Question

Question: $\mathbf{\int }\mathbf{\int }\mathbf{r}\mathbf{·}\mathbf{nd\sigma }$ over the surface in Problem 4, where r = ix + jy + kz. Hint: See Problem 10.9.

Short answer

The solution is derived to be $\int {\int }_{\mathrm{indind}\mathrm{plan}}\mathrm{r}·\mathrm{nd\sigma }=36$.

Step-by-step solution

Given Information.

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

Definition of vector.

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.

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

$\int {\int }_{\mathrm{indind}\mathrm{plan}}\mathrm{r}·\mathrm{nd\sigma }={\oint }_v\nabla ·\mathrm{rdt}-\int {\int }_{\mathrm{yz}}\mathrm{r}·\mathrm{nd\sigma }-\int {\int }_{\mathrm{xz}}\mathrm{r}·\mathrm{nd\sigma }-\int {\int }_{\mathrm{xy}}\mathrm{r}·\mathrm{nd\sigma }$

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,

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

The second term is :${\mathrm{l}}_2=-\int \int \mathrm{xd\sigma }\mathrm{but}\mathrm{x}=0,\mathrm{then}{\mathrm{l}}_2=0$

The third term is:${\mathrm{l}}_3=-\int \int \mathrm{xd\sigma }\mathrm{but}y=0,\mathrm{then}{\mathrm{l}}_3=0$

The fourth term is:${\mathrm{l}}_4=-\int \int \mathrm{xd\sigma }\mathrm{but}z=0,\mathrm{then}{\mathrm{l}}_4=0$

Hence, the solution is derived to be$\int {\int }_{\mathrm{indind}\mathrm{plan}}\mathrm{r}·\mathrm{nd\sigma }=36$