<s>[INST] Solid of revolution problem I would like to check with you guys if this exercise is right. The problem is:
"Let be $R$ be the region bounded by the graph of $y=2x$", $y=\frac{x^2}{4}$ and $y=2$. Find the volume of the solid generated by the rotation of the region $R$ about the $y$-axis. Make a sketch of the region. Show your calculations."
This is the graph I made:

This is what I did:
Since we are integrating with respect to $y$ I solve for $x$ both equations. $y=\frac{x^2}{4}$ becomes $x=2\sqrt{y}$ and $y=2x$ becomes $x=\frac{1}{2}y$. So for the volume:
$$V=\pi\int_{0}^{2} ((2\sqrt{y})^2-(\frac{1}{2}y)^2) \,dy$$
$$V=\pi\int_{0}^{2} (4y-\frac{1}{4}y^2) \,dy$$
$$V=\pi(4\cdot\frac{y^2}{2}-\frac{1}{4}\cdot\frac{y^3}{3})\bigg|_{0}^2$$
$$V=\pi(2y^2-\frac{1}{12}y^3)\bigg|_{0}^2$$
$$V=\pi[(2(2)^2-\frac{1}{12}(2)^3)-(2(0)^2-\frac{1}{12}(0)^3)]$$
$$V=\pi[(2(4)-\frac{1}{12}(8)]$$
$$V=\pi[(8-\frac{2}{3}]$$
$$V=\pi[(\frac{24}{3}-\frac{2}{3}]$$
$$V=\pi[\frac{22}{3}]$$
$$V=\frac{22}{3}\pi$$
Therefore the volumen of the solid of revolution is $\frac{22}{3}\pi$ cubic units.
Thanks in advance for the help.
 [/INST] Yes, this is exactly correct.
You can also verify your solution by using the method of cylindrical shells.  For instance, the volume of the parabolic region including the cone is just $$\int_{x=0}^{2 \sqrt{2}} 2\pi x \left(2 - \frac{x^2}{4}\right) \, dx = \left[2x^2 - \frac{x^4}{8}\right]_{x=0}^{2\sqrt{2}} \pi = 8\pi.$$  Then subtract out the volume of the cone, which has a circular base of radius $1$ and height $2$, thus volume $\frac{1}{3}\pi r^2 h = \frac{2}{3}\pi$.  This gives the desired volume $$\left(8 - \frac{2}{3}\right) \pi = \frac{22}{3}\pi.$$  We could also have computed the parabolic region's volume via disks, which is
$$\int_{y=0}^2 4\pi y \, dy = 8\pi.$$
</s>