Applying Stokes' Theorem to the part of a Sphere Above a Plane

  • #1

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.
 
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Answers and Replies

  • #2
Office_Shredder
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It feels like you're forgetting to include the Jacobian determinant when you reparametrize these integrals, which is why they're all evaluating to random numbers.
 
  • #3
It feels like you're forgetting to include the Jacobian determinant when you reparametrize these integrals, which is why they're all evaluating to random numbers.
Thanks for pointing that out. Although with line integrals, do we still have to use the Jacobian? In the final three options I was trying to use Stokes' Theorem and solve something of the form ∫c F⋅ds. It feels like there should be a good method to use a line integral, but I don't know if any of those are fruitful.

For the first option, I could see using a Jacobian if I'd like to solve the double integral straight up. For ∫∫s -6 dA, that would then be ∫∫s -6 r*drdθ. I would presume the limits being r: (0,1), θ: (0, 2π). But that would just get me -6π, which is still not quite right. It seems I would need my r bound to be sqrt(2)/3^0.25, so my guess is that's not the r I'm looking for given the complexity.
 
  • #4
Office_Shredder
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I think you still need the Jacobian determinant for line integrals. The Jacobian describes how one unit of measure changes between your new coordinates and your old coordinates. In idea 2, you were supposed to be integrating over a circle with radius sqrt (6)/3 ( double check this before you do anything crazy with that number) and replaced it with a triangle whose sides have length sqrt (2). So the shape you're integrating over went from having a perimeter of 6/9*pi to 3sqrt(2). And this is why you lost the pi you think you're supposed to have in your final answer.

I think you should try to parameterize exactly the circle on that plane directly to at least get a little practice with the geometry. What is the center of the circle? You can write out any circle as center+cos(theta)*radial vector + sin(theta)*orthogonal radial vector

I think your Jacobian for part 1 is still missing the point. When you project a hemisphere onto a disc in a plane, if you take an area of one unit of measure near the top of the hemisphere it will get projected into an area of about one unit of measure on the plane. But near the equator the hemisphere is almost orthogonal to the plane, so when you project a unit of measure down, you get a teeny tiny sliver on the plane. That's what the Jacobian determinant is for, to make sure you count that tiny sliver as the full unit of measure it was before the projection.

When you try to evaluate that integral in the plane, if you switch to polar coordinates to do so you will need a second Jacobian determinant.
 
  • #5
LCKurtz
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I think you have picked a random very tricky problem. First of all, the problem is stated poorly. I presume they are talking about the portion of the sphere above the plane. Note that the circular intersection of the plane and sphere is not a great circle and does not stay in the first octant. When you solve for the xy projection of that circle, you might recognize that equation as an ellipse whose center is not at the origin and which has been rotated. If you hope to use polar type coordinates to get a tractable integral you are going to have to translate and rotate in x and y axes to get a standard form like ##\frac {u^2}{a^2} +\frac {v^2}{b^2} = 1## which you could parameterize as ##u = a\cos t, ~v = b\sin t##. I don't see how to avoid that for either the line or surface integral. You are going to have to want to solve it more than I do. Good luck with it.
 
  • #6
pasmith
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I think it's much simpler than that.

The equation [tex]ax + by + cz = d[/tex] can be expressed as [tex]
\mathbf{r} \cdot \mathbf{n} = \frac{d}{\sqrt{a^2 + b^2 + c^2}}[/tex] where [tex]\mathbf{n} = \frac 1{\sqrt{a^2 + b^2 + c^2}} \begin{pmatrix} a \\ b \\ c \end{pmatrix}[/tex] is the unit normal. It follows that the plane defined by this equation can be parametrized as [tex]
\mathbf{r}(X,Y) = \frac{d}{\sqrt{a^2 + b^2 + c^2}}\mathbf{n} + X\mathbf{v}_1 + Y\mathbf{v}_2[/tex] where [itex]\mathbf{v}_1[/itex] and [itex]\mathbf{v}_2[/itex] are orthogonal unit vectors lying in the plane such that [itex]\mathbf{v}_1 \times \mathbf{v}_2 = \mathbf{n}[/itex].

