Let $W$ be the ice cream cone

$$W= \Big\{(x,y,z) | \sqrt{3x^2+3y^2} \leq z \leq \sqrt{1-x^2-y^2}\Big\}$$ in spherical coordinates.

Is the correct answer

$$\int_{0}^{2\pi} \int_{0}^{\pi/6}\int_{0}^{1} \rho^2 \sin\phi d\rho d\phi d\theta$$

or is the answer

$$\int_{0}^{2\pi} \int_{0}^{\pi/6}\int_{1}^{0} \rho^2 \sin\phi d\rho d\phi d\theta$$

Since the region looks like this: enter image description here

And the radius goes from 1 to 0?

I can show my work for the remainder of the problem, but essentially I am just stuck on this one part for the bounds of $\rho$

  • $\begingroup$ Is the integral you're trying to evaluate $$ \iiint_W\sqrt{x^2+y^2+z^2}\,dxdydz\ ? $$ That's what your first integral in polar coordinates is equal to. Your second integral would evaluate to its negative. $\endgroup$ Jun 7, 2022 at 6:13

1 Answer 1


Using $\rho$ for radius, $\phi$ for latitude and $\theta$ for longitude:

$$x = \rho \cos\phi \cos\theta$$ $$y = \rho \cos\phi \sin\theta$$ $$z = \rho \sin\phi$$ $$ \rho\cos\phi \sqrt{3} \leq \rho\sin\phi\leq\sqrt{1-\rho^2\cos^2\phi} \Longleftrightarrow $$ $$ \Longleftrightarrow \tan \phi \geq \sqrt{3} \wedge \rho \leq 1 \Longleftrightarrow \frac{\pi}{3} \leq \phi \leq \frac{\pi}{2} \wedge \rho \leq 1 $$ And since in spehrical coordinates $dV = \rho^2 \cos\phi \;d\rho \;d\phi\;d\theta$ we get:

$$ V = \int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_0^1 \rho^2 \cos\phi \;d\rho \;d\phi\;d\theta \;\;=\pi\frac{2-\sqrt{3}}{3} $$

  • $\begingroup$ Thanks, actually a part of the question was to set it up in cylindrical and I got the same answer for that. But the other part of the question was to represent in spherical, which is why I was asking about that $\endgroup$ Jun 7, 2022 at 0:56
  • $\begingroup$ Sorry, my mistake. I've posted the result in spherical coordinates. $\endgroup$
    – blamocur
    Jun 7, 2022 at 12:13

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