Answer
$a)$ $\frac{\pi^{2}}{2}$
$b)$ $\pi$
Work Step by Step
$a)$
$A(x)$ = $\frac{\pi}{4}(diameter)^{2}$
$A(x)$ = $\frac{\pi}{4}(\frac{2}{\sqrt[4] {1-x^{2}}}-0)^{2}$
$A(x)$ = $\frac{\pi}{\sqrt{1-x^{2}}}$
$V$ = $\int_{{\,-\frac{\sqrt 2}{2}}}^{{\,\frac{\sqrt 2}{2}}}$$\frac{\pi}{\sqrt{1-x^{2}}}$$dx$
$V$ = $\pi[sin^{-1}(x)]$ $|_{{\,-\frac{\sqrt 2}{2}}}^{{\,\frac{\sqrt 2}{2}}}$
$V$ = $\pi[sin^{-1}(\frac{\sqrt 2}{2})-sin^{-1}(-\frac{\sqrt 2}{2})]$
$V$ = $\pi(\frac{\pi}{4}-(\frac{-\pi}{4}))$
$V$ = $\pi(\frac{\pi}{4}+\frac{\pi}{4})$
$V$ = $\frac{\pi^{2}}{2}$
$b)$
$A(x)$ = $\frac{(diagonals)^{2}}{2}$
$A(x)$ = $\frac{1}{2}(\frac{2}{\sqrt[4] {1-x^{2}}}-0)^{2}$
$A(x)$ = $\frac{2}{\sqrt{1-x^{2}}}$
$V$ = $\int_{{\,-\frac{\sqrt 2}{2}}}^{{\,\frac{\sqrt 2}{2}}}$$\frac{2}{\sqrt{1-x^{2}}}$$dx$
$V$ = $2[sin^{-1}(x)]$ $|_{{\,-\frac{\sqrt 2}{2}}}^{{\,\frac{\sqrt 2}{2}}}$
$V$ = $2[sin^{-1}(\frac{\sqrt 2}{2})-sin^{-1}(-\frac{\sqrt 2}{2})]$
$V$ = $2(\frac{\pi}{4}-(\frac{-\pi}{4}))$
$V$ = $2(\frac{\pi}{4}+\frac{\pi}{4})$
$V$ = $\pi$
