<s>[INST] 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.
 [/INST] 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?
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