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Find the value of $$ \iint_{\Sigma} <x, y^3, -z>. d\vec{S} $$ where $ \Sigma $ is the sphere $ x^2 + y^2 + z^2 = 1 $ oriented outward by using the divergence theorem.

So I calculate $ div\vec{F} = 3y^2 $ and then I convert $ x, y, z $ into $ x = p\sin \phi \cos \theta, y = p\sin \phi \sin \theta, z = p\cos \phi $ but then I got stuck from that point.

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  • $\begingroup$ Did you try and write down the volume integral: $0 \leq p \leq 1, 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi$ $\endgroup$ – Simon S Dec 17 '15 at 8:10
  • $\begingroup$ The divergence is $1+3y^2-1$. $\endgroup$ – Emilio Novati Dec 17 '15 at 8:17
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You therefore want the volume integral

$$\iiint_V 3y^2 \ dV = \int_0^1 \int_0^{2\pi} \int_0^\pi 3(p\sin\phi \sin\theta)^2 . p^2 \sin\phi \ dp \ d\theta \ d\phi = \ \cdots$$

Since the volume of integration is a rectangular box in $(p,\theta, \phi)$ coordinates, you can separate out the integrating variables to make it easy to evaluate the overall integral.

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  • $\begingroup$ Got it thanks. I was stuck on evaluate $ \sin^3 \phi \; d\phi $ so I thought there is something wrong in my work, but I figure it out now. $\endgroup$ – user298251 Dec 17 '15 at 8:20
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You need to evaluate: $$\int_0^1 3 p^4 \ dp\int_0^{2\pi} \sin^2\theta \ d\theta \int_0^\pi \sin^3\phi \ d\phi $$

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$$\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^1 \left(\rho^2 \sin \phi \right)\underbrace{(\sqrt{3}\rho \sin \phi \sin \theta)^2}_{3y^2 \ \textrm{in spherical}} \ d\rho \ d\phi \ d\theta$$

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