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.
 A: 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?
A: $y = 3$ intersects $xy = 2$ at $x = \frac 23$ and $xy = 4$ at $x = \frac 43$
By shells I get.
$2\pi\int_\frac {2}{3}^{\frac 43} (3-\frac 2x)x \ dx + 2\pi\int_{\frac 43}^2 (\frac 4x -\frac 2x)x \ dx\\
2\pi\int_\frac {2}{3}^{\frac 43} 3x-2 \ dx + 2\pi\int_{\frac 43}^2 2\ dx\\   
2\pi(\frac{3}{2}x^2 -2x|_\frac {2}{3}^{\frac 43} + 2x| _{\frac 43}^2)\\
2\pi(\frac 83 - \frac 23 - \frac 83 + \frac 43 + 4 - \frac 83) \\
4\pi$
