# Flux integral with vector field in spherical coordinates

I have a vector field $\vec{A}$ that is given in spherical coordinates. $$\vec{A}=\frac{1}{r^2}\hat{e}_{r}$$ I need to calculate the flux integral over a unit sphere in origo (radius 1). I cannot use Gauss theorem since there exist a singularity in the volume. I have instead attempted to calculate it in the standard way with the integral below. $$\int_{\phi}^{ } \int_{\theta}^{ }\vec{A}(\vec{r}(\theta, \phi))\cdot(\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}})\space d\theta d\phi$$

So in this calculation i need a parameterization of the surface. I came up with this but have been informed that it is not a correct parametrization. $$\vec{r}(\theta, \phi)=1\hat{e}_{r}+\theta\hat{e}_{\theta}+\phi\hat{e}_{\phi}$$ $$\theta:0\rightarrow\pi$$ $$\phi:0\rightarrow2\pi$$

Why is that? And how should you describe a sphere in spherical coordinates? Or should I instead transform the vector field into cartesian coordinates? How is that done?

This is the rest of the calculation and my final answer. $$\frac{\partial{\vec{r}}}{\partial{\theta}}=1\hat{e}_{\theta}$$ $$\frac{\partial{\vec{r}}}{\partial{\phi}}=1\hat{e}_{\phi}$$ $$\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}}=1\hat{e}_{r}$$ $$\vec{A}(\vec{r}(\theta, \phi))=\frac{1} {(1)^2}\hat{e}_{r}=1\hat{e}_{r}$$ $$\vec{A}(\vec{r}(\theta, \phi))\cdot(\frac{\partial{\vec{r}}}{\partial{\theta}}\times \frac{\partial{\vec{r}}}{\partial{\phi}})=1\hat{e}_{r}\cdot1\hat{e}_{r}=1$$ $$\int_{0}^{2\pi} \int_{0}^{\pi}1\space d\theta d\phi=2\pi^2$$

The parameterization is not correct. The position vector has neither a $\theta$ component nor a $\phi$ component. Note that both of those compoents are normal to the position vector. Therefore, the sperical coordinate vector parameterization of a surface would be in general

$\vec r=\hat r(\theta,\phi)r(\theta,\phi)$.

For a spherical surface of unit radius, $r(\theta,\phi)=1$ and

$$\vec r=\hat r(\theta,\phi)$$

where the unit vector $\hat r(\theta,\phi)$ can be expressed on Cartesian coordinates as

$$\hat r(\theta,\phi)=\hat x\sin \theta \cos \phi+\hat y\sin \theta \sin \phi+\hat z\cos \theta$$

Now, we can show that the unit normal to the sphere is $\hat r(\theta,\phi)$ since $\frac{\partial \hat r}{\partial \theta}\times \frac{\partial \hat r}{\partial \phi}=\hat \theta \times \hat \phi=\hat r$.

Can you finish now?

• Im still not sure how I should apply it to solve the whole problem, but I will give it some more thought. Not sure how $\vec{A}(\vec{r}(\theta, \phi))$ should be handled. Do you think it might be easier to transform the vector field into cartesian coordinates? – user300219 Dec 21 '15 at 21:51

I found a very similar problem and decided to solve my problem with the same method. I belive that this yields the correct answer.

The vector field: $\vec{A}=\frac{1}{r^2}\hat{e}_{r}$

The surface: $S = Unit \space sphere \space centered \space in \space origo$

The flux through the surface $S$ is given by: $\int_{S}\vec{A}\space\cdot\space d\vec{S}$

$$d\vec{S}=r^2sin\theta d\theta d\phi\hat{e}_{r}$$

$$\int_{S}\vec{A}\space\cdot\space d\vec{S}=\int_{s}(\frac{1}{r^2}\hat{e}_{r})\cdot(r^2sin\theta d\theta d\phi\hat{e}_{r})=\int_{S}sin\theta \space d\theta d\phi=\int_{0}^{2\pi}\int_{0}^{\pi}sin\theta \space d\theta d\phi=4\pi$$