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.

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