Question

Area of the region in the first quadrant enclosed by the $x$-axis, the line $y=x$ and the circle $x^2+y^2=32$ is

• A. $16\pi$ sq. units
• B. $4\pi$ sq. units
• C. $32\pi$ sq. units
• D. $24\pi$ sq. units

Answer 1

Answer: (B)

We have, $x^2+y^2=32...\left(i\right)$, a circle with centre $\left(0,0\right)$ and radius $4\sqrt{2}$ and $y=x\dots \left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get point of intersection $\left(4,4\right)$

Required area, $A=\underset{0}{\overset{4}{\int }}xdx+\underset{4}{\overset{4\sqrt{2}}{\int }}\sqrt{32-x^2}dx$
$={\left[\frac{x^2}{2}\right]}_0^4+{\left[\frac{x}{2}\sqrt{32-x^2}+\frac{32}{2}si{n}^{-1}\frac{x}{4\sqrt{2}}\right]}_4^{4\sqrt{2}}$
$=8+\left[16\times \frac{\pi }{2}\right]-\left[2\sqrt{32-4^2}+16si{n}^{-1}\frac{1}{\sqrt{2}}\right]$
$=8+8\pi -8-4\pi$
$=4\pi$ sq. units