Math Forum

by **aaron-math** » Sun Oct 16, 2011 10:17 pm

Find the volume V of the described solid S.
A frustum of a right circular cone with height h, lower base radius R, and top radius r

This one I am not sure about. I guess they are just asking for a general formula because they dont give any numbers?

My best guess is to integrate the volume of a cylinder from R to r:

$V = \int_r^R \, 2\pi r^2 h dr$

but I think that is reaching at best.

by **MarkFL** » Sun Oct 16, 2011 10:43 pm

Think of it as a volume by slicing, where we slice the frustum into a stack of disks:

$V=\pi\int_0\,^h x^2\,dy$

x will be a linear function of y, passing through the points (R,0) and (r,h):

$x^2=$\frac{r-R}{h}y+R$^2$ thus we have:

$V=\pi\int_0\,^h $\frac{r-R}{h}y+R$^2\,dy$

Let $u=\frac{r-R}{h}y+R\:\therefore\:du=\frac{r-R}{h}\,dy$ giving:

$V=\frac{\pi h}{r-R}\int_R\,^{r}u^2\,du=\frac{\pi h}{3(r-R)}\[u^3\]_R^r=\frac{\pi h}{3(r-R)}$r^3-R^3$=\frac{\pi h}{3}$r^2+rR+R^2$$

by **aaron-math** » Sun Oct 16, 2011 10:58 pm

Impressive. The solution is so much more involved that I expected.

by **skipjack** » Wed Oct 19, 2011 3:10 am

aaron-math wrote:Find the volume V of the described solid S.

Can the standard formula base area × height/3 for the volume of a cone be used? It would make this problem easy.

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