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$\iint_S$ r.n $dS$

Over the surface of the sphere with radius $a$ centered at the origin

Now this is obviously trivial and the answer is $4\pi a^3$ but I want to do it the hard way because there's something I don't understand

The surface is $x^2 + y^2 + z^2 = a^2$ , then the normal vector $n = \nabla S$

$\hat n$ = $\frac{\nabla S}{|\nabla S|}$ = $\frac{x \hat i + y \hat j + z \hat k}{a}$

$dS = \frac{dxdy}{|\hat n . \hat k|} = \frac{dxdy}{a/z}$

Then $\iint_S$ r.n $dS$ = $\iint_S \frac{x^2 + y^2}{\sqrt{a^2 -x^2 -y^2}} + \sqrt{a^2 -y^2 -x^2}$ $dxdy$

Switching to polar coordinates, $x=\rho cos\phi , y =\rho \sin\phi$

Then $\iint_S$ r.n $dS$ = $\iint_S \frac{\rho^2}{\sqrt{a^2 -\rho^2}} + \sqrt{a^2 - \rho^2}$ $\rho d\rho d\phi$

Integrating $\rho$ from $0$ to $a$ and $\phi$ from $0$ to $2\pi$ , we get:

$\iint_S$ r.n $dS$ = $2\pi a^3$ which is half the required answer $4\pi a^3$ , is it because I only took into account that $dS = \frac{dxdy}{|\hat n . \hat k|} = \frac{dxdy}{a/z}$ and should have changed this surface element starting from a specific point? If so, how? Thanks

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  • $\begingroup$ When you take the square root, aren’t you only giving the correct value of $z$ in upper half space? $\endgroup$ – Charlie Frohman Dec 25 '18 at 2:10
  • $\begingroup$ Yes, but didn't I integrate from 0 to $2\pi$ anyway? $\endgroup$ – khaled014z Dec 25 '18 at 2:25
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    $\begingroup$ This seems to be full of errors. Why are you varying $\rho$. It's constant, you're on a sphere. The surface element should be $a^2\sin\theta\operatorname d\theta\operatorname d\varphi $. You need spherical coordinates, not polar. Etc... $\endgroup$ – Chris Custer Dec 25 '18 at 3:07
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notice that $$\vec r \cdot \vec n = \frac{x^2+y^2+z^2} a = \frac {a^2} a = a$$

which is a constant so can be taken outside the integral

so $$\iint_S \vec r \cdot \vec n \;dS = a \iint_S \;dS $$

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You get $a\int\int\operatorname dS=a\int\int a^2\sin\theta\operatorname d\theta\operatorname d\varphi=a^3\int_0^{2\pi}\int_0^{\pi}\sin\theta\operatorname d\theta\operatorname d\varphi=2\pi a^3[-\cos\theta]_0^{\pi}=4\pi a^3$.

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