Question

do it with explanation​

Answer 1

Given to evaluate:-

$\displaystyle\longrightarrow I=\int\limits_0^a\int\limits_0^{\sqrt{a^2-y^2}}\sqrt{a^2-x^2-y^2}\ dx\ dy$

or,

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\int\limits_0^{\sqrt{a^2-y^2}}\sqrt{(a^2-y^2)-x^2}\ dx\right]\ dy$

Let $\sqrt{a^2-y^2}=c$ which is treated as a constant wrt $x.$ Then,

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\int\limits_0^c\sqrt{c^2-x^2}\ dx\right]\ dy$

$\displaystyle\longrightarrow I=\int\limits_0^a\left[\dfrac{x}{2}\sqrt{c^2-x^2}+\dfrac{c^2}{2}\sin^{-1}\left(\dfrac{x}{c}\right)\right]_0^c\ dy$

$\displaystyle\longrightarrow I=\int\limits_0^a\left(\dfrac{\pi c^2}{4}\right)\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\int\limits_0^ac^2\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\int\limits_0^a\left(a^2-y^2\right)\ dy$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\left[a^2y-\dfrac{y^3}{3}\right]_0^a$

$\displaystyle\longrightarrow I=\dfrac{\pi}{4}\left(a^3-\dfrac{a^3}{3}\right)$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{\pi a^3}{6}}}$

Answer 2

☘️ See the attachment picture for Explanation.

$\Large\mathbb\pink{THANKS}$

$\Large\mathbb\green{HOPE\: IT'S\: HELPFUL}$