# Trouble with surface integral

1. Mar 27, 2013

### Log

1. The problem statement, all variables and given/known data
Given is the vector field, $\overline{A}$ = (x2-y2, (x+y)2, (x-y)2). The surface: $\overline{B}$ = (u+v, u-v, uv). The restrictions are the following: -1≤ u, v≤ 1, and the z-component of the normal has to be positive.
Calculate I, I = ∫∫$\overline{A}$$\cdot$$\overline{n}$dS

2. Relevant equations

3. The attempt at a solution
What I did was I first tried calculating the normal by using the cross product between the two vectors tangent to the surface. This however got me a normal vector with a z component equal to zero. Then I tried expressing u and v in terms of x and y, plugging that into the parameter for z, z = uv, got me z as a function of x and y. I then defined a potential, Φ = z - f(x,y), where z = f(x,y). Calculating n=∇Φ got me a normal vector with non-zero z-component. Easy.

Next I calculated the dot product between the vector field, A, and the normal, n, in terms of u and v. This is where I got stuck. I'm not sure how to calculate the integral given the boundaries above.

-1≤ u, v≤ 1 is the same as -1≤ u< ∞, -∞< v ≤ 1. I tried integrating with respect to one of the variables but what happens with the infinites when evaluating the result?

Maybe I missed something, my calculus skills are a little bit rusty at the moment.

Oh, I was wondering, how come the normal vectors calculated from the cross product and gradient are different? aren't they supposed to be the same?

2. Mar 27, 2013

### LCKurtz

Show us your steps. I don't get zero in the 3rd component.

You want to leave it all in terms of u and v. And use the formula$$\iint_S \vec F\cdot \hat n dS = \pm\iint_{(u,v)}\vec F(u,v)\cdot \vec R_u\times \vec R_v\, dudv$$using the appropriate sign to agree with your orientation (plus if $\vec R_u\times \vec R_v$ is in the right direction).

3. Mar 29, 2013

### Log

I calculated the normal again and got it to be:
n = (-(u + v), u - v, 2).
This has a z-component > 0.

Then I expressed A in terms of u and v:
A = (uv, u2, v2).

The dot product of A and n is:
4(2v2 + u3 - 2u2v - uv2)

I can't see how the integral of this is going to be finite, given the boundaries.

4. Mar 29, 2013

### vela

Staff Emeritus
I think the problem intends for you to take $-1 \le u \le 1$ and $-1 \le v \le 1$.

5. Mar 29, 2013

### Log

Ahhh, you're right!

I thought the comma was there to separate.

Thanks!