The curve of $\log x$ cuts the $x$-axis at $x=1$ and move to infinity. The line $y=0$ is the $x$-axis itself and $x=2$ is a line parallel to the $y$-axis and it cuts the $\log x$ curve.
So I get a shape with boundaries $\log x$, $x=2$ and the $x$-axis.
If I rotate it about the $x$-axis the formula for the volume is $$ V=\int_1^2 \pi y^2\, dx $$
I came this far, but I don't know what the integral of $(\log x)^2$ is.
Please help me guys :)
The answer is also given: it is $2\pi(1-\log 2)^2$ cubic units.
You can use the following : $$\begin{align}\int(\log x)^2dx&=\int (x)^\prime (\log x)^2dx\\&=x(\log x)^2-\int x\cdot 2\log x\cdot(1/x)dx\\&=x(\log x)^2-2\int (x)^\prime \log xdx\\&=x(\log x)^2-2\left(x\log x-x\right)+C.\end{align}$$