# Flux of radial vector field through sphere

Let $$x^2+y^2+z^2=R^2$$ be the equation of the sphere we want to calculate the flux through, and $$\vec{r}=x\hat i+y\hat j+z\hat k$$ be the position vector field. We can compute it via divergence theorem: $$\Phi=\iiint_V(\nabla\cdot \vec r)dV=3\iiint_V dV=3\dfrac{4}{3}\pi R^3=4\pi R^3.$$

Or like this too: $$\Phi=\oint_S \vec r\cdot d\vec S=\oint_S R(\hat r\cdot\hat r)dS=R\oint_S dS=4\pi R^3.$$

But when I try to calculate it using the fact that $$dS=||\partial_x \vec r\times \partial_y \vec r||dxdy$$ ($$\vec r(x,y)$$ being the parametrized surface), I fail:

Let's parametrize the surface like $$\vec r(x,y)= \left(x,y,\pm\sqrt{R^2-x^2-y^2}\right).$$ Now we can calculate the cross product, $$\partial_x \vec r\times \partial_y \vec r=\left|\begin{array} \hat{i} & \hat j& \hat k\\ 1 & 0 & -\frac{x}{z}\\ 0 & 1 & -\frac{y}{z} \end{array}\right|=\dfrac{x}{z}\hat i+\dfrac{y}{z}\hat j+\hat k,$$ and thus we'll get the norm: $$\Longrightarrow ||\partial_x \vec r\times \partial_y \vec r||=\sqrt{1+\dfrac{x^2}{z^2}+\dfrac{y^2}{z^2}}=\sqrt{\dfrac{x^2+y^2+z^2}{z^2}}=\dfrac{R}{|z|}.$$

We can now apply these equations to the surface integral: $$\Longrightarrow \Phi=\oint_S R(\hat r\cdot \hat r)dS=R\oint_S \dfrac{R}{|z|}dxdy.$$

And changing into spherical coordinates we'd get that $$R^2\int_0^{2\pi}\int_0^\pi\dfrac{R^2\sin\theta}{R|\cos\theta|} d\theta d\Phi=2\pi R^3 I,$$ and $$I$$ does not converge...

## 2 Answers

First, why not parametrize in spherical coordinates? :)

Switching an integral in $$(x,y)$$ over the disk to spherical coordinates is, of course, going to involve $$\rho$$ and $$\phi$$. But it’s silly. Switching to polar coordinates is what you want. So $$\iint_D \frac{R^2}{|z|}\,dx\,dy = R^2 \int_0^{2\pi}\int_0^R \frac r{\sqrt{R^2-r^2}}dr\,d\theta = 2\pi R^2\Big[{-}\sqrt{R^2-r^2}\Big]_0^R = 2\pi R^3.$$ And you get another for the integral over the lower hemisphere.

• The cross product and norm were easier to compute using cartesian parametrization and then I switched to spherical coords to solve the integral easily. About changing to polar, you're right, it makes more sense since we're integrating over the projection of the sphere in the X-Y plane, so the zenith angle does not intervene thus reducing to polar coordinates. Thank you! Jun 4 at 0:06

Just a general fact, if you want to calculate the flux $$\Phi$$ of a vector field $$X$$ through a (piece of a ) surface parametrized by $$D \ni (u,v) \mapsto r(u,v)$$, a convenient formula is

$$\Phi=\int_{D} \det (X, \frac{\partial r}{\partial u}, \frac{\partial r}{\partial v}) \, d u\, d v$$

(or minus that, depending on the orientation of the surface). It bypasses the calculation of $$n(u,v)$$, which involves a cross product ( a determinant) then a normalization ( dividing by a factor that reappears in the numerator anyways). Check with any parametrization you would like.

• Thanks, I'll take that into account. Jun 4 at 1:07