Find the volume of the solid obtained by rotating the region between the curves around the line $y = -1$ I'm stuck on this volume problem - everytime I calculate the volume I get zero. Here is the problem: 
Find the volume of the solid obtained by rotating the region between the curves around the line $y = -1$. 
$$
\begin{cases}
y = \sin x  \\[1ex]
y = \cos x\\[1ex]
\pi/4 \leq x \leq 5\pi/4
\end{cases}
$$
Heres my work https://drive.google.com/file/d/0B29hT1HI-pwxSHVwVWxsNk11S00/view?usp=sharing- once I get my volume integral, I use a calculator (symbolab) to plug in my bounds, and I keep getting zero. 
If anyone could show me the steps to solving this problem, that'd be amazing. 
Thank you 

Your set up is not completely correct. You do have $\pi \int _{\pi / 4}^{5\pi/4} R^2-r^2 dx$ where $R=1+\sin (x)$ and $r=1+\cos (x)$ but then you claim that $R=\cos (x) $ and $r= \sin (x)$ and this is incorrect. To see why (I believe you confused a trigonometric identity because you were correct until that simplification), just remember that $$\sin (x)^2+\cos (x)^2=1$$
If you calculate the above integral with the given $R$ and $r$, you should get $4\sqrt{2}\pi$
