Stokes theorem over a hemisphere

1. Apr 25, 2012

hughb

1. The problem statement, all variables and given/known data
The vector ﬁeld F is deﬁned in 3-D Cartesian space as
F = y(z^2−a^2)i + x(a^2− z^2)j,
where i and j are unit vectors in the x and y directions respectively, and a is a
real constant.
Evaluate the integral

Integral:(∇ ×F)·dS, where S is the open surface of the hemisphere
x^2+ y^2+ z^2= a^2, z ≥ 0 :
(i) by direct integration over the surface S
(ii) using Stokes’ theorem

2. Relevant equations
Stokes Theorem

3. The attempt at a solution
I've calculated the integral over the surface and the line integral around the boundary curve, and both answers are 2*pi*a^4. However, my problem is that when i try to use stokes theorem to switch the integral from the surface of the hemisphere to the surface of the circle on the bottom of the hemisphere, via stokes theorem, my answer is half the correct answer.

(∇ ×F).dS = 2a^2 (as z = 0 on circle). 2a^2 = 2x^2+2y^2.

The integral of 2x^2+2y^2 over the surface of the circle is 2*pi*integral(r^3 dr) between 0 and a, which just comes out as pi*a^4.

I'm not sure where i'm going wrong here, maybe the integrand should be a^2 r dr?? But the field varies over the surface of the circle so i don't think this is right??

2. Apr 25, 2012

HallsofIvy

Staff Emeritus
I don't understand what you mean by "However, my problem is that when i try to use stokes theorem to switch the integral from the surface of the hemisphere to the surface of the circle on the bottom of the hemisphere, via stokes theorem, my answer is half the correct answer." The given surface is "the open surface of the hemisphere". There is no "bottom" to be included. And when you say "my answer is half the correct answer", what is the "correct answer".

"I've calculated the integral over the surface and the line integral around the boundary curve, and both answers are 2*pi*a^4." Isn't that all you are asked to do?

3. Apr 25, 2012

hughb

Ok, so I can answer the question, that's fine. But for peace of mind, i tried to check my solutions by switching the integral of the curl of F from the surface of the hemisphere, to the circlular base of the hemisphere as allowed by Stokes theorem. The two integrals should be the same, however, I can't get them to match up.

4. Apr 25, 2012

Dick

As you correctly got 2a^2 for the function you want to integrate over the disk, why don't you just use that? Don't change it to 2x^2+2y^2. That not correct.

5. Apr 25, 2012

hughb

But 2a^2 is the same as 2x^2+2y^2, but the integral over the area of the circle isn't, if i stick with the 2a^2 and integrate that, then i wouldn't be integrating the field over the disk properly would I? As the field varies over the disk, surely I can't just integrate the constant 2a^2 over the disk, but have to switch to 2x^2+2y^2?

6. Apr 25, 2012

Dick

2a^2=2x^2+2y^2+2z^2 is only true on the surface of the hemisphere. It's not true inside of the hemisphere where the surface of the disk is. And in fact it's really easy to integrate 2a^2 over the surface of the disk. It's a constant. Just factor it out and multiply by the area of the disk, pi*a^2.

7. Apr 25, 2012

hughb

Ok, if they aren't the same thing, then when I calculate the direct integral of the curl of F over the hemisphere via projection onto the xy plane, my integral is: 2*pi∫2x^2+2y^2+2a^2 dxdy between 0 and a. The way i calculated this was to let 2a^2 = 2x^2+2y^2 and integrate, giving the correct answer, however, is this a valid method, to replace the 2a^2 this time, but not when i calculate the integral over the disk? If i don't replace the 2a^2, i get the wrong answer.

8. Apr 25, 2012

Dick

When you are doing the line integral around the boundary circle, you can use that a^2=x^2+y^2. But that's the only place that it's true. It's not true in the disk and even on the hemisphere it's only true if z=0.

9. Apr 25, 2012

hughb

But this means that the line integral and the integral of the curl over the surface give different results, 2*pi*a^4 for the line integral and 3*pi*a^4 for the surface integral.

10. Apr 25, 2012

Dick

The quantity I get to integrate dxdy is 2x^2+2y^2+2a^2-2z^2. That reduces to 4x^2+4y^2 which gives the the correct answer. You are right that 2x^2+2y^2+2a^2 would not. Can you show how you got to that point?

11. Apr 25, 2012

hughb

Aha, I think i see where i'm going wrong, I had what you had for the integral, but then assumed z^2=0 as i was projecting onto the plane, is this not the case then? Even if the integral is on the xy plane, does the z component still have to be accounted for?

12. Apr 25, 2012

Dick

You are integrating over a region in the xy plane, but the quantity you are integrating is evaluated on the hemisphere, where z is not zero.

13. Apr 25, 2012

hughb

Brilliant, I see it, thank you so much - that had been bothering me for days!!