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

$\vec r(t)=\langle\sqrt{13}\cos t,\sqrt{13}\sin t,0\rangle$

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

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle\sqrt{13}\cos t,\sqrt{13}\sin t,0\rangle\cdot\langle-\sqrt{13}\sin t,\sqrt{13}\cos t,0\rangle\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}0\,\mathrm dt=\boxed 0$

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 $S$:

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

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