# Thread: Vector Surface Integral

1. ## Vector Surface Integral

I have this vector-field:

F(x; y; z) = (x*exp(xy)cos(exp(xy)) ; y*exp(xy)cos(exp(xy)) ; x^2 + y^2)

And I'm supposed to take the surface integral over half a torus (over the positive z-axis).

Now, my book says that for this surface integral, I won't have to do it over the torus,
if I take it over its projection on the xy-plane (an annlus), it will give the same answer.

So, my question is: how is that the same thing? I figured maybe the vector field didn't have any z-component or something, but obviously it does. My book does not give an explanation, so I thought I'd ask here.

Thanks!

2. ## Re: Vector Surface Integral

Your vector field has a z component, but it doesn't depend on z. So F is the same for z=0 as z=whatever on the surface of the torus. I think there's a factor for the area that changes, too, but it ends up being the same thing. If you break up the integral into components, you should see the z terms drop out.

If you still have trouble, post your integral here and we'll work through it.

- Hollywood

3. ## Re: Vector Surface Integral

Thank you for your answer! I think I might get it now. The torus is symmetrical above the xy-plane. So every vector component not in the direction of the z-axis will cancel each other out, and we are left with just the vertical components. Now, the length of the semi-circle is longer than just a line across the annulus, but when we are taking the dot product of the vertical components with the normal vector to the semi-circle, it should compensate for that.

Not a very rigorous explanation, but is this basically correct?

Another thing I might mention, is that the exercise before this one, was to show that this vector field is the curl of the vector field:

G = (0 ; xy^2 + (1/3)*x^3 ; sin(exp(xy)))

So perhaps the easiest way would be to calculate two line integrals over the inner and outer boundaries of the annulus using the divergence theorem?

Thanks again!

4. ## Re: Vector Surface Integral

Originally Posted by gralla55
Now, the length of the semi-circle is longer than just a line across the annulus, but when we are taking the dot product of the vertical components with the normal vector to the semi-circle, it should compensate for that.
That is correct.

Originally Posted by gralla55
The torus is symmetrical above the xy-plane. So every vector component not in the direction of the z-axis will cancel each other out, and we are left with just the vertical components.
I don't think this is correct. What I was trying to say is that the vector field at (x,y,z) is the same as at (x,y,0). That, together with your observation about the dot product, means you can integrate over the annulus instead of the half-torus.

You have the integral of the curl of a vector field, so I think you might want Stoke's Theorem.

- Hollywood

5. ## Re: Vector Surface Integral

You're right. The vector field is of course not symmetrical in the middle of the torus, so my second statement is false.

But anyway, stokes theorem is what I meant with the line integral-thing. However, for it to work, I need a continuous curve along the boundary of the surface.

I could make one path where I first trace the inner circle, move to the outer circle, and then trace the outer circle. But in that case I need to break the path into 3 pieces.

Will it not also work if I take the line integral around the outer boundary, minus the line integral around the inner boundary? Both in counterclockwise directions I think.

Thank you so much for your help!

6. ## Re: Vector Surface Integral

Yes, that will work.

- Hollywood