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.

\(\displaystyle{\int_{{{1}}}^{{{e}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left[{x}{\ln{{x}}}−{x}\right]}{e}{1}={1}\)

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:

\(\displaystyle\pi{\int_{{{0}}}^{{{1}}}}{\left({e}+{2}\right)}^{{{2}}}-{\left({e}^{{{y}}}+{2}\right)}^{{2}}{\left.{d}{y}\right.}=π{\left[{\left({e}+{2}\right)}^{{2}}-{\frac{{{e}^{{{2}}}}}{{{2}}}}-{4}{e}+{\frac{{{1}}}{{{2}}}}\right]}={\frac{{π{\left({e}^{{2}}+{9}\right)}}}{{{2}}}}\)

You should notice that we are essentially finding the area of the curve \(\displaystyle{\ln{{\left({x}\right)}}}\) from 11 to e.

\(\displaystyle{\int_{{{1}}}^{{{e}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left[{x}{\ln{{x}}}−{x}\right]}{e}{1}={1}\)

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:

\(\displaystyle\pi{\int_{{{0}}}^{{{1}}}}{\left({e}+{2}\right)}^{{{2}}}-{\left({e}^{{{y}}}+{2}\right)}^{{2}}{\left.{d}{y}\right.}=π{\left[{\left({e}+{2}\right)}^{{2}}-{\frac{{{e}^{{{2}}}}}{{{2}}}}-{4}{e}+{\frac{{{1}}}{{{2}}}}\right]}={\frac{{π{\left({e}^{{2}}+{9}\right)}}}{{{2}}}}\)