Find the volume of the function $y=\frac{6}{x}$

Question

Consider the function $y=\frac{6}{x}$ bounded by $y=0$, $x=1$,and $x=3$ and is rotated around the $x$-axis.

Using the disk method I setup the integral as $$\pi\int_1^33^2-\left(\frac{6}{x} \right)^2dx $$ solving gave me $42\pi$ but the answer should be $24\pi$, my question is where did I mess up?

Answer 1

The integral should be

$$\pi\int_1^3 \left(\frac{6}{x} \right)^2dx =24\pi $$

where $6/x$ is the radius of the disk at $x$. Also note the volume revolves around $x$-axis, not $y$.

Answer 2

The integrand is incorrect. It should be just $(\frac{6}{x})^2$. There's no reason to subtract this from $3^2$.