So I have to evaluate $$\iiint_S \! (y^2+z^2)\ \mathrm{d}x\ \mathrm{d}y\ \mathrm{d}z$$ where $S$ is a right circular cone of altitude $h$ with its base, of radius $a$, in the $xy$-plane and its axis along the $z$-axis, in cylindrical coordinates. Switching to cylindrical coordinates, we clearly have $0 \leq \theta \leq 2\pi$ and $0 \leq z \leq h$. I imagine a cross section of the cone as a right triangle with $\rho$ from the origin to the hypotenuse and a horizontal line at $z$ intersecting $\rho$ at the hypotenuse. Then $\rho=\sqrt{z^2+x^2}$, where $x$ is the length of the horizontal line. The horizontal line forms another triangle similar to the first, and so its length should be $$\frac{x}{h-z}=\frac{a}{h}$$ $$x=\frac{a(h-z)}{h}$$ so I have the inequality $$0 \leq \rho \leq \sqrt{\left(\frac{a(h-z)}{h}\right)^2+z^2}$$ but when I try to evaluate the integral with this bound, it ends up very wrong. I assume I've done something wrong finding the bounds for $\rho$, but I can't figure out what it is so I was hoping someone could tell me what I'm doing wrong. I tried doing another substitution that mapped the cone to the cone with $h=1$ and $a=1$, but that didn't really help. The book has the answer as $\frac{1}{60}\pi a^2h(3a^2+2h^2)$, but I'd like to know how to get there. Thanks for the help!

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  • $\begingroup$ I'm not quite understanding what $\rho$ is representing. In fact, the geometry of the whole problem is a little vague. Could you draw a picture? $\endgroup$ – dfeuer May 31 '13 at 22:32
  • $\begingroup$ You say $0 < z < 1$. Don't you mean $0 < z < h$? $\endgroup$ – dfeuer May 31 '13 at 22:36
  • $\begingroup$ Yeah I meant $0<z<h$ $\endgroup$ – desi May 31 '13 at 22:41
  • $\begingroup$ $\rho$ is the distance from the origin to a point in the cone. I'm trying to maximize $\rho$ when $z$ and $\theta$ are constant, but maybe I'm not supposed to be doing that and just don't understand triple integrals. I added a link to my picture of the triangle $\endgroup$ – desi May 31 '13 at 22:45
  • $\begingroup$ I don't really think $\rho$ is going to be useful. $\endgroup$ – dfeuer May 31 '13 at 23:11

Usually in cylindrical coordinates you would have $r =\sqrt{x^2+y^2}$ (rather than dealing with $\rho$) and that looks like a good choice for the problem. Then we have $$\begin {align}\iiint_S \! (y^2+z^2)\,dx\,dy\,dz&=\int_0^h\int_0^{a(1-z/h)}\int_0^{2\pi}(r^2\sin^2\theta+z^2)r\;d\theta\;dr\; dz\\ &=\int_0^h\int_0^{a(1-z/h)} \left( \pi r^3+2\pi z^2r\right)\;dr\; dz\\ &=\int_0^h\left. \left(\frac \pi 4 r^4+\pi z^2r^2\right)\right|_0^{a(1-z/h)}\; dz\\ &=\int_0^h \left(\frac \pi 4ha^4(1-\frac zh)^4 +\pi z^2a^2(1-\frac zh)^2\right) \;dz \end {align}$$

  • $\begingroup$ Where'd you get that value for the integral of $r^2\sin^2\theta$? $\endgroup$ – dfeuer May 31 '13 at 23:29
  • $\begingroup$ You're also missing an $a^2$ somewhere, I believe. $\endgroup$ – dfeuer May 31 '13 at 23:31
  • $\begingroup$ The average of $\sin^2 \theta$ is $\frac 12$ and the length of the interval is $2 \pi$ so the $\theta$ integral has value $\pi$. You can use the identity $\sin^2 \theta=\frac 12 (1+\cos 2 \theta)$ to get it. $\endgroup$ – Ross Millikan May 31 '13 at 23:32
  • $\begingroup$ @dfeuer: yes, it got dropped in the eval in the last line. Fixed. $\endgroup$ – Ross Millikan May 31 '13 at 23:34
  • $\begingroup$ How did you get that value for the bound of $r$? $\endgroup$ – desi May 31 '13 at 23:35

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