Trouble with surface integral

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Homework Statement


Given is the vector field, [itex]\overline{A}[/itex] = (x2-y2, (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?
 

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  • #2
LCKurtz
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Homework Statement


Given is the vector field, [itex]\overline{A}[/itex] = (x2-y2, (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.

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

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?

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
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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
vela
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I think the problem intends for you to take ##-1 \le u \le 1## and ##-1 \le v \le 1##.
 
  • #5
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I think the problem intends for you to take ##-1 \le u \le 1## and ##-1 \le v \le 1##.

Ahhh, you're right!

I thought the comma was there to separate.

Thanks!
 

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