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Q.
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
Application of Integrals
Solution:
We have, $x^{2}+ y^{2} = 32\, ...\left(i\right)$, a circle with centre $\left(0,0\right)$ and radius $4\sqrt{2}$ and $y=x\quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get point of intersection $\left(4,4\right)$
Required area, $A = \int\limits_{0}^{4} x\, dx+\int\limits_{4}^{4\sqrt{2}}\sqrt{32-x^{2}}\, dx$
$=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x}{2}\sqrt{32-x^{2}}+\frac{32}{2} sin^{-1} \frac{x}{4\sqrt{2}}\right]_{4}^{4\sqrt{2}}$
$=8+\left[16\times\frac{\pi}{2}\right]-\left[2\sqrt{32-4^{2}}+16\, sin^{-1}\frac{1}{\sqrt{2}}\right]$
$=8+8\pi-8-4\pi$
$=4\pi $ sq. units
