# Volume bound by surface using cylindrical coordinates

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?

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 $$r<0$$
• 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? Feb 12 '20 at 21:00
• 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 $\rho$ Feb 12 '20 at 21:08
• 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? Feb 12 '20 at 21:34