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Evaluate the surface integral of the field $A(x,y,z)=(xy, yz, x^2)$ over the sphere $S$ givn by $x^2 + y^2 + z^2$ with the normal vector pointing to the exterior of the sphere

I've tried doing this using "spherical coordinates" to parametrize the sphere, but it doesnt have a good form. Any suggetion?

Thanks!

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  • $\begingroup$ $x^2+y^2+z^2$ equals what? You didn't finish the equation. $\endgroup$
    – David H
    Jun 5 '14 at 19:42
  • $\begingroup$ Do you know the Divergence Theorem or are you expected to do this explicitly? If you recognize that the normal $n$ to the sphere $x^2+y^2+z^2=R^2$ is a scalar multiple of $(x,y,z)$, then you get $A \cdot n = x^2y+y^2z+x^2z$ and you can, without bludgeoning it in spherical coordinates, employ symmetry (oddness) to see that each term integrates to $0$. $\endgroup$ Jun 5 '14 at 20:16
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I'll assume the sphere is given by the equation $x^2+y^2+z^2 = R^2$ but the answer does not depend at all on $R$ - the answer is zero.

By the divergence theorem, the surface integral is the volume integral of the divergence: $$ \int_S A(x,y,z) ds = \int_V \nabla \cdot A dv = \int_V (y + z ) dv = \int_V y dv + \int_V z dv $$ where V is the volume of the sphere. Since ofr the function $z$ each point in the lower hemisphere cancels the corresponding point in the upper hemisphere, $\int_V z dv = 0$. Similarly, $\int_V y dv = 0$.

So the volume integral of $\nabla \cdot A$ is zero and by the divergence theorem, the surface integral of $A$ is zero.

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