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.
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 $$\int_{y=0}^2 4\pi y \, dy = 8\pi.$$