6.6 Surface integrals  (Page 14/27)

If surface S is given by $\left\{\left(x,y,z\right):0\le x\le 1,0\le y\le 1,z=10\right\},$ then ${\displaystyle {\iint }_{S}f\left(x,y,z\right)}dS={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}f\left(x,y,10\right)}}dxdy.$

If surface S is given by $\left\{\left(x,y,z\right):0\le x\le 1,0\le y\le 1,z=x\right\},$ then ${\displaystyle {\iint }_{S}f\left(x,y,z\right)}dS={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{1}f\left(x,y,x\right)}}dxdy.$

Surface $\text{r}=⟨v\text{cos}u,v\text{sin}u,v^2⟩,\text{for}0\le u\le \pi ,0\le v\le 2,$ is the same as surface $\text{r}=⟨\sqrt{v}\text{cos}2u,\sqrt{v}\text{sin}2u,v⟩,$ for $0\le u\le \frac{\pi }{2},0\le v\le 4.$

Given the standard parameterization of a sphere, normal vectors ${\text{t}}_u^{}\times {\text{t}}_v$ are outward normal vectors.

Plane $3x-2y+z=2$

Paraboloid $z=x^2+y^2,$ for $0\le z\le 9.$

Plane $2x-4y+3z=16$

The frustum of cone $z^2=x^2+y^2,\text{for}2\le z\le 8$

The portion of cylinder $x^2+y^2=9$ in the first octant, for $0\le z\le 3$

A cone with base radius r and height h , where r and h are positive constants

[T] Half cylinder $\left\{\left(r,\theta ,z\right):r=4,0\le \theta \le \pi ,0\le z\le 7\right\}$

[T] Plane $z=10-x-y$ above square $\left|x\right|\le 2,\left|y\right|\le 2$

${\displaystyle {\iint }_{S}zdS}$

${\displaystyle {\iint }_S(x-2y)dS}$

${\displaystyle {\iint }_S\left(x^2+y^2\right)zdS}$

$\text{F}\left(x,y,z\right)=x\text{i}+2y\text{j}=3z\text{k},$ and S is that part of plane $15x-12y+3z=6$ that lies above unit square $0\le x\le 1,0\le y\le 1.$

$\text{F}(x,y)=x\text{i}+y\text{j},$ and S is hemisphere $z=\sqrt{1-x^2-y^2}.$

$\text{F}(x,y,z)=x^2\text{i}+y^2\text{j}+z^2\text{k},$ and S is the portion of plane $z=y+1$ that lies inside cylinder $x^2+y^2=1.$

[T] S is surface $z=4-x-2y,\text{with}z\ge 0\text{,}x\ge 0\text{,}y\ge 0\text{;}\xi =x.$

[T] S is surface $z=x^2+y^2,\text{with}z\le 1\text{;}\xi =z.$

[T] S is surface $x^2+y^2+x^2=5,\text{with}z\ge 1\text{;}\xi ={\theta }^2.$

Evaluate ${\displaystyle {\iint }_S\left(y^{2}z\text{i}+y^3\text{j}+xz\text{k}\right)·d\text{S}}\text{,}$ where S is the surface of cube $\mathrm{-1}\le x\le 1,\mathrm{-1}\le y\le 1,\text{and}0\le z\le 2.$ in a counterclockwise direction.

Evaluate surface integral ${\displaystyle {\iint }_{S}gdS},$ where $g(x,y,z)=xz+2x^2-3xy$ and S is the portion of plane $2x-3y+z=6$ that lies over unit square R : $0\le x\le 1\text{,}0\le y\le 1.$

Evaluate ${\displaystyle {\iint }_S(x+y+z)d\text{S}}\text{,}$ where S is the surface defined parametrically by $\text{R}(u,v)=(2u+v)\text{i}+(u-2v)\text{j}+(u+3v)\text{k}$ for $0\le u\le 1,\text{and}0\le v\le 2.$

[T] Evaluate ${\displaystyle {\iint }_S(x-y^2+z)d\text{S}}\text{,}$ where S is the surface defined by $\text{R}(u,v)=u^2\text{i}+v\text{j}+u\text{k}\text{,}0\le u\le 1\text{,}0\le v\le 1.$

[T] Evaluate where S is the surface defined by $\text{R}(u,v)=u\text{i}-u^2\text{j}+v\text{k}\text{,}0\le u\le 2\text{,}0\le v\le 1.$ for $0\le u\le 1\text{,}0\le v\le 2.$

Evaluate ${\displaystyle {\iint }_S\left(x^2+y^2\right)d\text{S}},$ where S is the surface bounded above hemisphere $z=\sqrt{1-x^2-y^2},$ and below by plane $z=0.$

Evaluate ${\displaystyle {\iint }_S\left(x^2+y^2+z^2\right)d\text{S}},$ where S is the portion of plane $z=x+1$ that lies inside cylinder $x^2+y^2=1.$

[T] Evaluate ${\displaystyle {\iint }_S{x}^{2}zdS,}$ where S is the portion of cone $z^2=x^2+y^2$ that lies between planes $z=1$ and $z=4.$

[T] Evaluate ${\displaystyle {\iint }_S\left(xz\text{/}y\right)dS},$ where S is the portion of cylinder $x=y^2$ that lies in the first octant between planes $z=0,z=5,y=1,$ and $y=4.$