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

Skilled2021-01-29Added 88 answers

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.

${\int}_{1}^{e}\mathrm{ln}\left(x\right)dx=[x\mathrm{ln}x-x]e1=1$

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:

$\pi {\int}_{0}^{1}{(e+2)}^{2}-{({e}^{y}+2)}^{2}dy=\pi [{(e+2)}^{2}-\frac{{e}^{2}}{2}-4e+\frac{1}{2}]=\frac{\pi ({e}^{2}+9)}{2}$

You should notice that we are essentially finding the area of the curve

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:

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