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.
Drawing looks ok to me. If axis of rotation s $x$ axis, rotated volume comes from section area between two disks:

$$ \pi\int_0^{\pi/4} [(3\sin x)^2-(3\cos x)^2]dx $$

EDIT1/2

But if axis of rotation is $y=3$ , then

$$ \pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx=18 \pi\int_0^{\pi/4} (1-\sin x -\cos x) dx = \pi/4 - 1 $$