I have the problem: Assuming $y = ln(x)$, and $y = 0$, find the volume bound by these two lines and the point $x = 2$ if the area were rotated around the $x$-axis.
I ended up with $2\pi\int_1^2 \ln(x) dx $
I would like to know if this is the correct way to set this problem up given the bounds.
If it is the finite region bounded by the lines $y=0$, $x=2$, and the curve $y=\ln x$, and we are rotating about the $x$-axis, then the volume is $$\int_1^2 \pi(\ln x)^2\,dx.$$ For take a slice of the solid at $x$. Then the radius of cross-section is $y$, that is, $\ln x$, so the area of cross-section is $\pi(\ln x)^2$.
Remark: If you want to use cylindrical shells instead, then we get the expression $$\int_0^{\ln 2}2\pi y(2-e^y)\,dy$$ for the volume.