$$\int_S (x^2 + y^2)d\sigma,$$

where S is the sphere of radius 1 centered at (0,0,0) and $\sigma$ is surface area.

I would like some hints on how to proceed. This is tricky, since I am not being asked for a volume integral computation, so I can't use spherical coordinates, I think.


  • $\begingroup$ By symmetry, it is $\frac23 \int_{S} d\sigma$. $\endgroup$ – achille hui Jun 22 '15 at 3:42
  • $\begingroup$ Can you elaborate, @achillehui? Thanks, $\endgroup$ – User001 Jun 22 '15 at 3:44
  • $\begingroup$ @matthewlevy, - should I be thinking about an n*ds factor in the integrand? $\endgroup$ – User001 Jun 22 '15 at 3:49
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    $\begingroup$ The surface $S$ is symmetric under any permutation of the 3 coordinates $x,y,z$, so $$\int_S x^2 d\sigma = \int_S y^2 d\sigma = \int_S z^2 d\sigma \implies \int_S (x^2+y^2)d\sigma = \frac23 \int_S (x^2+y^2+z^2) d\sigma = \frac23\int_S d\sigma$$ because $x^2+y^2+z^2 = 1$ on $S$. To proceed, you either use the fact the surface area of unit sphere is $4\pi$ or perform the integration in spherical polar coordinates directly. $$\int_S d\sigma = \int_0^{\pi} \int_0^{2\pi} \sin\theta d\phi d\theta = 2\pi \int_0^{\pi}\sin\theta d\theta = 4\pi$$ $\endgroup$ – achille hui Jun 22 '15 at 3:51
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    $\begingroup$ $\vec{n} = \frac{\vec{x}_u \times \vec{x}_v}{| \vec{x}_u \times \vec{x}_v|}$, $d\vec{S} = \left( \vec{x}_u \times \vec{x}_v\right) du dv$, so $d\sigma = \vec{n}\cdot d \vec{S} = |\vec{x}_u \times \vec{x}_v| du dv$. $\endgroup$ – achille hui Jun 24 '15 at 3:11

Use spherical coordinates with $\rho =1$

In detail:

if you parameterize the sphere by setting

$$\textbf u=\cos \theta \sin \phi \textbf i+\sin \theta \sin \phi \textbf j+\cos \phi \textbf k$$ and then compute the Jacobian (surface element) by taking $\vert \textbf u_{\theta }\times \textbf u_{\phi }\vert $ you get $\sin \phi $.

Also $$x^{2}+y^{2}=1-z^{2}=1-\cos ^{2}\phi =\sin^{2}\phi $$

So your integral is

$$\int_{0}^{\pi }\int_{0}^{2\pi }\sin^{3} \phi d\theta d\phi=2\pi \int_{0}^{\pi }\sin^{3} \phi d\phi =2\pi \left ( \frac{4}{3} \right )=\frac{8}{3}\pi$$

  • $\begingroup$ Hi @chilango, I am getting 4pi as my answer. As I mentioned above to RobJohn, I see a 1/3 and -1/3 canceling from the integration of sin^3 -- instead of 1/3 + 1/3, which would then agree with your answer. $\endgroup$ – User001 Jun 22 '15 at 7:30
  • $\begingroup$ I actually (finally) got the right answer of 8pi/3 -- thanks so much, @chilango! $\endgroup$ – User001 Jun 22 '15 at 8:26
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    $\begingroup$ @LebronJames: glad to help! $\endgroup$ – Matematleta Jun 22 '15 at 12:31

As described in this answer, the area of an annulus on a sphere of radius $r$ between $z=a$ and $z=b$ is $$ 2\pi r(b-a) $$ This is also mentioned in this Wikipedia article.

Thus, on the the surface of a sphere, the integral of a function dependent only on $z$ is $$ 2\pi r\int_{-r}^rf(z)\,\mathrm{d}z $$ In the case here, we have $r=1$ and $f(z)=1-z^2$, so we get $$ 2\pi\int_{-1}^1(1-z^2)\,\mathrm{d}z=\frac{8\pi}3 $$

  • $\begingroup$ Hi @RobJohn, I've checked my computations several times and am getting 4pi as my answer, which agrees with Achille Hui's answer, too. What do you think? The issue seems to be a 2/3 missing, but my computations show that a 1/3 and a -1/3 cancel -- and doesn't get added to contribute the 2/3. $\endgroup$ – User001 Jun 22 '15 at 7:28
  • $\begingroup$ Another thing is that the inner integral has inverted upper and lower limits, but this is changed back to normal because of a minus sign coming out from the substitution u = cos (phi or theta, depending on your notation). Maybe that's where the extra factor of 2/3 is coming from, but I think it might be wrong... $\endgroup$ – User001 Jun 22 '15 at 7:40
  • $\begingroup$ Hi @RobJohn - I got the right answer of 8pi/3. Now I can go to bed. Thanks so much for your simple solution! $\endgroup$ – User001 Jun 22 '15 at 8:29
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    $\begingroup$ @LebronJames: sorry for being gone so long. $4\pi$ is the area of the sphere, so it is $\int_{S^2}(x^2+y^2+z^2)\,\mathrm{d}\sigma=\int_{S^2}1\,\mathrm{d}\sigma=4\pi$. Your integral is $\frac23$ of that; that is, $\int_{S^2}(x^2+y^2)\,\mathrm{d}\sigma=\frac23\cdot4\pi=\frac{8\pi}3$. Sleep well. $\endgroup$ – robjohn Jun 22 '15 at 8:36

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