi$

Question 4: Choose the correct answer in the following.

Area of the region bounded by the curve $\small y^2=4x$ , $\small y$ -axis and the line $\small y=3$ is

(A) $\small 2$

(B) $\small \frac{9}{4}$

(C) $\small \frac{9}{3}$

(D) $\small \frac{9}{2}$

Answer:

The area bounded by the curve $y^2=4x$ and y =3

The required area = OAB

$= \int_0^3 x\,dy$

$= \int_0^3 \frac{y^2}{4}\,dy$

$= \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3$

$= \frac{9}{4}$

Hence, The correct answer is $\small \frac{9}{4}$