Answer
$\displaystyle{V=\frac{\pi }{2}\left(4\sqrt2-3\right)}\\ $
Work Step by Step
$\displaystyle{A\left(x\right)=\pi \left(1+\cos x\right)^2-\pi \left(1+\sin x\right)^2}\\
\displaystyle{A\left(x\right)=\pi \left(1+\cos^2x+2\cos x-1-\sin^2x-2\sin x\right)}\\
\displaystyle{A\left(x\right)=\pi \left(\cos^2x-\sin^2x+2\cos x-2\sin x\right)}\\
\displaystyle{A\left(x\right)=\pi \left(\cos{2x}+2\cos x-2\sin x\right)}\\$
$\displaystyle{V=\int_0^{\frac{\pi}{4}}A\left(x\right)\ dx}\\
\displaystyle{V=\int_0^{\frac{\pi}{4}}\pi \left(\cos{2x}+2\cos x-2\sin x\right)\ dx}\\
\displaystyle{V=\pi \int_0^{\frac{\pi}{4}}\cos{2x}+2\cos x-2\sin x\ dx}\\
\displaystyle{V=\pi\left[\frac{1}{2}\sin{2x}+ 2\sin x+ 2\cos x\right]_0^{\frac{\pi}{4}}}\\
\displaystyle{V=\pi\left(\left(\frac{1}{2}\sin{\left(2\times\frac{\pi}{4}\right)}+ 2\sin \left(\frac{\pi}{4}\right)+ 2\cos \left(\frac{\pi}{4}\right)\right)-\left(\frac{1}{2}\sin{2(0)}+ 2\sin (0)+ 2\cos (0)\right)\right)}\\
\displaystyle{V=\frac{\pi }{2}\left(4\sqrt2-3\right)}\\ $
