Recent content by hughb

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

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?

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.

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...

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...

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...

Homework Statement 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...