Example 1: Find the flux of $\vec{v}=2\vec{i}+4\vec{j}-3\vec{k}$ through the square plate of area 16 in the $xy$-plane oriented in the positive $z$-direction.

Solution: Since we know that the surface is oriented in the positive $z$-direction, a unit normal vector for the surface is then just $\vec{n}=\vec{k}$. Thus, we have

\[\int_S\vec{v}\cdot d\vec{A}=\int_S(2\vec{i}+4\vec{j}-3\vec{k})\cdot\vec{k}dA=\int_S(-3)dA.\]

Since we may pull the $-3$ out in front of the integral, and since we know the area of the surface to be 16, we have


Example 2: Find the flux of the electric field $\vec{E}(\vec{r})=q\frac{\vec{r}}{||\vec{r}||^3}$ ($\vec{r}\neq 0$) for $q$ is a positive constant experienced by the sphere of radius 2 centered at the origin.

Solution: We have

\[\int_S\vec{E}\cdot d\vec{A}=\int_Sq\frac{\vec{r}}{||\vec{r}||^3}\cdot d\vec{A}.\]

We know that (on the surface of the sphere of radius 2 centered at the origin) $||\vec{r}||=2$, and so this integral becomes

\[\frac{q}{8}\int_S\vec{r}\cdot d\vec{A}.\]

Now, considering $d\vec{A}$, we notice that the normal vectors of the surface of the sphere can be represented using


Notice then that both of the vectors inside the integral are pointing in the same direction, and are hence parallel. Thus, we simply have

\[\frac{q}{8}\int_S\vec{r}\cdot d\vec{A}=\frac{q}{8}\int_S ||\vec{r}||dA=\frac{q}{8}\int_S 2dA=\frac{q}{4}\int_S dA.\]

Now, if you remember the equation for the surface area of a sphere of radius $r$ ($4\pi r^2$), then you’ll immediately recognize this as

\[\frac{q}{4}\int_S dA=4q\pi.\]