Originally Posted by

**yzc717** Let C be the intersection of the sphere $\displaystyle x^2 +y^2 +z^2 = 2$ and the plane z=1, oriented counter-clockwise as viewed from high above the z-axis. Use Stokes' Theorem to evaluate.

$\displaystyle \int_{c} (-y^3 +z) dx + (x^3 +2y) dy + (y - x) dz$

Is my solution right?

Set $\displaystyle \vec F=(-y^3 +z) \vec i + (x^3 +2y) \vec j + (y - x) \vec k $

$\displaystyle curl(\vec F)= \vec i + \vec 2j + (3x^2 + 3y^2) \vec k$

By Stokes' Theorem, the line integral is equal to the surface integral

$\displaystyle \int \int_{\Sigma} (\nabla \times F) dS$

$\displaystyle \int \int_{\Sigma} (\nabla \times F) dS = \int \int_{\Sigma}' (\nabla \times F) \cdot \vec n ds

\int \int_{\Sigma}' (\nabla \times F) \cdot \vec k ds$

Sigma': $\displaystyle { (x,y,z): x^2 +y^2 \leq 1, z=1} $ flat disk

so $\displaystyle (x, y, z) \rightarrow (cos \theta, sin \theta, 1)$

so

$\displaystyle \int \int_{\Sigma} ' 3x^2 +3y^2 ds$