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

• Hi Megan, welcome to mathSE. Could you type up your solutions here on math.SE rather than linking to them. I know diagrams are tricky, but you could take a picture of your diagrams and upload them as a jpeg or png file, there are some explanations on how to do that on your left when you ask a question. You'll find a really helpful tutorial on using MathJax here: meta.math.stackexchange.com/questions/5020/… – Daniel Buck Sep 18 '16 at 17:40
• So sorry, I tried to upload my solutions directly to the post but unfortunately the file was too large. – Megan Byers Sep 18 '16 at 17:41
• Your outer and inner radius are not correct. The outer radius is $R= \sin x +1$ and the inner radius is $r=\cos x + 1$. Use the original functions to calculate the radii, not the functions you get after rotating. – Miguel Landeros Sep 18 '16 at 17:50
• No worries, the accepted method of posting questions is to use MathJax, as then users don't have large photo-files to download and can see your question at source, as well as correct it if necessary by editing it, something that cannot be done with a photo. I suggest you get familiar with MathJax from the tutorial, and then post the rest of your question once it is all in MathJax, as this will be a great way to learn. – Daniel Buck Sep 18 '16 at 17:53

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 $4\sqrt{2}\pi$