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