It follows that [tex]\|\mathbf{r}(X,Y)\|^2 = \frac{d^2}{a^2 + b^2 + c^2} + X^2 + Y^2[/tex] and as [itex]\mathbf{v}_1[/itex], [itex]\mathbf{v_2}[/itex] and [itex]\mathbf{n}[/itex] form a right-handed triple we have [tex]
\mathbf{v_1} \times \mathbf{n} = - \mathbf{v}_2, \qquad \mathbf{v}_2 \times \mathbf{n} = \mathbf{v}_1.[/tex]
 
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  • #7
LCKurtz
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Well, yes, but, playing Devil's advocate for the OP, assuming he knows how to make vectors ##\vec v_1,~\vec v_2##, the real problem is setting up the appropriate line and surface integrals with correct limits.
 
  • #8
vela
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I'd change to a rotated set of coordinates ##(x',y',z')## so that the vector ##(x,y,z)=(1,1,1)## now points along the ##z'##-axis, becoming ##(x',y',z')=(0,0,\sqrt{3})##.

In terms of the new coordinates, the equation of the sphere will be ##x'^2 + y'^2 + z'^2=1##, and it's straightforward to show that the equation of the plane is ##z' = 1/\sqrt 3##. It's much easier to evaluate the integrals then.
 
  • #9
Thanks everyone who contributed here. It's all very helpful. I think I'm seeing what I missed the first time around (parameterization was backward, logically) and am playing with a few of the suggestions. But as some have said, I think I may have picked a tricky problem (though interesting!).
 
  • #10
LCKurtz
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I'd change to a rotated set of coordinates ##(x',y',z')## so that the vector ##(x,y,z)=(1,1,1)## now points along the ##z'##-axis, becoming ##(x',y',z')=(0,0,\sqrt{3})##.

In terms of the new coordinates, the equation of the sphere will be ##x'^2 + y'^2 + z'^2=1##, and it's straightforward to show that the equation of the plane is ##z' = 1/\sqrt 3##. It's much easier to evaluate the integrals then.
I think you may be helping me make my point. No third semester calculus course I ever taught covered 3d rotations. I wouldn't expect a typical 3rd semester calculus student to know how to find the equations for your ##x',~y',~z'##.
 
  • #11
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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

F= r x (i + j+ k) = (y-z, z-x, x-y)
What is the ##\nabla \cdot \mathbf F## ?
Might that suggest something easier ?
 
  • #12
What is the ##\nabla \cdot \mathbf F## ?
Might that suggest something easier ?
The divergence would be zero, which probably indicates something meaningful! I know the divergence of the curl must be zero, and that a curl of zero means that F is conservative. If it were the opposite (curl zero, divergence non-zero), I would use the concept that f=∫c F⋅ds = f(b)-f(a) for conservative forces.

Zero divergence would indicate the fluid/gas is not expanding or being compressed, but I'm not sure what that suggests here.
 
  • #13
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Well it means the flow into any closed surface equals the flow out (net flux zero) . So you can probably calculate on an easy part of the surface.
 
  • #14
vela
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What is the ##\nabla \cdot \mathbf F## ?
Did you mean ##\mathbf F## here or ##\nabla \times \mathbf F## since that's what appears in the surface integral?
 
  • #15
pasmith
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I think you may be helping me make my point. No third semester calculus course I ever taught covered 3d rotations. I wouldn't expect a typical 3rd semester calculus student to know how to find the equations for your ##x',~y',~z'##.
Would they be expected to realize that the problem is equivalent to integrating [itex]\nabla \times ( \mathbf{r} \times \sqrt{3} \mathbf{k})[/itex] over the part of the unit sphere above the plane [itex]z = 1/\sqrt{3}[/itex]?

Inh niether my nor vela's methods is it actually necessary to find the new basis vectors; one just has to know that they exist, form a right-handed triple, and one of them is the normal.
 
