# 6.8 The divergence theorem  (Page 7/12)

 Page 7 / 12

Work the previous example for surface S that is a sphere of radius 4 centered at the origin, oriented outward.

$\approx 6.777\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}$

## Key concepts

• The divergence theorem relates a surface integral across closed surface S to a triple integral over the solid enclosed by S . The divergence theorem is a higher dimensional version of the flux form of Green’s theorem, and is therefore a higher dimensional version of the Fundamental Theorem of Calculus.
• The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
• The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.

## Key equations

• Divergence theorem
${\iiint }_{E}\text{div}\phantom{\rule{0.2em}{0ex}}\text{F}dV=\underset{S}{\iint }\text{F}·d\text{S}$

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral ${\int }_{S}^{}\text{F}·\text{n}ds$ for the given choice of F and the boundary surface S. For each closed surface, assume N is the outward unit normal vector.

[T] $\text{F}\left(x,y,z\right)=x\text{i}+y\text{j}+z\text{k};$ S is the surface of cube $0\le x\le 1,0\le y\le 1,0

[T] $\text{F}\left(x,y,z\right)=\left(\text{cos}\phantom{\rule{0.2em}{0ex}}yz\right)\text{i}+{e}^{xz}\text{j}+3{z}^{2}\text{k}\text{;}$ S is the surface of hemisphere $z=\sqrt{4-{x}^{2}-{y}^{2}}$ together with disk ${x}^{2}+{y}^{2}\le 4$ in the xy -plane.

${\int }_{S}^{}\text{F}·\text{n}ds=75.3982$

[T] $\text{F}\left(x,y,z\right)=\left({x}^{2}+{y}^{2}-{x}^{2}\right)\text{i}+{x}^{2}y\text{j}+3z\text{k};$ S is the surface of the five faces of unit cube $0\le x\le 1,0\le y\le 1,0

[T] $\text{F}\left(x,y,z\right)=x\text{i}+y\text{j}+z\text{k}\text{;}$ S is the surface of paraboloid $z={x}^{2}+{y}^{2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}0\le z\le 9.$

${\int }_{S}^{}\text{F}·\text{n}ds=127.2345$

[T] $\text{F}\left(x,y,z\right)={x}^{2}\text{i}+{y}^{2}\text{j}+{z}^{2}\text{k}\text{;}$ S is the surface of sphere ${x}^{2}+{y}^{2}+{z}^{2}=4.$

[T] $\text{F}\left(x,y,z\right)=x\text{i}+y\text{j}+\left({z}^{2}-1\right)\text{k}\text{;}$ S is the surface of the solid bounded by cylinder ${x}^{2}+{y}^{2}=4$ and planes $z=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=1.$

${\int }_{S}^{}\text{F}·\text{n}ds=37.6991$

[T] $\text{F}\left(x,y,z\right)=x{y}^{2}\text{i}+y{z}^{2}\text{j}+{x}^{2}z\text{k}\text{;}$ S is the surface bounded above by sphere $\rho =2$ and below by cone $\phi =\frac{\pi }{4}$ in spherical coordinates. (Think of S as the surface of an “ice cream cone.”)

[T] $\text{F}\left(x,y,z\right)={x}^{3}\text{i}+{y}^{3}\text{j}+3{a}^{2}z\text{k}\phantom{\rule{0.2em}{0ex}}\text{(constant}\phantom{\rule{0.2em}{0ex}}a>0\right)\text{;}$ S is the surface bounded by cylinder ${x}^{2}+{y}^{2}={a}^{2}$ and planes $z=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=1.$

${\int }_{S}^{}\text{F}·\text{n}ds=\frac{9\pi {a}^{4}}{2}$

[T] Surface integral ${\iint }_{S}\text{F}·d\text{S},$ where S is the solid bounded by paraboloid $z={x}^{2}+{y}^{2}$ and plane $z=4,$ and $\text{F}\left(x,y,z\right)=\left(x+{y}^{2}{z}^{2}\right)\text{i}+\left(y+{z}^{2}{x}^{2}\right)\text{j}+\left(z+{x}^{2}{y}^{2}\right)\text{k}$

Use the divergence theorem to calculate surface integral ${\iint }_{S}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)=\left({e}^{{y}^{2}}\right)\text{i}+\left(y+\text{sin}\left({z}^{2}\right)\right)\text{j}+\left(z-1\right)\text{k}$ and S is upper hemisphere ${x}^{2}+{y}^{2}+{z}^{2}=1\text{,}\phantom{\rule{0.2em}{0ex}}z\ge 0\text{,}$ oriented upward.

${\iint }_{S}\text{F}·d\text{S}=\frac{\pi }{3}$

Use the divergence theorem to calculate surface integral ${\iint }_{S}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)={x}^{4}\text{i}-{x}^{3}{z}^{2}\text{j}+4x{y}^{2}z\text{k}$ and S is the surface bounded by cylinder ${x}^{2}+{y}^{2}=1$ and planes $z=x+2$ and $z=0.$

Use the divergence theorem to calculate surface integral ${\iint }_{S}\text{F}·d\text{S}$ when $\text{F}\left(x,y,z\right)={x}^{2}{z}^{3}\text{i}+2xy{z}^{3}\text{j}+x{z}^{4}\text{k}$ and S is the surface of the box with vertices $\left(±1,±2,±3\right).$

${\iint }_{S}\text{F}·d\text{S}=0$

Use the divergence theorem to calculate surface integral ${\iint }_{S}\text{F}·d\text{S}$ when $\text{F}\left(x,y,z\right)=z\phantom{\rule{0.2em}{0ex}}{\text{tan}}^{-1}\left({y}^{2}\right)\text{i}+{z}^{3}\text{ln}\left({x}^{2}+1\right)\text{j}+z\text{k}$ and S is a part of paraboloid ${x}^{2}+{y}^{2}+z=2$ that lies above plane $z=1$ and is oriented upward.

[T] Use a CAS and the divergence theorem to calculate flux ${\iint }_{S}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)=\left({x}^{3}+{y}^{3}\right)\text{i}+\left({y}^{3}+{z}^{3}\right)\text{j}+\left({z}^{3}+{x}^{3}\right)\text{k}$ and S is a sphere with center (0, 0) and radius 2.

${\iint }_{S}\text{F}·d\text{S}=241.2743$

Use the divergence theorem to compute the value of flux integral ${\iint }_{S}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)=\left({y}^{3}+3x\right)\text{i}+\left(xz+y\right)\text{j}+\left[z+{x}^{4}\text{cos}\left({x}^{2}y\right)\right]\text{k}$ and S is the area of the region bounded by ${x}^{2}+{y}^{2}=1,x\ge 0,y\ge 0,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0\le z\le 1.$

Use the divergence theorem to compute flux integral ${\iint }_{S}\text{F}·d\text{S},$ where $\text{F}\left(x,y,z\right)=y\text{j}-z\text{k}$ and S consists of the union of paraboloid $y={x}^{2}+{z}^{2},0\le y\le 1,$ and disk ${x}^{2}+{z}^{2}\le 1,y=1,$ oriented outward. What is the flux through just the paraboloid?

${\iint }_{D}\text{F}·d\text{S}=\text{−}\pi$

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