6.8 The divergence theorem  (Page 7/12)

[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] $\text{F}(x,y,z)=(\text{cos}yz)\text{i}+e^{xz}\text{j}+3z^2\text{k}\text{;}$ 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.

[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 $z=x^2+y^2\text{for}0\le z\le 9.$

[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 $z=0\text{and}z=1.$

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

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

Use the divergence theorem to calculate surface integral ${\displaystyle {\iint }_S\text{F}·}d\text{S},$ 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 $x^2+y^2+z^2=1\text{,}z\ge 0\text{,}$ oriented upward.

Use the divergence theorem to calculate surface integral ${\displaystyle {\iint }_S\text{F}·}d\text{S},$ 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 ${\displaystyle {\iint }_S\text{F}·d\text{S}}$ 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 $(\mathrm{\pm 1},\mathrm{\pm 2},\mathrm{\pm 3}).$

Use the divergence theorem to calculate surface integral ${\displaystyle {\iint }_S\text{F}·d\text{S}}$ when $\text{F}(x,y,z)=z{\text{tan}}^{\mathrm{-1}}\left(y^2\right)\text{i}+z^3\text{ln}\left(x^2+1\right)\text{j}+z\text{k}$ 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 ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ 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.

Use the divergence theorem to compute the value of flux integral ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ 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 $x^2+y^2=1,x\ge 0,y\ge 0,\text{and}0\le z\le 1.$

Use the divergence theorem to compute flux integral ${\displaystyle {\iint }_S\text{F}·d\text{S}},$ 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?