True or False The divergence of a vector field $\mathbf{ F}$ is a vector.

If $\mathbf{F} (x,y,z)=P(x,y,z) \mathbf{i}+ Q(x,y,z)\mathbf{j}+R(x,y,z)\mathbf{k}$ is a vector field, where $\dfrac{\partial P}{\partial x}$, $\dfrac{\partial Q}{ \partial y}$, and $\dfrac{\partial R}{\partial z}$ each exist, then $\rm{ div}\mathbf{F}=$__________.

$\mathbf{F}(x,y,z)=x^{2}\mathbf{i}+y^{2}\mathbf{j}+z^{2}\mathbf{k}$

$\mathbf{F}(x,y,z)=x\mathbf{i}+xy\mathbf{j}+xyz\mathbf{k}$

$\mathbf{F}(x,y,z)=(x+\cos x)\mathbf{i}+(y+y\sin x)\mathbf{j}+2z\mathbf{k}$

$\mathbf{F}(x,y,z)=xye^{z}\mathbf{i}+x^{2}e^{z}\mathbf{j}+x^{2}ye^{z}\mathbf{k}$

$\mathbf{F}(x,y,z)= \sqrt{x^2+y^2} \mathbf{i}+ \sqrt{x^2+y^2} \mathbf{j} +z\mathbf{k}$

$\mathbf{F}(x,y,z) = xy^2 \mathbf{i} + x^2y \mathbf{j} + xyz{\bf k}$

$\mathbf{F}=(2xy+2z)\mathbf{i}+(y^{2}+1)\mathbf{j} -(x+y)\mathbf{k}$; $S$ is the surface of the solid enclosed by $x+y+z=4$, $x=0$, $y=0$, and $z=0$.

$\mathbf{F}=(2xy+z)\mathbf{i}+y^{2}\mathbf{j}-(x+4y)\mathbf{k}$ ; $S$ is the surface of the solid enclosed by $2x+2y+z=6$, $x=0$, $y=0$, and $z=0$.

$\mathbf{F}=x^{2}\mathbf{i}+y^{2}\mathbf{j}+z^{2}\mathbf{k}$; $S$ is the surface of the solid enclosed by $x=0$, $x=1$, $y=0$, $y=1$, $z=0$ , and $z=1$.

$\mathbf{F}=(x-y)\mathbf{i}+(y-z)\mathbf{j}+(x-y)\mathbf{k}$; $S$ is the surface of the solid cube with its center at the origin and faces in the planes $x=\pm 1$, $y=\pm 1$, and $z=\pm 1$.

$\mathbf{F}=x^{2}\mathbf{i}+2y\mathbf{j}+4z^{2} \mathbf{k}$; $S$ is the surface of the solid cylinder $x^{2}+y^{2}\leq 4$, $0\leq z\leq 2$.

$\mathbf{F}=x\mathbf{i}+2y^{2}\mathbf{j}+3z^{2}\mathbf{k}$; $S$ is the surface of the solid cylinder $x^{2}+y^{2}\leq 9$, $0\leq z\leq 1$.

$\mathbf{F}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$; $S$ is the sphere $x^{2}+y^{2}+z^{2}=1$.

$\mathbf{F}=2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}$; $S$ is the sphere $x^{2}+y^{2}+z^{2}=2$.

$\mathbf{F}=(x+\cos x)\mathbf{i}+(y+y\sin x)\mathbf{j}+2z \mathbf{k}$; $S$ is the surface of the solid enclosed by the tetrahedron with vertices $(0,0,0),(1,0,0),(0,1,0)$, and $(0,0,1)$.

$\mathbf{F}=yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}$; $S$ is the surface of the solid enclosed by the tetrahedron with vertices $(0,0,0),(1,0,0),(0,1,0)$, and $(0,0,1)$.

Volume Let $\mathbf{F}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$; let $S$ be the surface of a solid $E$ satisfying the Divergence Theorem; and let $\mathbf{n}$ be the outer unit normal vector to $S$. Show that the volume $V$ of $E$ is given by the formula $V=\dfrac{1}{3}\iint\limits_{S} \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS$.

A rectangular parallelepiped with sides of length $a,b$, and $c$

A right circular cone with height $h$ and base radius $R$

(Hint: The calculation is simplified with the cone oriented as shown in the figure.)

A sphere of radius $R$

$\mathbf{F}(x,y,z)=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$; $S$ is the surface of the sphere $x^{2}+y^{2}+z^{2}=100.$

$\mathbf{F}(x,y,z)=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$; $S$ is the closed cylindrical surface $x^{2}+y^{2}=1$ between $z=0$ and $z=2.$

Show that if $\mathbf{F}$ is a constant vector field and $S$ is the surface of a solid $E$ satisfying the assumptions of the Divergence Theorem, then $\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=0$.

Let $\mathbf{F}=3x\mathbf{i}+4y\mathbf{j}+(7z+2x)\mathbf{k}$ and $\mathbf{G}=2x\mathbf{i}+3y\mathbf{j}+(9z+6y)\mathbf{k}$. Let $S$ be the surface of a solid $E$ satisfying the assumptions of the Divergence Theorem. Show that $\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n} \,dS=\iint\limits_S \mathbf{G}\,{\bf\cdot}\, \mathbf{n}\,dS$.

Flux Suppose the velocity of a fluid flow in space is constant.

Let $\mathbf{F}(x,y,z)=\dfrac{1}{\rho }(x\mathbf{i}+y\mathbf{j}+ z\mathbf{k})$, where $\rho =\sqrt{x^{2}+y^{2}+z^{2}}$. Show that ${\rm{div}} \mathbf{F}=\dfrac{2}{\rho }$.

Assume that the hypotheses of the Divergence Theorem hold for $S$ and $E.$ Show that if $f$ satisfies the Laplace equation $(f_{xx}+f_{yy}+f_{zz}=0)$ in a closed, bounded solid $E$ with boundary $S$ with outer unit normal $\mathbf{n}$, then $\iint\limits_S {\bf \nabla} f\,{\bf\cdot}\, \mathbf{n}\,dS=0$.

Let $\mathbf{F}=x^{2}\mathbf{i}+y^{2}\mathbf{j}+z^{2}\mathbf{k}$ and let $\mathbf{n}$ be the outer unit normal to the surface $S$. Use the Divergence Theorem to show that $$\iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=\dfrac{8\pi q^{4}}{3}$$

if $S$ is the surface $x^{2}+y^{2}+z^{2}=2qz$, $q>0$.

What is the value of the integral in Problem 31 if $S$ is the surface of the cube $x=0$, $x=q$, $y=0$, $y=q$, $z=0$, and $z=q$, $q>0$?

Let $f$ and $g$ be two scalar functions and let $S$ be the surface of a solid $E$ satisfying the assumptions of the Divergence Theorem. Show that $$\iint\limits_{S}f( {\bf\nabla}g) \,{\bf\cdot}\, \mathbf{n\,} dS=\iiint\limits_E[ f( {\bf\nabla}^{2}g) + {\bf\nabla}\! f\,{\bf\cdot}\, {\bf\nabla}g] \,dV$$

[Hint: Let $\mathbf{F}=f( {\bf\nabla }g)$ in the Divergence Theorem and use ${\bf\nabla}^{2}g={\bf\nabla} \,{\bf\cdot}\, {\bf\nabla} g=g_{xx}+g_{yy}+g_{zz}$.]