Write a triple integral in spherical coordinates that expresses the volume of the solid formed when a sphere with radius $a$ tangent to the $xy$ plane at the origin intersects at the plane z = a.

(Equation of the sphere is $x^2 + y^2 + (z-a)^2 = a^2$)

I recognize that this is just half the volume of the sphere, but I need to just write the integral expressing this volume.

I'm having some trouble because it is translate from the origin... Not sure how that affects the values for phi and rho. Some clarity would be appreciated.

  • $\begingroup$ Is the 'solid formed' the portion of the sphere above the plane or below the plane...? $\endgroup$
    – icurays1
    Nov 20, 2012 at 7:17
  • $\begingroup$ above the plane $\endgroup$ Nov 20, 2012 at 7:19
  • $\begingroup$ ....any ideas? :( $\endgroup$ Nov 20, 2012 at 7:58
  • $\begingroup$ Translate it back. $\endgroup$ Nov 20, 2012 at 8:45

1 Answer 1


I'll be using $(\rho,\theta,\phi)$ to represent respectively the radius, azimuth angle (the one in the $xy$ plane) and zenith angle (the one measured from the $z$-axis). Also, since the problem is trivial when $a=0$, assume $a>0$.

First, notice that the sphere is symmetrical with respect to $\theta$, so we can fix a vertical cross section to establish our $\rho$ and $\phi$ integration limits. Here is a picture of the sphere in the $zx$-plane (sorry about the glare):

enter image description here

Now, notice right away that we must have $0\leq\phi\leq\pi/4$. $\phi=0$ represents the coordinate point $(0,0,2a)$, while $\phi=\pi/4$ represents the coordinate point $(a,0,a)$. If you're not quite convinced of this second one just draw a triangle. So the only difficult part really is establishing the $\rho$ limits, which will be in terms of $\phi$. In other words, our integral will look like: $$ \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\pi/4}\int_{\rho=\rho_1(\phi)}^{\rho=\rho_2(\phi)}\rho^2\sin\phi d\rho d\phi d\theta $$

I've labelled the $\rho$ limits $\rho_1$ and $\rho_2$ in the picture ($\rho_2$ is a little hard to see - its the distance from the origin to the point on the surface of the sphere).

For $\rho_1$: the triangle drawn illustrates how to find the distance from the origin to the plane $z=a$ in terms of $\phi$. This is our $\rho_1$: $$\rho_1=\frac{a}{\cos\phi}$$

For $\rho_2$, we need to find a point on the surface of the sphere. For that, we use the equation of the sphere, which is re-written at the top left of the picture, and make our substitutions $\rho^2=x^2+y^2+z^2$ and $z=r\cos\phi$ to arrive at:


and thus


We're done! The integral is now ready for you to integrate:

$$ \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\pi/4}\int_{\rho=\frac{a}{\cos\phi}}^{\rho=2a\cos\phi}\rho^2\sin\phi d\rho d\phi d\theta $$

I've checked this answer with Mathematica, and indeed we arrive at the expected result $\frac{2\pi a^2}{3}$.

Hope this helps. I didn't work out the integration because I assume you can do that step. It should be straightforward - you might need to look up/re-derive a couple simple trigonometric integrals.

  • $\begingroup$ great answer, thanks. I really see the geometric intuition behind it now $\endgroup$ Nov 20, 2012 at 13:44
  • 1
    $\begingroup$ Very late comment, would the bounds $[0,\frac{\pi}{2}], [0,2\pi].[0, a \sec \phi]$ be correct for $\phi , \theta , \rho$ respectively? $\endgroup$ May 4, 2017 at 10:59

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