스토크스 정리

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\,\text{F}\cdot \text{N}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

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

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\,\text{F}\cdot \text{N}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

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

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\,\text{F}\cdot \text{N}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

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

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\,\text{F}\cdot \text{N}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

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

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\,\text{F}\cdot \text{N}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

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

$\text{F}(x,y,z)=2y\text{i}-6z\text{j}+3x\text{k};$ 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.

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

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

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

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

[T] Use a CAS and Stokes’ theorem to approximate line integral $\int_{C}(3ydx+2zdy-5xdz),$ 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 $\int_{C}[(1+y)zdx+(1+z)xdy+(1+x)ydz],$ where C is a triangle with vertices $(1,0,0),$ $(0,1,0),$ and $(0,0,1)$ oriented counterclockwise.

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

[T] Use a CAS and Stokes’ theorem to evaluate ${\iint }_{S}(\text{curl}\,\text{F}\cdot \text{N})dS,$ where $\text{F}(x,y,z)={x}^{2}y\text{i}+x{y}^{2}\text{j}+{z}^{3}\text{k}$ and S is the curve of the part of plane $3x+2y+z=6$ above cylinder ${x}^{2}+{y}^{2}=4,$ oriented clockwise when viewed from above.

[T] Use a CAS and Stokes’ theorem to evaluate $\iint_{S}\text{curl}\,\text{F}\cdot d\text{S},$ where $\text{F}(x,y,z)=\left(\sin (y+z)-y{x}^{2}-\frac{{y}^{3}}{3}\right)\text{i}+x\,\cos (y+z)\text{j}+\cos (2y)\text{k}$ and S consists of the top and the four sides but not the bottom of the cube with vertices $(\pm 1,\pm 1,\pm 1),$ oriented outward.

[T] Use a CAS and Stokes’ theorem to evaluate $\iint_{S}\text{curl}\,\text{F}\cdot d\text{S},$ where $\text{F}(x,y,z)={z}^{2}\text{i}-3xy\text{j}+{x}^{3}{y}^{3}\text{k}$ and S is the top part of $z=5-{x}^{2}-{y}^{2}$ above plane $z=1,$ and S is oriented upward.

Use Stokes’ theorem to evaluate ${\iint }_{S}(\text{curl}\,\text{F}\cdot \text{N})dS,$ where $\text{F}(x,y,z)={z}^{2}\text{i}+{y}^{2}\text{j}+x\text{k}$ and S is a triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) with upward orientation.

Use Stokes’ theorem to evaluate line integral $\int_{C}(zdx+xdy+ydz),$ where C is a triangle with vertices (3, 0, 0), (0, 0, 2), and (0, 6, 0) traversed in the given order.

Use Stokes’ theorem to evaluate $\int_{C}\left(\frac{1}{2}{y}^{2}dx+zdy+xdz\right),$ where C is the curve of intersection of plane $x+z=1$ and ellipsoid ${x}^{2}+2{y}^{2}+{z}^{2}=1,$ oriented clockwise from the origin.