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The problem is actually to verify the divergence theorem by computing both $\iiint_E \text{div } \mathbf{F\space} dV$, which was relatively easy to compute and gives $\frac{256\pi}{3}$. To find $\iint_S \mathbf{F}\cdot d\mathbf{S}$, I parametrized the surface with spherical coordinates: $\mathbf{r}=\langle 4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi\rangle$ with $0\leq\phi\leq\pi,0\leq\theta\leq2\pi$. Now, noting that $\iint_S\mathbf{F}\cdot d\mathbf{S}=\iint_S \mathbf{F}\cdot \mathbf{n}\space dS$ where $\mathbf{n}$ is the normal vector to the $S$, and since $S$ is a sphere, we have $\mathbf{n}=\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle$, and $$\mathbf{F\cdot n}=4\sin\phi\cos\phi\cos\theta+4\sin^2\phi\sin\theta\cos\theta+4\sin\phi\cos\phi\cos\theta$$

But the integral of this is 0.

Where am I going wrong?

user161997
  • 99

1 Answers1

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On the surface of the sphere $\langle x,y,z \rangle = \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$.

Hence, $F = \langle z,y,x \rangle = \langle 4\cos\phi, 4\sin\phi\sin\theta, 4\sin\phi\cos\theta \rangle$.

Also, $n = \frac{1}{4}\langle x,y,z \rangle = \langle \sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi \rangle$.

Therefore, $F \cdot n = 4\cos\phi\sin\phi\cos\theta + 4\sin^2\phi\sin^2\theta + 4\cos\phi\sin\phi\cos\theta$.

This shouldn't integrate to $0$.

JimmyK4542
  • 54,331