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?
 A: 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.
A: You have been offered three solutions, one of which is incorrect. But nobody has shown you what to in the event that you have another such problem. First of all, these short-cut methods have their origin in Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$.
Now, we'll need the centroid as the vertical distance relative to the axis of rotation at $y=3$, therefore
$$R=\frac{\int_0^{\pi/4} \int_{3\sin x}^{3\cos x}(3-y) dydx}{\int_0^{\pi/4} \int_{3\sin x}^{3\cos x} dydx}=\frac{1}{A}{\int_0^{\pi/4} \int_{3\sin x}^{3\cos x}(3-y) dydx}{A}\\=3-\frac{1}{2A}{\int_0^{\pi/4} y^2\big|_{3\sin x}^{3\cos x} dx}\\=3-\frac{9}{2A}{\int_0^{\pi/4} (\cos^2 x-\sin^2 x) dx}=3\left(1-\frac{1}{4(\sqrt{2}-1)} \right)$$
And, of course
$$A=\int_0^{\pi/4} \int_{3\sin x}^{3\cos x} dydx=3\int_0^{\pi/4} (\cos x-\sin x) dx=3(\sqrt{2}-1)$$
And finally,
$$V=2\pi RA=18\pi \left(1-\frac{1}{4(\sqrt{2}-1)} \right)(\sqrt{2}-1)\approx 9.286 $$
This result is in agreement with the others and has been verified numerically, as well.
A: 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
$$ \pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx=18 \pi\int_0^{\pi/4} (1-\sin x -\cos x) dx = \pi/4 - 1 $$
