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?

  • $\begingroup$ @Arthur, I've changed my question to be clearer. $\endgroup$ – Magela Nov 20 '20 at 13:21
  • $\begingroup$ $\boldsymbol F$ and $d\boldsymbol S$ are collinear, clearly the dot product cannot be zero. We have $\boldsymbol F = 11 \boldsymbol r/r^3$ and $\hat {\boldsymbol n} = \boldsymbol r/r$, so $$\iint_{r = R} \boldsymbol F \cdot d\boldsymbol S = \frac {11} {R^2} \iint_{r = R} dS.$$ $\endgroup$ – Maxim Nov 20 '20 at 16:25

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 $\pi$}$

  • $\begingroup$ Thank you for your answer. Could you, please, explain why $\hat{n} = \frac{1}{r}(x, y, z)$? $\endgroup$ – Magela Nov 20 '20 at 14:27
  • $\begingroup$ An outward normal vector to a spherical surface with center at the origin will be $(x-0)i + (y-0)j + (z-0)k$ where $(x, y, z)$ is point on the surface of the sphere. To find unit normal vector, you need to divide it by its magnitude which is $\sqrt{x^2+y^2+z^2} = r$ $\endgroup$ – Math Lover Nov 20 '20 at 14:33
  • $\begingroup$ I got it, thank you. Regarding my approach, why am I getting the flux equals to 0? Isn't it right what I'm doing? $\endgroup$ – Magela Nov 20 '20 at 14:36
  • $\begingroup$ @Magela you should not get dot product as zero. What is the normal vector you are getting? That must be wrong. $\endgroup$ – Math Lover Nov 20 '20 at 15:08
  • 1
    $\begingroup$ Surface integral is directly calculating flux over the surface of the sphere so it works. Divergence theorem says this surface integral (flux) is same as volume integral of the divergence inside the surface. Now you can see why continuity issue at origin matters when we want to use divergence theorem method to find flux. $\endgroup$ – Math Lover Nov 20 '20 at 17:02

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.

  • $\begingroup$ Thank you for answering. I'm afraid that I did not understand how you can get $F(r) = \frac{11}{r^2}$. Did you see that the vector field has the coordinates $ F = \left\{ \frac{11x}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{11y}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{11z}{\left(x^2+y^2+z^2\right)^{3/2}} \right\}$ $\endgroup$ – Magela Nov 20 '20 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.