Stokes' Theorem: Evaluating a Surface Integral on a Hemisphere

[JaysFan31]

Here's my problem:
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?

 

[cristo]

Here's my problem:
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.

Is this what you mean: \int\int\nabla u \times\nabla v \cdot \bold{n}dS?

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

Where does the minus sign in front of the j component come from? \nabla u=\frac{\partial u}{\partial x}\bold{i}+\frac{\partial u}{\partial y}\bold{j}+\frac{\partial u}{\partial z}\bold{k}

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

These look correct with the amended form for grad(u)

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.

What's F? You should define anything you introduce. I imagine you've used Stokes' Theorem here, but where?

[As an aside, it would be helpful if you could learn LaTex; it's very simple to learn! Click on one of the images in my text, and follow the link to the tutorial]

 

[Dick]

I think I would agree the total flux is zero, but I don't see how you are getting it. What does the stated problem have to do with Stokes Theorem? Where did the curl come from? What is F?

 

[JaysFan31]

Yes, the integral that cristo wrote is what I'm looking for.

Can't I just let F=grad(u) x grad (v)= (-3y^2-2z)i+(2z-3x^2)j+(3x^2-3y^2)k?

I made a mistake. u should equal (x^3)-(y^3)+(z^2). That's why I have a negative j component.

If I let F equal the above, isn't this correct?

 

[Dick]

You are integrating F. Not curl(F). So you can't take it to a line integral.

 

[Dick]

See my post to your latest question. grad(u)xgrad(v) is the curl of something. What?