Question

Verify that the line integral and the surface integral of​ Stokes' Theorem are equal for the following vector​ field, surface​ S, and closed curve C. Assume that C has counterclockwise orientation and S has a consistent orientation. Fequals=left angle y comma font size decreased by 6 minus x comma font size decreased by 6 14 right angley, −x, 14​; S is the upper half of the sphere x squaredx2plus+y squaredy2plus+z squaredz2equals=44 and C is the circle x squaredx2plus+y squaredy2equals=44 in the​ xy-plane.

Answer 1

As near as I can tell, you're given the vector field

`\vec F(x,y,z)=\langle y,-x,14\rangle`

and that
`S` is the part of the upper half of the sphere with equation

`x^2+y^2+z^2=4`

with boundary
`C` the circle in the plane
`z=0`.

• Line integral:

Parameterize
`C` by

`\vec r(t)=\langle2\cos t,2\sin t,0\rangle`

with
`0\le t\le2\pi`. Then the line integral of
`\vec F(x,y,z)` along
`C` is

`\displaystyle\int_C\vec F(x,y,z)\cdot\mathrm d\vec r=\int_0^(2\pi)\langle2\sin t,-2\cos t,14\rangle\cdot\langle-2\sin t,2\cos t,0\rangle\,\mathrm dt`

`=\displaystyle-4\int_0^(2\pi)(\sin^2t+\cos^2t)\,\mathrm dt=\boxed{-8\pi}`

• Surface integral:

Parameterize
`S` by

`\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle`

with
`0\le u\le2\pi` and
`0\le v\le\frac\pi2`. We have

`\\abla*\vec F(x,y,z)=\langle0,0,-2\rangle`

Take the normal vector to
`S` to be

`\vec s_v*\vec s_u=\langle4\cos u\sin^2v,4\sin u\sin^2v,2\sin2v\rangle`

Then the surface integral of the curl of
`\vec F(x,y,z)` across
`S` is

`\displaystyle\iint_S(\\abla*\vec F(x,y,z))\cdot\mathrm d\vec S=\iint_S(\\abla*\vec F(x(u,v),y(u,v),z(u,v)))\cdot(\vec s_v*\vec s_u)\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^(\pi/2)\int_0^(2\pi)\langle0,0,-2\rangle\cdot\langle4\cos u\sin^2v,4\sin u\sin^2v,2\sin2v\rangle\,\mathrm du\,\mathrm dv`

`=\displaystyle-4\int_0^(\pi/2)\int_0^(2\pi)\sin2v\,\mathrm du\,\mathrm dv=\boxed{-8\pi}`