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Work the previous example for surface S that is a sphere of radius 4 centered at the origin, oriented outward.
$\approx 6.777\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}$
For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral ${\int}_{S}^{}\text{F}\xb7\text{n}ds$ for the given choice of F and the boundary surface S. For each closed surface, assume N is the outward unit normal vector.
[T] $\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k};$ S is the surface of cube $0\le x\le 1,0\le y\le 1,0<z\le 1.$
[T] $\text{F}(x,y,z)=(\text{cos}\phantom{\rule{0.2em}{0ex}}yz)\text{i}+{e}^{xz}\text{j}+3{z}^{2}\text{k}\text{;}$ 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.
${\int}_{S}^{}\text{F}\xb7\text{n}ds}=75.3982$
[T] $\text{F}(x,y,z)=\left({x}^{2}+{y}^{2}-{x}^{2}\right)\text{i}+{x}^{2}y\text{j}+3z\text{k};$ S is the surface of the five faces of unit cube $0\le x\le 1,0\le y\le 1,0<z\le 1.$
[T] $\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k}\text{;}$ S is the surface of paraboloid $z={x}^{2}+{y}^{2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}0\le z\le 9.$
${\int}_{S}^{}\text{F}\xb7\text{n}ds}=127.2345$
[T] $\text{F}(x,y,z)={x}^{2}\text{i}+{y}^{2}\text{j}+{z}^{2}\text{k}\text{;}$ S is the surface of sphere ${x}^{2}+{y}^{2}+{z}^{2}=4.$
[T] $\text{F}(x,y,z)=x\text{i}+y\text{j}+\left({z}^{2}-1\right)\text{k}\text{;}$ S is the surface of the solid bounded by cylinder ${x}^{2}+{y}^{2}=4$ and planes $z=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=1.$
${\int}_{S}^{}\text{F}\xb7\text{n}ds}=37.6991$
[T] $\text{F}(x,y,z)=x{y}^{2}\text{i}+y{z}^{2}\text{j}+{x}^{2}z\text{k}\text{;}$ 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] $\text{F}(x,y,z)={x}^{3}\text{i}+{y}^{3}\text{j}+3{a}^{2}z\text{k}\phantom{\rule{0.2em}{0ex}}\text{(constant}\phantom{\rule{0.2em}{0ex}}a>0)\text{;}$ S is the surface bounded by cylinder ${x}^{2}+{y}^{2}={a}^{2}$ and planes $z=0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z=1.$
${\int}_{S}^{}\text{F}\xb7\text{n}ds}=\frac{9\pi {a}^{4}}{2$
[T] Surface integral ${\iint}_{S}\text{F}\xb7}d\text{S},$ where S is the solid bounded by paraboloid $z={x}^{2}+{y}^{2}$ and plane $z=4,$ and $\text{F}(x,y,z)=\left(x+{y}^{2}{z}^{2}\right)\text{i}+\left(y+{z}^{2}{x}^{2}\right)\text{j}+\left(z+{x}^{2}{y}^{2}\right)\text{k}$
Use the divergence theorem to calculate surface integral ${\iint}_{S}\text{F}\xb7}d\text{S},$ where $\text{F}(x,y,z)=\left({e}^{{y}^{2}}\right)\text{i}+\left(y+\text{sin}\left({z}^{2}\right)\right)\text{j}+\left(z-1\right)\text{k}$ and S is upper hemisphere ${x}^{2}+{y}^{2}+{z}^{2}=1\text{,}\phantom{\rule{0.2em}{0ex}}z\ge 0\text{,}$ oriented upward.
${\iint}_{S}\text{F}\xb7}d\text{S}=\frac{\pi}{3$
Use the divergence theorem to calculate surface integral ${\iint}_{S}\text{F}\xb7}d\text{S},$ where $\text{F}(x,y,z)={x}^{4}\text{i}-{x}^{3}{z}^{2}\text{j}+4x{y}^{2}z\text{k}$ 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 $\iint}_{S}\text{F}\xb7d\text{S$ when $\text{F}(x,y,z)={x}^{2}{z}^{3}\text{i}+2xy{z}^{3}\text{j}+x{z}^{4}\text{k}$ and S is the surface of the box with vertices $(\mathrm{\pm 1},\mathrm{\pm 2},\mathrm{\pm 3}).$
${\iint}_{S}\text{F}\xb7d\text{S}}=0$
Use the divergence theorem to calculate surface integral $\iint}_{S}\text{F}\xb7d\text{S$ when $\text{F}(x,y,z)=z\phantom{\rule{0.2em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left({y}^{2}\right)\text{i}+{z}^{3}\text{ln}\left({x}^{2}+1\right)\text{j}+z\text{k}$ 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 ${\iint}_{S}\text{F}\xb7d\text{S}},$ where $\text{F}(x,y,z)=\left({x}^{3}+{y}^{3}\right)\text{i}+\left({y}^{3}+{z}^{3}\right)\text{j}+\left({z}^{3}+{x}^{3}\right)\text{k}$ and S is a sphere with center (0, 0) and radius 2.
${\iint}_{S}\text{F}\xb7d\text{S}}=241.2743$
Use the divergence theorem to compute the value of flux integral ${\iint}_{S}\text{F}\xb7d\text{S}},$ where $\text{F}(x,y,z)=\left({y}^{3}+3x\right)\text{i}+\left(xz+y\right)\text{j}+\left[z+{x}^{4}\text{cos}\left({x}^{2}y\right)\right]\text{k}$ 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 ${\iint}_{S}\text{F}\xb7d\text{S}},$ where $\text{F}(x,y,z)=y\text{j}-z\text{k}$ 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}\xb7d\text{S}}=\text{\u2212}\pi $
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