Sketch the region bounded by the curves: y=ln⁡x,y=0y=ln⁡x,y=0 and x=ex=e, then find, the area of...

The sketch is easily obtainable through a graphing software/website, it’s also very easy to sketch it from basic principles. As for the area of the figure,
You should notice that we are essentially finding the area of the curve ${\displaystyle \ln \left(x\right)}$ from 11 to e.
${\displaystyle {\int }_1^e\ln \left(x\right)dx=[x\ln x-x]e1=1}$
As for the solid of revolution, we have to use the washer method, which has a ring of constant outer radius (e+2)(e+2) and a ring of inner radius ey+2
Therefore, we can just compute:
${\displaystyle \pi {\int }_0^1{(e+2)}^2-{(e^y+2)}^{2}dy=\pi [{(e+2)}^2-\frac{e^2}{2}-4e+\frac{1}{2}]=\frac{\pi (e^2+9)}{2}}$