Re: joey e's questions from Yahoo! Answers regarding solids of revolution
Hello joey e,
I am going to work both problems using the disk/washer method and the shell method so you can see both in action and as a means of checking my work, plus it's good practice when practical.
1.) Let's first look at a plot of the region to be revolved about the $x$-axis:
https://www.physicsforums.com/attachments/780._xfImport
Washer method:
Let's find the volume $dV$ of an arbitrary washer:
$$dV=\pi\left(R^2-r^2 \right)\,dx$$
$R$ is the outer radius, and it is given by:
$$R=8\sqrt{x}$$
$r$ is the inner radius, and it is given by:
$$r=8x^3$$
and so we have:
$$dV=64\pi\left(x-x^6 \right)\,dx$$
Summing the washers by integrating, we find:
$$V=64\pi\int_0^1 x-x^6\,dx=64\pi\left[\frac{x^2}{2}-\frac{x^7}{7} \right]_0^1=64\pi\left(\frac{1}{2}-\frac{1}{7} \right)=\frac{160\pi}{7}$$
Shell method:
The volume $dV$ of an arbitrary shell is:
$$dV=2\pi rh\,dy$$
where:
$$r=y$$
$$h=\frac{y^{\frac{1}{3}}}{2}-\frac{y^2}{64}$$
and so we have:
$$dV=\pi \left(y^{\frac{4}{3}}-\frac{y^3}{32} \right)\,dy$$
Summing the shells by integrating, we find:
$$V=\pi\int_0^8 y^{\frac{4}{3}}-\frac{y^3}{32}\,dy=\pi\left[\frac{3}{7}y^{\frac{7}{3}}-\frac{1}{128}y^4 \right]_0^8=\pi\left(\frac{3}{7}8^{\frac{7}{3}}-\frac{1}{128}8^4 \right)=\frac{160\pi}{7}$$
2.) Here is a plot of the region to be revolved about the $y$-axis:
https://www.physicsforums.com/attachments/781._xfImport
Washer method:
The volume $dV$ of an arbitrary washer is:
$$dV=\pi\left(R^2-r^2 \right)\,dy$$
$$R=4$$
$$r=\frac{1}{2}y^3$$
and so we have:
$$dV=\pi\left(16-\frac{1}{4}y^6 \right)\,dy$$
Summing the washers by integrating, we find:
$$V=\pi\int_0^2 16-\frac{1}{4}y^6\,dy=\pi\left[16y-\frac{1}{28}y^7 \right]_0^2=\pi\left(16\cdot7-\frac{1}{28}2^7 \right)=\frac{192\pi}{7}$$
Shell method:
The volume $dV$ of an arbitrary shell is:
$$dV=2\pi rh\,dx$$
$$r=x$$
$$h=(2x)^{\frac{1}{3}}$$
and so we have:
$$dV=2^{\frac{4}{3}}\pi x^{\frac{4}{3}}\,dx$$
Summing the shells by integration, we find:
$$V=2^{\frac{4}{3}}\pi\int_0^4 x^{\frac{4}{3}}\,dx=2^{\frac{4}{3}}\pi\left[\frac{3}{7}x^{\frac{7}{3}} \right]_0^4=2^{\frac{4}{3}}\pi\cdot\frac{3}{7}4^{\frac{7}{3}}=\frac{192\pi}{7}$$
To joey e and any other guests viewing this topic, I invite and encourage you to post other solids of revolution problems here in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.
