# Sketch the region bounded by the curves: y=lnx,y=0y=lnx,y=0 and x=ex=e, then find, the area of this region, and find the volume of the solid generated by revolving this area about the line x=−2?x=−2?

Question
Integrals
Sketch the region bounded by the curves: $$\displaystyle{y}={\ln{{x}}},{y}={0}{y}={\ln{{x}}},{y}={0}$$ and $$\displaystyle{x}={e}{x}={e}$$, then find, the area of this region, and find the volume of the solid generated by revolving this area about the line $$\displaystyle{x}=−{2}?{x}=−{2}$$?

2021-01-29
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:
$$\displaystyle\pi{\int_{{{0}}}^{{{1}}}}{\left({e}+{2}\right)}^{{{2}}}-{\left({e}^{{{y}}}+{2}\right)}^{{2}}{\left.{d}{y}\right.}=π{\left[{\left({e}+{2}\right)}^{{2}}-{\frac{{{e}^{{{2}}}}}{{{2}}}}-{4}{e}+{\frac{{{1}}}{{{2}}}}\right]}={\frac{{π{\left({e}^{{2}}+{9}\right)}}}{{{2}}}}$$

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