 Cabiolab

2021-01-28

Sketch the region bounded by the curves: $y=\mathrm{ln}x,y=0y=\mathrm{ln}x,y=0$ 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 $x=-2?x=-2$? 2abehn

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 $\mathrm{ln}\left(x\right)$ from 11 to e.
${\int }_{1}^{e}\mathrm{ln}\left(x\right)dx=\left[x\mathrm{ln}x-x\right]e1=1$
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:
$\pi {\int }_{0}^{1}{\left(e+2\right)}^{2}-{\left({e}^{y}+2\right)}^{2}dy=\pi \left[{\left(e+2\right)}^{2}-\frac{{e}^{2}}{2}-4e+\frac{1}{2}\right]=\frac{\pi \left({e}^{2}+9\right)}{2}$

Do you have a similar question?