Set up an integral to find the volume of the solid generated by revolving the region from x=1 to x=2 about the line x = 3 using the disk method. Do not evaluate the integral. Your answer should not contain absolute value signs. (hint: the inverse function of $e^{ax}$ is $\frac{lnx}{a}$ and the inverse function of ln(ax) is $\frac{e^x}{a}$.)
Here is a graph of the problem
With a = 48, my two equations that make the region are ln(48x) (top) and $e^{-48x}$ (bottom). My initial approach was to split the area into three integrals, ln(48) to ln(96), $e^{-48}$ to ln(48), and $e^{-96}$ to $e^{-48}$. Now, I don't know what to put in the integrands.
Any help is appreciated.
I figured it out myself. Here's the answer:
$$\pi\int_{\ln48}^{\ln96}\left(\left(3-\frac{e^x}{48}\right)^2-1\right)dx+\pi\int_{e^{-48}}^{\ln48}(2^2-1)dx+\pi\int_{e^{-96}}^{e^{-48}}\left(\left(3-\frac{\ln x}{48}\right)^2-1\right)dx$$