Find volume of region bounded by two graphs I need some assistance on how to finish this problem.
The region enclosed by the graph of $y=x^2$, and the line $x=2$, and the $x$ axis is revolved about the $y$ axis. I need to find the volume of the are bounded by the two graphs.  I believe I need to find the integral of $\pi r^2$, but I am not sure how to find the radius, or the boundaries since it is a $x=2$ and $y$ axis revolution. Can someone help me and please explain how to find these and the answer? Thanks 
 A: If you think about this a bit, you'll find that all you're being asked to do in the two dimensional part is a simple integral of $f(x)=x^2$ from 0 to 2.
Once you've done that, you can consider summing the volumes of cylindrical shells of infinitesimal thickness - sort of like using circular strips of paper of varying heights to fill the space. 
Partition the interval [0,2] into n subintervals $[x_{i-1}, x_i]$. Then the volume of one cylinder shell is 
$(*)\space \pi x_i^2h_i^* - \pi x_{i-1}^2h_i^*$
where $h_i^*$ is $f(x_i^*)$ for some $x_i^* \in [x_{i-1}, x_i]$, (or in this case $x_i^*{^2}$). 
The sum of these volumes is the total volume: V=$\sum_{i=1}^{n}\pi (x_i^2-x_{i-1}^2) f(x_i^*)$ $=\sum_{i=1}^{n}\pi (x_i+x_{i-1})f(x_i^*)(x_i-x_{i-1}) $$=2\pi\int_{x_0}^{x_n}xf(x)dx$.
I got from my second to third step by making my partition finer and finer, that is $n\to \infty$. Then $x_i\approx x_{i-1}$ and $x_i-x_{i-1}\to 0$ (you've probably seen this argument as $\Delta x \to 0$ or $\Delta x = dx$.
In your case:

$V=2\pi\int_{0}^{2}x\cdot x^2dx$

A: The method of shells is the simplest way to solve the problem. But if you want to do it by slicing perpendicular to the $y$-axis, you will be integrating from $y=0$ to $y=4$. The cross-section has outer radius $2$, and inner radius $x$, which is $\sqrt{y}$. So the volume is
$$\int_0^4 \pi(2^2 -(\sqrt{y})^2)\,dy.$$
Thus we simply want $\int_0^4(4-y)\,dy$. This is $8\pi$. 
