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I'm trying to compute $\iint_S F\cdot$ dS where $F=\langle x,y,z^2\rangle$ and $S$ is the unit sphere centered at the origin.

Here's my attempt:

On the sphere we can describe any point by $r(\phi,\theta)=\langle \sin\phi \cos\theta,\sin\phi \cos\theta, \cos\theta\rangle$ and the outward normal vector to $S$ is given by $n=r.$ Therefore,

$$\iint_S F\cdot dS = \iint_D F(r(\phi,\theta))\cdot n dA$$ $$=\iint_D \langle \sin\phi\cos\theta,\sin\phi\sin\theta,\cos^2\theta\rangle\cdot \langle \sin\phi\cos\theta,\sin\phi\sin\theta,\cos\theta\rangle dA$$ $$=\iint_D \sin^2\phi(\cos^2\theta + \sin^2\theta)+\cos^3d\theta dA = \int_0^{\pi}\int_0^{2\pi} \sin^2\phi +\cos^3\theta d\theta d\phi = \pi^2$$

however the answer is $\frac{8}{3}\pi$ and the last equals sign is correct (used Wolfram to confirm), so I guess I did something wrong in the setup.

user351797
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2 Answers2

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The position vector should be $$r(\phi,\theta)=\langle \sin\theta \cos\phi,\sin\theta \sin\phi, \cos\theta\rangle.$$ And the volume element should be $$ dA = \langle \sin\theta \cos\phi,\sin\theta \sin\phi, \cos\theta\rangle \sin \theta \ d\theta \ d\phi.$$ So the integral should be $$\iint_S F\cdot dA = \int_{\theta = 0}^{\theta = \pi} \int_{\phi = 0}^{\phi = 2\pi} \left( \sin^2 \theta + \cos^3 \theta\right) \sin \theta \ d\theta \ d\phi.$$

Kenny Wong
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  • dA has a cross product in it..... Did you intend that? Or is that ordinary multiplication? – user351797 Dec 29 '18 at 00:54
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    @Clclstdnt I meant ordinary multiplication. Sorry, will change the notation. – Kenny Wong Dec 29 '18 at 01:09
  • I'm confused how you got dA. This looks like dV for a triple integral in spherical coordinates. But this isn't a triple integral....... How did you get dA? – user351797 Dec 29 '18 at 01:13
  • $\sin \theta \ d\theta d\phi$ is the area element on the surface of a unit sphere. [… while $r^2 \sin \theta dr \theta d\phi$ is the volume element in $\mathbb R^3$] – Kenny Wong Dec 29 '18 at 01:14
  • Yes but how did you arrive at this? Should this just be common knowledge? I haven't come across the area element yet. – user351797 Dec 29 '18 at 01:16
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    @Clclstdnt It's a good thing to learn off by heart. You can also derive it using $d\vec A = \frac{d\vec r}{d\theta} \times \frac{d\vec r}{d\phi}d\theta d\phi$. (And this time, I really do mean the cross product.) – Kenny Wong Dec 29 '18 at 01:18
  • @Clclstdnt See here too: https://math.stackexchange.com/questions/131735/surface-element-in-spherical-coordinates – Kenny Wong Dec 29 '18 at 01:20
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You can use the divergence theorem to simplify things. We have $$\operatorname{div} \vec{F}(x,y,z) = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} = 2+2z$$

so $$I = \iint\limits_{\text{unit sphere}} \vec{F}\cdot d\mathbf{S} = \iiint\limits_{\text{unit ball}} \operatorname{div} \vec{F} \,dV = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi \int_{r=0}^1 2(1+r\cos\theta)\underbrace{r^2\sin\theta\,dr\,d\theta\,d\phi}_{dV}$$

The integrand is independent of $\phi$ so it factors out as $2\pi$, and then the second integral vanishes:

$$I = 4\pi\left(\int_{\theta = 0}^\pi \int_{r=0}^1 r^2\sin\theta\,dr\,d\theta + \int_{\theta = 0}^\pi \int_{r=0}^1 r^3\cos\theta\sin\theta\,dr\,d\theta\right) = \frac83\pi$$

mechanodroid
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