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

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.

Thanks,

Martin Sleziak
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User001
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  • By symmetry, it is $\frac23 \int_{S} d\sigma$. – achille hui Jun 22 '15 at 03:42
  • Can you elaborate, @achillehui? Thanks, – User001 Jun 22 '15 at 03:44
  • @matthewlevy, - should I be thinking about an n*ds factor in the integrand? – User001 Jun 22 '15 at 03:49
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    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. $$\int_S d\sigma = \int_0^{\pi} \int_0^{2\pi} \sin\theta d\phi d\theta = 2\pi \int_0^{\pi}\sin\theta d\theta = 4\pi$$ – achille hui Jun 22 '15 at 03:51
  • very cool, @achillehui. I have never used spherical-polar coordinates before. So, by dropping the integration over the radius variable, r, it's essentially switching from a volume integral to a surface integral, but you keep the sin$\theta$ factor of the Jacobian determinant (while dropping the r^2 factor).? – User001 Jun 22 '15 at 04:30
  • @LebronJames: Symmetry is the way to go here but tt's not too much harder to do it from scratch, if you haven't seen these symmetry arguments before. – Matematleta Jun 22 '15 at 04:36
  • http://math.stackexchange.com/questions/1073275/calculate-surface-area-of-a-f-using-the-surface-integral/1073837#1073837 – Alec Teal Jun 22 '15 at 04:44
  • @LebronJames essentially yes, you just need to keep the $\sin\theta$ for the Jacobian determinant. In any event, if you parametrize the points on your surface as $\vec{x} = (\sin\theta\cos\phi, \sin\theta,\sin\phi,\cos\theta)$, you can compute the surface area element directly $$d\sigma =\left |\frac{\partial \vec{x}}{\partial\theta} \times \frac{\partial\vec{x}}{\partial\phi}\right|d\theta d\phi = \sin\theta d\theta d\phi$$. – achille hui Jun 22 '15 at 05:08
  • thanks so much, @achillehui. my answer agrees with yours: 4pi. but mysteriously, robjohn and chilango both got 8pi/3, and also an old student solution also has 8pi/3. i've computed many, many times already tonight to verify the jacobian computation and the integration of sin^3, etc, and am still getting 4pi. :( – User001 Jun 22 '15 at 07:33
  • and thanks so much for the symmetry argument @achillehui - it's very cool to learn a bit more geometrical methods, along the way. – User001 Jun 22 '15 at 07:36
  • @LebronJames my answer is also $8\pi/3$, you forget one need to multiply the $4\pi$ by the $\frac23$ factor from symmetry. – achille hui Jun 22 '15 at 07:37
  • That is strange, @achillehui, because, yes I didn't want to use the symmetry argument and instead computed the integral explicitly, so I expanded out the x^2 and the y^2, in terms of phi and theta, computed the Jacobian correctly to get sin(phi), and fixed radius = 1, so that there is no differential, dr. This integration gives me 4pi. So, why doesn't it match the computation for the symmetric argument? hmm... – User001 Jun 22 '15 at 07:50
  • Oh, gosh. I had the answer all written up and was ready to post it for you to read, when I then did a final check on my computations and realized that the terms -1/3 and -1/3 were to be added - not cancelled! haha, I do indeed get the answer of 8pi/3. Thanks so much for your help and for your patience, @achillehui!! I need some sleep. Have a great day in HK!! – User001 Jun 22 '15 at 08:25
  • Hi @achillehui, I was just curious: how come I didn't have to compute a normal vector, n? There's always that tricky nds component to figure out in surface integrals, and I know that, based on the online MIT course that I watch when I want to review multivariable calculus, that the surface area element that we computed is not really exactly the nds component... – User001 Jun 24 '15 at 01:19
  • ...meaning that the Jacobian is not = n, and ds is not equal to $d\phi$$d\theta$ – User001 Jun 24 '15 at 01:21
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    $\vec{n} = \frac{\vec{x}_u \times \vec{x}_v}{| \vec{x}_u \times \vec{x}_v|}$, $d\vec{S} = \left( \vec{x}_u \times \vec{x}_v\right) du dv$, so $d\sigma = \vec{n}\cdot d \vec{S} = |\vec{x}_u \times \vec{x}_v| du dv$. – achille hui Jun 24 '15 at 03:11
  • ok, got it - thanks again, @achillehui. – User001 Jun 24 '15 at 03:22

2 Answers2

2

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

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

Matematleta
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  • 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. – User001 Jun 22 '15 at 07:30
  • I actually (finally) got the right answer of 8pi/3 -- thanks so much, @chilango! – User001 Jun 22 '15 at 08:26
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    @LebronJames: glad to help! – Matematleta Jun 22 '15 at 12:31
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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 $$ 2\pi\int_{-1}^1(1-z^2)\,\mathrm{d}z=\frac{8\pi}3 $$

Community
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robjohn
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  • 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. – User001 Jun 22 '15 at 07:28
  • 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... – User001 Jun 22 '15 at 07:40
  • Hi @RobJohn - I got the right answer of 8pi/3. Now I can go to bed. Thanks so much for your simple solution! – User001 Jun 22 '15 at 08:29
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    @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. – robjohn Jun 22 '15 at 08:36