How should I evaluate the flux?

Question

I have the following vector field

$F= \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{i} + \frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{j} + \frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{k}$

I need to calculate the flux, $\iint_{S} F \cdot \vec{n}\ \text{dS}$, through the surface $S: x^2+y^2+z^2 = 16$ (sphere). As $F$ has a singularity at origin, I can't use the Divergent Theorem (right?). So I try to calculate using the surface integral of a vector field, $\iint_{S} F \cdot \vec{n}\ \text{dS} = \iint_{D} F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})\ \text{dA}$.

I've used the parametric representation $\mathbf{r}(\phi,\theta) = (4 \sin\phi\cos\theta, 4 \sin\phi\sin\theta, 4 \cos\phi)$. When I perform $F(\mathbf{r}(\mathbf{u},\mathbf{v})) \cdot (\mathbf{r_u} \times \mathbf{r_v})$, I'm getting 0. With this result, my flux would be 0, but according to my teacher answer, the flux is $44\pi$. What I'm doing wrong?

Answer 1

In spherical coordinates,

$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r\cos \phi$

$F = \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{i} + \frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{j} + \frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \vec{k}$

$F = (\frac{11\cos \theta \sin \phi}{r^2}, \frac{11\sin \theta \sin \phi}{r^2}, \frac{11 \cos \phi}{r^2})$

Outward normal vector $\hat{n} = \frac{1}{r}(x, y, z) = (\cos \theta \sin \phi \, , \sin \theta \sin \phi \, , \cos \phi \,)$

In spherical coordinates, $dS = r^2 \sin \phi \ d \theta \, d \phi$

$Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS$

$\vec{F} \cdot \hat{n} dS = (\frac{11\cos \theta \sin \phi}{r^2}, \frac{11\sin \theta \sin \phi}{r^2}, \frac{11 \cos \phi}{r^2}) \cdot (\cos \theta \sin \phi \, , \sin \theta \sin \phi \, , \cos \phi \,) dS$

$ = 11(\cos^2 \theta \sin^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \phi) \sin \phi d\theta d\phi = 11 \sin \phi d\theta d\phi$

So, $Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS = 11 \int_0^\pi \int_0^{2\pi}\sin \phi d\theta d\phi = \fbox {44 $\pi$}$

Answer 2

There is no real need to use divergence theorem here. The polar coordinate representation should do.

\begin{gather*} Transferring\ to\ polar\ coordinates,\ we\ see\ that\ \\ F( r) =\frac{11}{r^{2}} \ \\ Since\ the\ field\ is\ spherically\ symmetric,\ \\ we\ can\ directly\ calculate\ the\ flux\ as\ \\ \oint F( r) dr=\int 4\pi r^{2} F( r) dr=\int 44\pi dr=176\pi \\ \\ \end{gather*} I am not sure whether your final answer is correct, because that would mean that the radius of the sphere should be 1 in that case.