I want to know how to calculate the volume bound by a surface $f(x,y)$ in cylindrical coordinates. As an example, for a sphere of radius $r$, I would use $f(x,y) = \sqrt{r^2 - x^2 - y^2}$, and $V = \int^r_{-r}\int^r_{-r} f(x,y)dxdy$ in Cartesian coordinates.

To do the same in cylindrical coordinates, I do $x\rightarrow x$, $y\rightarrow \rho\sin\theta$, $z\rightarrow \rho\cos\theta$. To calculate the volume, I can then do $V = \int_0^{2\pi}\int_{-r}^r \rho^2 (x,\theta)dxd\theta$, where $\rho(x,\theta)$ is the cylindrical radius corresponding to the surface $f(x,y)$: $x^2 + \rho^2 = r^2 \Leftrightarrow \rho^2 (x,\theta) = r^2 - x^2$. One $\rho$ comes from the $f(x,y)$, and the other from going to cylindrical coordinates. When I calculate this integral, however, the outcome is $V = \frac{8}{3}\pi r^3$, which is off by a factor 2. What am I doing wrong here?


1 Answer 1


This is because the limits $0<\theta <2\pi$ cover two times the volume: one for the values that gives $r\ge 0$ and the other for the values that gives $r<0$

  • $\begingroup$ I just discussed the same problem in a comment to my answer to math.stackexchange.com/questions/3543966/… $\endgroup$ Commented Feb 12, 2020 at 20:02
  • $\begingroup$ Thanks for the quick reply! I'm afraid I don't fully understand. I don't integrate over $r$, does that mean that I implicitly assume it can be both positive and negative? $\endgroup$
    – John Smith
    Commented Feb 12, 2020 at 21:00
  • $\begingroup$ Sorry, you use $\rho$ as variable, and, yes, using $0<\theta<2\pi$ you take also the values for which $\rho$ is negative and this means that, reversing the direction on the line oriented by $\theta$, you takes two fold the same values of $\rho$ $\endgroup$ Commented Feb 12, 2020 at 21:08
  • $\begingroup$ Ah thanks, that makes sense! However, when I calculate this integral numerically, I always only calculate $\rho > 0$, but I still get twice the volume. How does that work? $\endgroup$
    – John Smith
    Commented Feb 12, 2020 at 21:34

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