## Section 19.1 Examples

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

$\int_S(-3)dA=-3\int_SdA=-3*16=-48.$

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

$d\vec{A}=\vec{n}dA=\frac{\vec{r}}{||\vec{r}||}dA.$

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

$\frac{q}{4}\int_S dA=4q\pi.$