91Ó°ÊÓ

It has been asked to use problem 4.and r = ix + jy + kz.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

Apply Gauss' theorem, and the fact that $∇·\mathrm{r}=3$

$∫{∫}_{\mathrm{indind}\mathrm{plan}}\mathrm{r}·\mathrm{²Ô»åσ}={âˆ\circledR }_v∇·\mathrm{rdt}-∫{∫}_{\mathrm{yz}}\mathrm{r}·\mathrm{²Ô»åσ}-∫{∫}_{\mathrm{xz}}\mathrm{r}·\mathrm{²Ô»åσ}-∫{∫}_{\mathrm{xy}}\mathrm{r}·\mathrm{²Ô»åσ}$

The first term is simply reduced to be ${\mathrm{l}}_1=3\mathrm{V}$where V the volume of the entire shape is, calculate it,

$$\begin{gathered} \mathrm{V}=\frac{1}{2}{∫}_0^3\left(6-2\mathrm{z}\right)\left(4-\frac{4}{3}\mathrm{z}\right)\mathrm{dz}=12 \\ {\mathrm{l}}_1=36 \end{gathered}$$

The second term is :${\mathrm{l}}_2=-∫∫\mathrm{³\ae »åσ}\mathrm{but}\mathrm{x}=0,\mathrm{then}{\mathrm{l}}_2=0$

The third term is:${\mathrm{l}}_3=-∫∫\mathrm{³\ae »åσ}\mathrm{but}y=0,\mathrm{then}{\mathrm{l}}_3=0$

The fourth term is:${\mathrm{l}}_4=-∫∫\mathrm{³\ae »åσ}\mathrm{but}z=0,\mathrm{then}{\mathrm{l}}_4=0$

Hence, the solution is derived to be$∫{∫}_{\mathrm{indind}\mathrm{plan}}\mathrm{r}·\mathrm{²Ô»åσ}=36$