Forrest's question at Yahoo Answers regarding a solid of revolution

[MarkFL]

Here is the question:

Calculus Two Help Washers?

Find the volume of the solid by rotating the region bounded by the given curves
y= e^-x, y= 1 and x=2 about y=2
step by step would be very helpful thanks!

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Forrest,

I would begin by drawing a diagram of the region to be revolved, with the axis of rotation and the lengths of the inner and outer radius ($r$ and $R$ respectively):

View attachment 1315

The volume of an arbitrary washer is:

\(\displaystyle dV=\pi\left(R^2-r^2 \right)\,dx\)

where:

\(\displaystyle R=1+1-e^{-x}=2-e^{-x}\)

\(\displaystyle r=1\)

and so we have:

\(\displaystyle dV=\pi\left(\left(2-e^{-x} \right)^2-(1)^2 \right)\,dx=\pi\left(e^{-2x}-4e^{-x}+3 \right)\,dx\)

Summing the washers through integration, we obtain:

\(\displaystyle V=\pi\int_0^2 e^{-2x}-4e^{-x}+3\,dx\)

Applying the FTOC, we find:

\(\displaystyle V=\pi\left[-\frac{1}{2}e^{-2x}+4e^{-x}+3x \right]_0^2=\pi\left(\left(-\frac{1}{2}e^{-4}+4e^{-2}+6 \right)-\left(-\frac{1}{2}+4 \right) \right)=\frac{\pi}{2e^4}\left(5e^4+8e^2-1 \right)\)

If we wish to check our result by using the shell method, then refer to the following diagram:

View attachment 1316

The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dy\)

where:

\(\displaystyle r=2-y\)

\(\displaystyle h=2+\ln(y)\)

And so we have:

\(\displaystyle dV=2\pi (2-y)\left(2+\ln(y) \right)\,dy=2\pi\left(4+2\ln(y)-2y-y\ln(y) \right)\,dy\)

Summing the shells by integration, we have:

\(\displaystyle V=2\pi\int_{e^{-2}}^1 4+2\ln(y)-2y-y\ln(y)\,dy\)

Each of the terms can be readily integrated except for those involving the natural log function, so let's use integration by parts:

i) \(\displaystyle I=\int \ln(y)\,dy\)

\(\displaystyle u=\ln(y)\,\therefore\,du=\frac{1}{y}\,dy\)

\(\displaystyle dv=dy\,\therefore\,v=y\)

Hence:

\(\displaystyle I=y\ln(y)-\int\,dy=y\left(\ln(y)-1 \right)\)

ii) \(\displaystyle I=\int y\ln(y)\,dy\)

\(\displaystyle u=\ln(y)\,\therefore\,du=\frac{1}{y}\,dy\)

\(\displaystyle dv=y\,dy\,\therefore\,v=\frac{1}{2}y^2\)

Hence:

\(\displaystyle I=\frac{1}{2}y^2\ln(y)-\frac{1}{2}\int y\,dy\)

\(\displaystyle I=\frac{1}{2}y^2\ln(y)-\frac{1}{4}y^2=\frac{y^2}{4}\left(2\ln(y)-1 \right)\)

Now we may apply the FTOC to compute the volume:

\(\displaystyle V=2\pi\left[4y+2y\left(\ln(y)-1 \right)-y^2-\frac{y^2}{4}\left(2\ln(y)-1 \right) \right]_{e^{-2}}^1=2\pi\left(\left(\frac{5}{4} \right)-\left(-\frac{2}{e^2}+\frac{1}{4e^4} \right) \right)=\)

\(\displaystyle \frac{\pi}{2e^4}\left(5e^4+8e^2-1 \right)\)

And this checks with the washer method.