Flux Through Cylinders

Suppose we want to compute the flux through a cylinder of radius R, whose axis is aligned with the z-axis. An element of surface area for the cylinder is

$$dS = R d\theta dz,$$

as seen from the picture below.

What is the outward normal vector for this surface? It should point in the straight out from the z-axis, so it should be a unit vector in the direction of $x\hat{i} + y\hat{j}$. Thus,

$$\hat{n} = \frac{x\hat{i} + y\hat{j}}{\vert\vert x\hat{i} + y... ...\theta \hat{j}}{r} = \cos \theta \hat{i} + \sin \theta \hat{j}.$$

This gives us the result that the flux through a portion of the surface of a cylinder of radius R oriented along the z-axis is

$$\mbox{flux}\thinspace = \int_S \vec{F} \cdot \hat{n}dS = \in... ...z) \cdot (\cos \theta \hat{i} + \sin \theta \hat{j})Rd\theta dz$$

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

Here's a quick example: Compute the flux of the vector field $\vec{F}(x,y,z) = z\hat{i}$ through the piece of the cylinder of radius 3, centered on the z-axis, with $x \ge 0, y \ge 0,$ and $0 \le z \le 3$.The cylinder is oriented along the z-axis and has an inward pointing normal vector.

1.

Clearly, the flux is negative since the vector field points away from the z-axis and the surface is oriented towards the z-axis.

2.

The area element is $dS = R d\theta dz = 3 d\theta dz$.

3.

The unit outward normal is $\hat{n} = -\cos \theta \hat{i} -\sin \theta \hat{j}$.

4.

The vector field, in cylindrical coordinates, is $\vec{F}(R,\theta,z) = z\hat{i}$.

5.

The region T is defined by $0 \le \theta \le \frac{\pi}{2}, 0 \le z \le 3$.

6.

The flux is

$$\mbox{Flux} \thinspace = \int_S \vec{F} \cdot \hat{n} dS = -... ...cdot (\cos \theta \hat{i} + \sin \theta \hat{j}) 3 d\theta dz$$

$$= -\int^3_0 \int^{\pi/2}_0 3z \cos \theta d\theta dz = -3 \int_0^3 z [\sin \theta ]_0^{\pi/2} dz$$

$$= -3\int_0^3 z dz = -\left.\frac{3}{2}z^2\right\vert _0^3 = -\frac{27}{2}.$$