How to find a volume of a solid created by revolving an area bounded by two functions? The question says to find the volume of the figure generated by the following equations and rotations.
Bounded by: $y=x^2$ and $y=4x-x^2$
Rotated about $y=6$
When solved, x = 0 and 2.
I understand that the volume involves the outer function minus the inner function (as my teacher put it), and that in this case, $y=x^2$ is the outer and $y=4x-x^2$ is the inner.
It is in setting up the integral that I get confused. Originally, I did: $$V=\pi\int _0^2 (x^2)^2 - (4x-x^2)^2dx$$
My answer was $-10.6667\pi$, which did not seem right to me, so I tried it differently: $$V=\pi\int_0^2(x^2-4x+x^2)^2dx$$
This gave me an answer of $4.2667\pi$, which seems more likely, but I still do not feel that I did this correctly.
I am also notorious for simple algebraic mistakes, that could always be my issue.
 A: It is very helpful usually to draw the graphs of the functions and the rotation that the problem asks of us. 
To get the volume, we are rotating $y=x^2$ around $y=6$ and subtracting the volume obtained from rotating $y=4x-x^2$ around $y=6$. This leaves us with the volume of rotating the regions bounded by the two functions around $y=6$.
So, we have the following integral
$\pi\int_0^2(6-x^2)^2-(6-4x+x^2)^2dx=\dfrac{64\pi}{3}$ 
A: The first step is to visualize the volume.  Here you have (in the XY plane) a "fish body" with mouth at $x=y=0$ and tail at $x=2, y=4)$ -- these points are found by solving $x^2 = 4x-x^2$.
We are now going to rotate this fish around the barbecue spit at $y=6$.  So any infinitesimal area $dA$ of the fish at $(x,y)$ on the fish will contribute 
$2\pi (6-y) dA$ to the volume.
The easiest way for me to do this integral is to break it into vertical slivers first, and later integrate in the $x$ direction.  Thus:
$$
V = \int_{x=0}^2 dx \int _{y=4x-x^2}^{x^2} 2\pi (6-y)dy = 2\pi \int_{x=0}^2 dx \left[ 6y - \frac{y^2}{2}\right]^{y=x^2}_{y=4x-x^2}\\=2\pi \int_{x=0}^2 dx \left[ 6x^2-\frac{x^4}{2}-6(4x-x^2)+\frac{(4x-x^2)^2}{2}\right]\\=2\pi \int_{x=0}^2\left[ -24x+20x^2-4x^3\right]dx = \frac{32\pi}{3}
$$
I've tried to resist using Mathematica or something to check the answer...there were plenty of opportunities to drop a term or forget the factor of 2 on the outside.
A few slivers of ginger and some wasabi sauce would go good on this problem...
