Volume of the revolution solid Let $K \subset \mathbb R^2$ be area bounded by curves $x=2$, $y=3$, $xy=2$ and $xy=4$.
What is the volume of the solid we get after rotating $K$ around $y$-axis?
I tried to express the curves in terms of $y$ and solved the integral $\displaystyle \pi\left(\int_2^3 \frac{16}{y^2}dy - \int_1^3\frac{4}{y^2}dy \right)$ but it came out zero. I got the integral by subtracting the volume of the inner shape from the volume of the outer shape, if that makes any sense.

Do this in two parts. First find the section of the volume between $y=3$ and $y=2$, then the section of the volume between $y=2$ and $y=1$. For the first volume, use the integral
$$\pi\int_{2}^{3} \frac{16}{x^2}dx-\pi\int_{2}^{3} \frac{4}{x^2}dx$$
This is the difference of the volumes formed by rotating the regions between $xy=4$ and the y-axis and between $xy=2$ and the y-axis between $y=2$ and $y=3$. When you evaluate these integrals, you get
$$\pi\int_{2}^{3} \frac{12}{x^2}dx$$
$$\pi(-\frac{12}{3}+\frac{12}{2})$$
$$\pi(6-4)$$
$$2\pi$$
Now for the second region. For this I use the integrals
$$4\pi-\pi\int_{1}^{2} \frac{4}{x^2}dx$$
Which is the difference of the volumes of the cylinder made by rotating a rectangle about the x-axis and the volume of the area under $xy=2$ rotated about the x-axis. This gives us
$$4\pi-\pi(-\frac{4}{2}+\frac{4}{1})$$
$$4\pi-2\pi$$
$$2\pi$$
The total volume is the sum of the volumes, which is
$$4\pi$$
Is this the correct answer?
