Solid Revolution question I'm trying to do the following problem:
Find the volume of the solid obtained by rotating the region bounded by the given curves about the $y=3$  axis specified by means of  the circular arrow drawn :
$$ y = 3 \sin x, y = 3 \cos x, 0 \le x \le π/4 $$
First off, am I drawing the picture correctly?

Second, am I setting up the integral correctly?
$$
\pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx
$$
Did I correctly identify $3-3 \sin x$ as the outer radius? Because it makes sense for it to be that way, but I saw this same question worked out on like yahoo answers and it instead says that $y=3-3 \cos x $ should be the outer radius?

Yahoo!Answers is notoriously unreliable on math and technical matters; my informal count is that the answer is alright only about half the time. (I suspect the poster overlooked the rotation axis being $ \ y = 3 \ , $ rather than the $ \ x-$ axis.) Your integrand looks fine and reduces to 
$$ \ (9 - 18 \sin x + 9 \sin^2 x) \ - \ (9 - 18 \cos x + 9 \cos^2 x) $$
$$ = \ 18 \ (\cos x - \sin x) \ + \ 9 \ (\sin^2 x - \cos^2 x) \ = \ 18 \ (\cos x - \sin x) \ - \ 9 \ \cos 2x \ . $$
The evaluation of the volume is then
$$ \pi \ \left[ \ 18 \ (\sin x + \cos x) \ - \ \frac{9}{2}  \sin 2x \ \right]_0^{\pi/4} $$
$$ = \ \pi \ \left( \ [ \ 18 \ (\ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) \ - \ \frac{9}{2} \cdot 1  \ ] \ - \ [ \ 18 \ (0 + 1) \ - \ \frac{9}{2} \cdot 0 \ ] \ \right) \ $$
$$ = \ \pi \  ( \ 18 \sqrt{2}  \ - \ \frac{9}{2}   \ -  \ 18  \ ) \ = \ \pi \  ( \ 18 \sqrt{2}  \ - \ \frac{45}{2}    \ ) \ \ \text{or} \ \ \frac{9 \pi}{2} \ ( \ 4 \sqrt{2}  \ - \ 5    \ ) \ \ , $$
confirming JohnD's result in the comment above.
