Evaluate the flux $\mathbf{F} = (-x, -y, z)$ across the surface $S$ that is the part of the cone $z = \sqrt{x^2 + y^2}$ that lies between the planes $z = 1$ and $z = 2$ and has inward orientation. Use spherical coordinates.

I've graphed the cone and the two planes intersecting through it. I know how to calculate flux using the correct orientation, and I know spherical coordinates. What I can't figure out is how to find the limits of integration using spherical coordinates. For instance, I know that $\theta \in [0,2\pi]$ for this problem, but how do I find $\rho$ and $\phi$?

I would greatly appreciate it if people could please explain how to find the limits of integration in spherical coordinates for this problem after having graphed the object.


1 Answer 1


The conversion from Cartesian coordinates to spherical coordinates is:

$$\begin{align}{ \rho = \sqrt{x^{2}+y^{2}+z^{2}} \\ \theta = \arctan{\frac{y}{x}} \\ \phi = \arccos{\frac{z}{\rho}}}\end{align}$$

You've already ascertained that $\theta \in [0,2\pi]$ by rotational symmetry, but you can use the equation of the cone to work out the integration limits of $\rho$ and $\phi$ by expressing them in terms of $z$. Answer below.

$\rho = \sqrt{z^{2}+z^{2}} = \sqrt{2}z \,;\;\phi = \arccos{\frac{1}{\sqrt{2}}} = \pi/4$. So $\rho \in [\sqrt{2},2\sqrt{2}]$, and as $\phi$ is constant you wouldn't even integrate over it (which makes sense, since you're integrating over a surface, so there should only be two integration variables, which here are $\rho$ and $\theta$).

  • $\begingroup$ Thanks for the response. I am confused; $\phi = \arccos{\dfrac{1}{\sqrt{2}}} = \pi/4$? How did you get $\pi/2$? $\endgroup$ Commented Aug 9, 2017 at 16:24
  • 1
    $\begingroup$ Yep, I messed up. Corrected, thanks! $\endgroup$ Commented Aug 10, 2017 at 8:13

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