Question

This problem has been solved!See the answerVerify that the line intergral 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.F= < x,y,z>; S is the paraboloid z = 13 - x^2 - y^2, for 0 less than or equal z less than or equal 13 and C is the circle x^2 + y^2 = 13 in the xy plane.

Answer 1

Line integral: Parameterize
`C` by

`\vec r(t)=\langle√(13)\cos t,√(13)\sin t,0\rangle`

with
`0\le t\le2\pi`. Then

`\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^(2\pi)\langle√(13)\cos t,√(13)\sin t,0\rangle\cdot\langle-√(13)\sin t,√(13)\cos t,0\rangle\,\mathrm dt`

`=\displaystyle\int_0^(2\pi)0\,\mathrm dt=\boxed 0`

Surface integral: By Stokes' theorem, the line integral of
`\vec F` over
`C` is equivalent to the surface integral of the curl of
`\vec F` over
`S`:

`\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\\abla*\vec F)\cdot\mathrm d\vec S`

The curl of
`\langle x,y,z\rangle` is 0, so the value of the surface integral is 0, as expected.