I'm aware of the usual method for calculating the volume by expressing the integrals for $dr$ and $dz$ in terms of $z$ to get the correct answer but when I attempted to solve it expressing everything in terms of $r$ I can't quite seem to get it.

Trying to calculate the volume of a cone of radius $R$ and height $h$:

If we try to express everything in terms of $r$ then using similar triangles we obtain $r=\frac{zR}{h}$, now for integration limits $r:\frac{zR}{h}\to R$, $z: 0\to h$ and $\theta:0\to 2\pi$ so the integral becomes $$ \int_{0}^{2\pi}\int_{0}^{h}\int_{\frac{zR}{h}}^{R}r\, drdzd\theta=\pi\int_{0}^{h}(R^2-\frac{z^2R^2}{h^2})dz=\frac{\pi R^2}{h^2}\int_{0}^{h}(h^2-z^2)dz=\frac{\pi R^2}{h^2}(h^3-\frac{h^3}{3})\\ \int_C r\, drdzd\theta=\frac{2\pi R^2h}{3} $$

which is double the correct answer of $\frac{1}{3}\pi R^2h$


Let $(\rho,z,\phi)$ be the cylindrical coordinate of a point $(x,y,z)$. Let $r$ be the radius and $h$ be the height. Then $z \in [0,h], \phi \in [0,2\pi], \rho \in [0,rz/h]$. The volume is given by $$\begin{align*} \iiint_C\,dV &= \int_0^{2\pi} \int_0^h \int_0^{rz/h} \rho\,d\rho\,dz\,d\phi \\ &= 2\pi \int_0^h \left. \frac{\rho^2}{2} \right|_0^{rz/h}\,dz \\ &= \pi \int_0^h \frac{r^2 z^2}{h^2}\,dz \\ &= \frac{\pi r^2}{h^2} \frac{h^3}{3} = \frac{\pi r^2 h}{3} \end{align*}$$ as desired.

Your integral gives the volume of the inverse of a cone. That is, the part of a cylinder remained when a cone is removed from it.

  • $\begingroup$ Oh, yea that makes sense since it starts at $\rho=0$ for every integral. I'm confused now as to why if we integrate with respect to z first like here: math.washington.edu/~aloveles/Math324Fall2013/… they don't have 0 in the lower limit for z but instead rh/a $\endgroup$ – Craig Mar 25 '16 at 5:21
  • $\begingroup$ @Craig The order of integration does not matter. See Fubini's theorem. $\endgroup$ – Henricus V. Mar 25 '16 at 5:22
  • $\begingroup$ I'm referring to the limits they used for their integration, they have $\int_{0}^{2\pi}\int_{0}^{R}\int_{hr/R}^{R}r\, dzdrd\theta$ although I can't comprehend why the lower limit isn't 0 with respect to the $dz$ integration. $\endgroup$ – Craig Mar 25 '16 at 5:25
  • $\begingroup$ @Craig Try drawing an upside down cone with vertex at the origin. It might help clearing things up a bit. $\endgroup$ – Henricus V. Mar 25 '16 at 5:26

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