Question

# \text { 4. (12 points) Let } \vec{F}=\frac{1}{r^{2}} \hat{r} \text { and let } V \text { be open, bounded and contain }(0,0,0) \text { . } \text { (a) (5 points) By showing that } \operatorname{div}(\vec{F})=0 \forall \vec{x} \in \mathbb{R}^{3} \backslash\{(0,0,0)\} \text { , show that } \iint_{\partial V} \vec{F} \cdot d \vec{S}=4 \pi (b) (5 points) Letting S be the boundary of solid tetrahedron V with vertices at (0, 0,0),(2,0,0), (0, 1,0) and (0,0, 4), compute the outward flux of the vector field \vec{G}=-\vec{\nabla}\left(\frac{q_{1}}{4 \pi\left|\vec{r}-\vec{r}_{1}\right|}\right)-\vec{\nabla}\left(\frac{q_{2}}{4 \pi\left|\vec{r}-\vec{r}_{2}\right|}\right)-\vec{\nabla}\left(e^{x} \sin (y)+z\right) \text { where } q_{1}, q_{2} \text { are constants, } \vec{r}_{1}=(1 / 4,1 / 4,1 / 16) \text { and } \vec{r}_{2}=(2,2,2) \text { . } (c) (2 points) Does your calculation in part (b) violate the divergence theorem? Explain.  Fig: 1  Fig: 2  Fig: 3  Fig: 4  Fig: 5  Fig: 6  Fig: 7