  • #16
vela
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I think you may be helping me make my point. No third semester calculus course I ever taught covered 3d rotations. I wouldn't expect a typical 3rd semester calculus student to know how to find the equations for your ##x',~y',~z'##.
Fair point. I do think, however, the intent of the problem is probably to get students to picture what's going on and to realize the problem can be solved relatively simply conceptually. If one tries to brute-force it, it becomes a bit of mess because the symmetry in the problem is obscured by the choice of coordinates.
 
  • #17
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Did you mean F here or ∇×F since that's what appears in the surface integral?
Oops yeah I missed that. Need to look at this again later. Thanks for the help.
 
  • #18
Charles Link
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I like the use of the rotation of the coordinates and/or of the symmetry. The student should know the formula for the distance ##s ## from a point ##(x_o, y_o, z_o) ## to a plane ## ax+by+cz+d=0 ## is ## s=\frac{|ax_o+by_o+cz_o+d|}{\sqrt{a^2+b^2+c^2}} ##. They should also know that ## |\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin{\theta} ##. This one is easier than it first appears.
 
  • #19
LCKurtz
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Let me add that when I read the OP's title and post, I assumed he wanted to verify Stoke's theorem, i.e., work both sides and show they are equal. If you can use Stoke's theorem, there is indeed an easy method to evaluate the flux integral, appropriate for a 3rd semester calculus class, which I will agree was likely the intent of the problem. Some of the above posts have alluded to it. I will just outline it here for the OP:
1. Because the sphere and the plane have the circle as a common boundary, the flux integral through the sphere has the same value as through the plane surface that is cut off. (They are both equal to the corresponding line integral by Stoke's theorem.)
2. As the OP noted, ##\nabla \times \vec F = \langle -2,-2,-2\rangle ##.
3. For the plane ##\hat n =\frac 1 {\sqrt 3}\langle 1,1,1\rangle## so ##\nabla \times \vec F \cdot \hat n = -\frac 6 {\sqrt 3}##.
4. So the flux integral for the plane region P is ##\iint_P -\frac 6 {\sqrt 3}~dS = -\frac 6 {\sqrt 3}\text{Area}(P)##.
5. Observe both the sphere and the plane pass through the three points ##(1,0,0), (0,1,0), (0,0,1)##. Figure out the center, hence the radius and area of the circle ##P## and you are done.
 
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  • #20
Charles Link
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For "5" above, alternatively you can use the distance from the origin (0,0,0) to the plane x+y+z=1, along with the Pythagorean theorem to figure out the radius of the circle.
 
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  • #21
vela
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Here are parametric equations for the circle of the intersection:
\begin{align*}
x(t) &= \frac 13 ( 1 + \cos t + \sqrt{3} \sin t) \\
y(t) &= \frac 13 ( 1 + \cos t - \sqrt{3} \sin t) \\
z(t) &= \frac 13 ( 1 - 2\cos t)
\end{align*}
 
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  • #22
Charles Link
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@vela I found your expressions interesting. Because of the 3-fold symmetry axis for the cube, (along ## \vec{v}=\hat{i}+\hat{j}+\hat{k} ##), and the corresponding coordinate axes having a cubic geometry, using a trigonometric identity, one can get from your expressions the following:
##x(t)=\frac{1}{3}(1-2 \cos(t+2 \pi/3)) ##, and ## y(t)=\frac{1}{3}(1-2 \cos(t-2 \pi/3)) ##, with ## z(t)=\frac{1}{3}(1-2 \cos(t)) ##.
Edit: To add to this, the ## z(t) ## is centered at ## z=\frac{1}{3} ##,(##x=y=z ## with ## x+y+z=1 ##),with ## r=\sqrt{2/3} ##, and the cosine of the angle that the plane (going up in the steepest direction w.r.t. z) makes with the ## z ## axis is also ## \sqrt{2/3} ##, making for the ##2/3 ## factor on ## \cos{t} ##.
 
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