# Is there an easier way to solve this solid of revolution problem?

We just learned about solids of revolution today and one of the homework problems asks to find the volume obtained by revolving the region bounded by $x^2$ and $x^5$, and the revolution is around the $x$ axis. The easy solution is of course to find the volume made by $x^5$, the one made by $x^2$, and subtract. However, I really feel like there is another clever way (not necessarily simpler) to do this. I think that finding the volume created by $x^5-x^2$ in the same domain should also work. Any other ideas? Thanks!

• Not really. That's the standard way. – Potato Apr 2 '13 at 0:33
• I think that finding the volume created by x^5-x^2 in the same domain should also work, right? – Ovi Apr 2 '13 at 0:34
• @Ovi No. For example, think of the solid generated by rotating the region between $1$ and $2$ around the $x$-axis vs. the solid generated by rotating $2-1=1$ around the $x$-axis. The first has volume $3\pi r^2h$ while the second has volume $\pi r^2 h$. – A.S Apr 2 '13 at 0:37
• Oh yeah sorry I missed that. But I still think the general shape of x^5-x^2 should be the same, but only it has a smaller radius and therefore a smaller volume, right? – Ovi Apr 2 '13 at 0:40
• Haha never mind I just realized how stupid and wrong my idea was – Ovi Apr 2 '13 at 0:45

Let us do the volume generated by the first-quadrant part of region between the two curves.

Rewrite the curves as $x=y^{1/3}$ and $x=y^{1/5}$. Take a horizontal slice of our region, extending from height $y$ to height $y+dy$. Rotate this slice about the $x$-axis.

We get what you may soon be calling a cylindrical shell. The thickness of the shell is $dy$. The shell has perimeter $2\pi y$. And the width of the shell is $y^{1/5}-y^{1/3}$. So the volume of the shell is approximately $2\pi y(y^{1/5}-y^{1/3})\,dy$. For the full volume, "add up" (integrate) from $y=0$ to $y=1$. We end up with $$\int_0^1 2\pi y(y^{1/5}-y^{1/3})\,dy.$$

Remark: The method of cylindrical shells views the solid of revolution as being made up of thin cylindrical layers. The method you have been using views the solid as made up of stacked disks, possibly with holes in them.

For your problem, what you call the easy solution is quicker than cylindrical shells. Occasionally, cylindrical shells are more efficient.

And the answer to your "is there an easier solution" question is almost certainly no. That's why we supplied a solution based on a different idea, a solution that probably should be considered harder. The "harder" part comes from the fact that to use the idea, we had to solve for $x$ in terms of $y$. In this case, that posed little difficulty, but in general it could be the main hurdle.

• Thanks Andre, you're one of my favorite answerers haha – Ovi Apr 2 '13 at 22:21