Let $S$ be the unit sphere $x^2+y^2+z^2=1$ with the outward pointing normal vector n. Calculate the flux for the vector field $\mathbf{f}(\mathbf{r})=4\mathbf{r}$ through $S$.
What I have done so far:
I have rearranged the equation to get $z=\sqrt{1-x^2-y^2}$ and thus $N(x,y)=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$
Now to calculate the flux the equation is $\iint\limits_{S}\mathbf{f}\cdot\mathbf{n}dS$.
Hence I now have $$4\iint\limits_{S}(x,y, \sqrt{1-x^2-y^2} )\cdot (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)dS$$
$$=4\iint\limits_{S}\frac{1}{\sqrt{-x^2-y^2+1}}dxdy$$
However I don't know how to get the limits for the integral.
Best Answer
In spherical coordinates,
$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r \cos \phi$
Surface area element $dS = r^2 \sin \phi \ d \theta d \phi = \sin \phi \ d \theta d \phi \, $ (as $r = 1$)
Please note the outward normal vector should be a unit vector pointing directly away from the origin for this surface. So using, $\vec{r} = x \hat{i} + y\hat{j} + z\hat{k} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$
$\hat{n} = \frac{\vec{r}}{||r||} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$
$\vec{F} = 4r = 4(\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k})$
$Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS$
$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \cdot (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \sin \phi \, d\theta d\phi$
$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} \sin \phi d\theta d\phi = 16 \pi$