6.7 Stokes’ theorem

$\text{F}(x,y,z)=y^2\text{i}+z^2\text{j}+x^2\text{k}\text{;}$ S is the first-octant portion of plane $x+y+z=1.$

$\text{F}(x,y,z)=z\text{i}+x\text{j}+y\text{k}\text{;}$ S is hemisphere $z={\left(a^2-x^2-y^2\right)}^{1\text{/}2}.$

$\text{F}(x,y,z)=y^2\text{i}+2x\text{j}+5\text{k}\text{;}$ S is hemisphere $z={\left(4-x^2-y^2\right)}^{1\text{/}2}.$

$\text{F}(x,y,z)=z\text{i}+2x\text{j}+3y\text{k}\text{;}$ S is upper hemisphere $z=\sqrt{9-x^2-y^2}.$

$\text{F}(x,y,z)=\left(x+2z\right)\text{i}+\left(y-x\right)\text{j}+\left(z-y\right)\text{k}\text{;}$ S is a triangular region with vertices (3, 0, 0), (0, 3/2, 0), and (0, 0, 3).

$\text{F}(x,y,z)=2y\text{i}-6z\text{j}+3x\text{k}\text{;}$ S is a portion of paraboloid $z=4-x^2-y^2$ and is above the xy -plane.

$\text{F}(x,y,z)=xy\text{i}-z\text{j}$ and S is the surface of the cube $0\le x\le 1,0\le y\le 1,0\le z\le 1,$ except for the face where $z=0,$ and using the outward unit normal vector.

$\text{F}(x,y,z)=xy\text{i}+x^2\text{j}+z^2\text{k}\text{;}$ and C is the intersection of paraboloid $z=x^2+y^2$ and plane $z=y,$ and using the outward normal vector.

$\text{F}(x,y,z)=4y\text{i}+z\text{j}+2y\text{k}$ and C is the intersection of sphere $x^2+y^2+z^2=4$ with plane $z=0,$ and using the outward normal vector

Use Stokes’ theorem to evaluate ${\displaystyle \underset{C}{\int }\left[2x{y}^{2}zdx+2x^{2}yzdy+\left(x^2{y}^2-2z\right)dz\right]},$ where C is the curve given by $x=\text{cos}t,y=\text{sin}t,z=\text{sin}t,0\le t\le 2\pi ,$ traversed in the direction of increasing t .

[T] Use a computer algebraic system (CAS) and Stokes’ theorem to approximate line integral ${\displaystyle \underset{C}{\int }\left(ydx+zdy+xdz\right)},$ where C is the intersection of plane $x+y=2$ and surface $x^2+y^2+z^2=2\left(x+y\right),$ traversed counterclockwise viewed from the origin.

[T] Use a CAS and Stokes’ theorem to approximate line integral ${\displaystyle \underset{C}{\int }\left(3ydx+2zdy-5xdz\right)},$ where C is the intersection of the xy -plane and hemisphere $z=\sqrt{1-x^2-y^2},$ traversed counterclockwise viewed from the top—that is, from the positive z -axis toward the xy -plane.

[T] Use a CAS and Stokes’ theorem to approximate line integral ${\displaystyle \underset{C}{\int }\left[\left(1+y\right)zdx+\left(1+z\right)xdy+\left(1+x\right)ydz\right]},$ where C is a triangle with vertices $\left(1,0,0\right),$ $\left(0,1,0\right),$ and $\left(0,0,1\right)$ oriented counterclockwise.

Use Stokes’ theorem to evaluate ${\displaystyle {\iint }_S\text{curl}\text{F}·d\text{S}},$ where $\text{F}(x,y,z)=e^{xy}\text{cos}z\text{i}+x^{2}z\text{j}+xy\text{k},$ and S is half of sphere $x=\sqrt{1-y^2-z^2},$ oriented out toward the positive x -axis.