- #1

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## Homework Statement

Given is the vector field, [itex]\overline{A}[/itex] = (x

^{2}-y

^{2}, (x+y)

^{2}, (x-y)

^{2}). The surface: [itex]\overline{B}[/itex] = (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 = ∫∫[itex]\overline{A}[/itex][itex]\cdot[/itex][itex]\overline{n}[/itex]dS

## Homework Equations

## 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?