Surface integral of position vector over a sphere

Question

$\iint_S$ 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

Answer 1

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

Answer 2

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