- #1

Rippling Hysteresis

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- Homework Statement
- Evaluate the integral ∫∫s (∇xF)⋅dS where S is the portion of the surface of a sphere defined by x^2+y^2+z^2=1 and x+y+z>1, and where F=r x (i + j + k), r= xi + yj + zk.

- Relevant Equations
- ∫∫s (∇xF)*dS = ∫c F⋅ds

I've tried a few ways of solving this, both directly and by using Stokes' Theorem. I may be messing up what the surface is in the first place

F= r x (i + j+ k) = (y-z, z-x, x-y)

Idea 1: Solve directly. So ∇ x F = (-2,-2,-2). I was unsure on which surface I could use for the normal vector, but considered both the plane (1,1,1), as well as it's projection on the x-y plane (0,0,1). The latter is probably a poor choice, but so often we seem to use Stokes' to project a hemisphere or whatever onto x^2+y^2=1. Anyway, if I use the (1,1,1), I get ∫∫s -6 dA = -3√3. If I use the obviously wrong (0,0,1) it's just ∫∫s -2 dA = -√3

Idea 2: Use the path on the plane from (1,0,0) to (0,1,0) to (0,0,1). So the first part of the path parameterizes to c(t) = (1-t, t, 0), and c'(t) = (-1,1,0). F under this parameterization is (t, t-1, 1-2t), and the domain of c is (0,1). ∫c F⋅ds = (t, t-1, 1-2t)⋅(-1,1,0)= ∫c -1 dt = -1. It's evident there won't be any π terms in any of the paths, so I abandoned this pursuit.

Idea 3: Use find the intersection of the plane with the sphere by substitution. But I get a convoluted mess with x^2+y^2-x-y+xy=0. I could use x=cost, y=sint to make it 1-cost - sint +cost*sint. But I'm not sure if I have to get a precise path, and if so, this doesn't generalize what that is. Either way, I don't see how that gives me the solution of -4π/√3 with how I'd set it up.

Idea 4: Use the sphere to parameterize the plane and take the line integral. I'm thinking this may work since the intersection will be on the sphere, so the radius is always 2, but practically I'm not sure how to handle is the fact that I have two variables, whereas line integrals usually parameterize in terms of just one.

x=2cosθsinΦ

y=2sinθsinΦ

z=2cosΦ

So x+y+z=1 becomes 2cosθsinΦ + 2sinθsinΦ +2cosΦ=1. Also, any manipulation I do to solve for one in terms of the other seems fruitless.

Am I anywhere on the right track for any of these? I think my surface is all messed up and I am lacking intuition on when I can just simply lay it flat in a plane of my choosing. I'm trying to understand generally how to set-up these problems and when I can do what. This is just a problem I picked out of the book that stumped me, compared to all the others which seemed more straight-forward.

F= r x (i + j+ k) = (y-z, z-x, x-y)

Idea 1: Solve directly. So ∇ x F = (-2,-2,-2). I was unsure on which surface I could use for the normal vector, but considered both the plane (1,1,1), as well as it's projection on the x-y plane (0,0,1). The latter is probably a poor choice, but so often we seem to use Stokes' to project a hemisphere or whatever onto x^2+y^2=1. Anyway, if I use the (1,1,1), I get ∫∫s -6 dA = -3√3. If I use the obviously wrong (0,0,1) it's just ∫∫s -2 dA = -√3

Idea 2: Use the path on the plane from (1,0,0) to (0,1,0) to (0,0,1). So the first part of the path parameterizes to c(t) = (1-t, t, 0), and c'(t) = (-1,1,0). F under this parameterization is (t, t-1, 1-2t), and the domain of c is (0,1). ∫c F⋅ds = (t, t-1, 1-2t)⋅(-1,1,0)= ∫c -1 dt = -1. It's evident there won't be any π terms in any of the paths, so I abandoned this pursuit.

Idea 3: Use find the intersection of the plane with the sphere by substitution. But I get a convoluted mess with x^2+y^2-x-y+xy=0. I could use x=cost, y=sint to make it 1-cost - sint +cost*sint. But I'm not sure if I have to get a precise path, and if so, this doesn't generalize what that is. Either way, I don't see how that gives me the solution of -4π/√3 with how I'd set it up.

Idea 4: Use the sphere to parameterize the plane and take the line integral. I'm thinking this may work since the intersection will be on the sphere, so the radius is always 2, but practically I'm not sure how to handle is the fact that I have two variables, whereas line integrals usually parameterize in terms of just one.

x=2cosθsinΦ

y=2sinθsinΦ

z=2cosΦ

So x+y+z=1 becomes 2cosθsinΦ + 2sinθsinΦ +2cosΦ=1. Also, any manipulation I do to solve for one in terms of the other seems fruitless.

Am I anywhere on the right track for any of these? I think my surface is all messed up and I am lacking intuition on when I can just simply lay it flat in a plane of my choosing. I'm trying to understand generally how to set-up these problems and when I can do what. This is just a problem I picked out of the book that stumped me, compared to all the others which seemed more straight-forward.

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