- #1

JaysFan31

Take u=(x^3)+(y^3)+(z^2) and v=x+y+z and evaluate the surface integral

double integral of grad(u) x grad(v) ndS where x is the cross product and between the cross product and the ndS there should be a dot product sign. The region S is the hemisphere x^2+y^2+z^2=1 with z greater than or equal to 0 and n has non-negative k component.

Here's my work:

grad(u)=(3x^2)i-(3y^2)j+(2z)k

grad(v)= i+j+k

grad(u) x grad (v)= (-3y^2-2z)i+(2z-3x^2)j+(3x^2-3y^2)k

curl F * k = (-6x+6y)

Thus I=double integral of curl F * ndS = double integral of curl F *k dA = double integral of (-6x+6y)dA.

I used polar coordinates:

-6 integral from 0 to 2pi integral from 0 to 1 (rcost-rsint) rdrdt. However, this integral equals zero. Is this right?

Is this flux zero?