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
On
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 $8\pi$.
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: