6.8 The divergence theorem  (Page 9/12)

Use the divergence theorem to calculate the flux of $\text{F}(x,y,z)=x^3\text{i}+y^3\text{j}+z^3\text{k}$ through sphere $x^2+y^2+z^2=1.$

Find ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k}$ and S is the outwardly oriented surface obtained by removing cube $\left[1,2\right]\times \left[1,2\right]\times \left[1,2\right]$ from cube $\left[0,2\right]\times \left[0,2\right]\times \left[0,2\right].$

Consider radial vector field $\text{F}=\frac{\text{r}}{\left|\text{r}\right|}=\frac{⟨x,y,z⟩}{{\left(x^2+y^2+z^2\right)}^{1\text{/}2}}.$ Compute the surface integral, where S is the surface of a sphere of radius a centered at the origin.

Compute the flux of water through parabolic cylinder $S:y=x^2,$ from $0\le x\le 2,0\le z\le 3,$ if the velocity vector is $\text{F}(x,y,z)=3z^2\text{i}+6\text{j}+6xz\text{k}.$

[T] Use a CAS to find the flux of vector field $\text{F}(x,y,z)=z\text{i}+z\text{j}+\sqrt{x^2+y^2}\text{k}$ across the portion of hyperboloid $x^2+y^2=z^2+1$ between planes $z=0$ and $z=\frac{\sqrt{3}}{3},$ oriented so the unit normal vector points away from the z -axis.

[T] Use a CAS to find the flux of vector field $\text{F}(x,y,z)=\left(e^y+x\right)\text{i}+\left(3\text{cos}(xz)-y\right)\text{j}+z\text{k}$ through surface S , where S is given by $z^2=4x^2+4y^2$ from $0\le z\le 4,$ oriented so the unit normal vector points downward.

[T] Use a CAS to compute ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=x\text{i}+y\text{j}+2z\text{k}$ and S is a part of sphere $x^2+y^2+z^2=2$ with $0\le z\le 1.$

Evaluate ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=bx{y}^2\text{i}+b{x}^{2}y\text{j}+\left(x^2+y^2\right)z^2\text{k}$ and S is a closed surface bounding the region and consisting of solid cylinder $x^2+y^2\le a^2$ and $0\le z\le b.$

[T] Use a CAS to calculate the flux of $\text{F}(x,y,z)=\left(x^3+y\text{sin}z\right)\text{i}+\left(y^3+z\text{sin}x\right)\text{j}+3z\text{k}$ across surface S , where S is the boundary of the solid bounded by hemispheres $z=\sqrt{4-x^2-y^2}$ and $z=\sqrt{1-x^2-y^2},$ and plane $z=0.$

Use the divergence theorem to evaluate ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=xy\text{i}-\frac{1}{2}{y}^2\text{j}+z\text{k}$ and S is the surface consisting of three pieces: $z=4-3x^2-3y^2,1\le z\le 4$ on the top; $x^2+y^2=1,0\le z\le 1$ on the sides; and $z=0$ on the bottom.

[T] Use a CAS and the divergence theorem to evaluate ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=\left(2x+y\text{cos}z\right)\text{i}+\left(x^2-y\right)\text{j}+y^{2}z\text{k}$ and S is sphere $x^2+y^2+z^2=4$ orientated outward.

Use the divergence theorem to evaluate ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k}$ and S is the boundary of the solid enclosed by paraboloid $y=x^2+z^2-2,$ cylinder $x^2+z^2=1,$ and plane $x+y=2,$ and S is oriented outward.

$T(x,y,z)=100+x+2y+z;$ $D=\left\{\left(x,y,z\right):0\le x\le 1,0\le y\le 1,0\le z\le 1\right\}$

$T(x,y,z)=100+e^{\text{−}z};$ $D=\left\{\left(x,y,z\right):0\le x\le 1,0\le y\le 1,0\le z\le 1\right\}$

$T(x,y,z)=100e^{\text{−}{x}^2-y^2-z^2};$ D is the sphere of radius a centered at the origin.

Vector field $\text{F}(x,y)=x^{2}y\text{i}+y^{2}x\text{j}$ is conservative.

For vector field $\text{F}(x,y)=P(x,y)\text{i}+Q(x,y)\text{j},$ if $P_y(x,y)=Q_x(x,y)$ in open region $D,$ then ${\displaystyle {\int }_{\partial D}Pdx+Qdy=0}.$

The divergence of a vector field is a vector field.

If $\text{curl}\text{F}=0,$ then $\text{F}$ is a conservative vector field.

$\text{F}\left(x,y\right)=\frac{1}{2}\text{i}+2x\text{j}$

$\text{F}\left(x,y\right)=\sqrt{\frac{y\text{i}+3x\text{j}}{x^2+y^2}}$

$\text{F}\left(x,y\right)=y\text{i}+\left(x-2e^y\right)\text{j}$

$\text{F}\left(x,y\right)=\left(6xy\right)\text{i}+\left(3x^2-y{e}^y\right)\text{j}$

$\text{F}\left(x,y,z\right)=\left(2xy+z^2\right)\text{i}+\left(x^2+2yz\right)\text{j}+\left(2xz+y^2\right)\text{k}$

$\text{F}(x,y,z)=\left(e^{x}y\right)\text{i}+\left(e^x+z\right)\text{j}+\left(e^x+y^2\right)\text{k}$

${\displaystyle \underset{C}{\int }{x}^{2}dy+\left(2x-3xy\right)dx,}$ along $C:y=\frac{1}{2}x$ from (0, 0) to (4, 2)

${\displaystyle \underset{C}{\int }ydx+x{y}^{2}dy,}$ where $C:x=\sqrt{t},y=t-1,0\le t\le 1$

${\displaystyle {\iint }_{S}x{y}^{2}dS},$ where S is surface $z=x^2-y,0\le x\le 1,0\le y\le 4$

$\text{F}(x,y,z)=3xyz\text{i}+xy{e}^z\text{j}-3xy\text{k}$

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

${\displaystyle \underset{C}{\int }3xydx+2x{y}^{2}dy,}$ where C is a square with vertices (0, 0), (0, 2), (2, 2) and (2, 0)

${\displaystyle {\oint }_{C}3ydx+\left(x+e^y\right)dy,}$ where C is a circle centered at the origin with radius 3

$\text{F}(x,y,z)=y\text{i}-x\text{j}+z\text{k},$ where $S$ is the upper half of the unit sphere

$\text{F}(x,y,z)=y\text{i}+xyz\text{j}-2zx\text{k},$ where $S$ is the upward-facing paraboloid $z=x^2+y^2$ lying in cylinder $x^2+y^2=1$

$\text{F}(x,y,z)=\left(x^{3}y\right)\text{i}+\left(3y-e^x\right)\text{j}+\left(z+x\right)\text{k},$ over cube $S$ defined by $\mathrm{-1}\le x\le 1,$ $0\le y\le 2,$ $0\le z\le 2$

$\text{F}(x,y,z)=\left(2xy\right)\text{i}+\left(\text{−}{y}^2\right)\text{j}+\left(2z^3\right)\text{k},$ where $S$ is bounded by paraboloid $z=x^2+y^2$ and plane $z=2$

Find the amount of work performed by a 50-kg woman ascending a helical staircase with radius 2 m and height 100 m. The woman completes five revolutions during the climb.

Find the total mass of a thin wire in the shape of a semicircle with radius $\sqrt{2,}$ and a density function of $\rho \left(x,y\right)=y+x^2.$

Find the total mass of a thin sheet in the shape of a hemisphere with radius 2 for $z\ge 0$ with a density function $\rho \left(x,y,z\right)=x+y+z.$

Use the divergence theorem to compute the value of the flux integral over the unit sphere with $\text{F}(x,y,z)=3z\text{i}+2y\text{j}+2x\text{k}.$