Flux Through Spheres

Now suppose we want to calculate the flux of $\vec{F}$ through S where S is a piece of a sphere of radius R centered at the origin. The surface area element (from the illustration) is

$$dS = R^2 \sin \phi d\theta d\phi.$$

The outward normal vector should be a unit vector pointing directly away from the origin, so (using $\vec{r} = x\hat{i} + y\hat{j} +z\hat{k}$ and spherical coordinates) we find

$$\hat{n} = \frac{\vec{r}}{\vert\vert\vec{r}\vert\vert} = \sin... ...ta \hat{i} + \sin \phi \sin \theta \hat{j} + \cos \phi \hat{k},$$

and we are left with

$$\mbox{Flux} \thinspace = \int_S \vec{F} \cdot \hat{h} dS$$

$$= \int_T \vec{F}(R, \phi, \theta) \cdot (\sin \phi \cos \the... ...theta \hat{j} + \cos \phi \hat{k}) R^2 \sin \phi d\phi d\theta,$$

where T is the $\phi \theta$-region corresponding to S.

As an example, let's compute the flux of $\vec{F}(x,y,z) = z\hat{k}$through S, the upper hemisphere of radius 2 centered at the origin, oriented outward.

1.

Flux is positive, since the vector field points in the same direction as the surface is oriented.

2.

The area element is $dS = R^2 \sin \phi d\phi d\theta = 4\sin \phi d\phi d\theta$.

3.

The unit outward normal is $\hat{n} = \vec{r}/r = \sin \phi \cos \theta \hat{i} + \sin \phi \sin \theta \hat{j} + \cos \phi \hat{k}$.

4.

The region T is given by $0 \le \theta \le 2\pi, 0 \le \phi \le \pi/2$.

5.

The vector field (in spherical coordinates) is $\vec{F} (R,\phi, \theta) = R\cos \phi \hat{k} = 2\cos \phi \hat{k}$.

6.

Thus, the flux is

$$\mbox{Flux} \thinspace = \int_S \vec{F} \cdot \hat{n}dS$$

$$= \int_0^{2\pi} \int_0^{\pi/2} (2\cos \phi \hat{k}) \cdot (\... ...sin \theta \hat{j} + \cos \phi \hat{k})4\sin \phi d\phi d\theta$$

$$= \int_0^{2\pi} \int_0^{\pi/2} 8\cos^2 \phi \sin \phi d\phi d\theta$$

$$= 8 \int_0^{2\pi} -\frac{1}{3} [\cos^3 \theta]_0^{\pi/2} d\theta$$

$$= -\frac{8}{3} \int_0^{2\pi} (-1)d\theta = \frac{16\pi}{3}.$$