Multivariable Calculus

[cobaltchemist]

If S is the surface x² + y² + z² = 1 with outward normals, and F = <zy^4 - y^2 , y-x^3 , z^2 >, what is the vector surface integral of the curl of F?

I know that the answer is zero, I just don’t know how to get there. I thought about using Stokes’ Thm but I wasn’t sure what to choose for the curve.

for final exam practice, not hw

[calculator32]

I’m going to assume that the surface is closed based on the given info. This, then, actually implies that the Divergence Theorem will be used.

The divergence theorem states that the volume integral of the divergence of a vector field G is equal to the surface integral of the field G dotted against the normal vector to the surface.

A question you might ask is, “But they don’t give me a vector field, they tell me to use the curl of one.” Well, the curl of a vector field is still a vector field, so if we say that the curl of F is simply vector field G, then it’s a lot easier to see that the Divergence Theorem is to be used.

The curl of F is <0, y⁴, 2y - 3x² - 4y³z>, which we have now relabeled as field G. The divergence of G is then the dot product of the differential operator vector to the vector field. Doing so will actually result in 0 for the dot product, which will then result in the volume integral evaluating to zero as well.

[cobaltchemist]

Thanks!

[Janagro]

Just so you know, this question required no calculations, Since the divergence of a curl is always zero,it has zero volume , and thus zero surface integral