Using Stokes Theorem to $\int_{L}^{} y dx + z dy + x dx$

In summary: Thanks for the help.In summary, the use of Stokes Theorem is demonstrated to solve an integral involving a circle and a sphere, with the final answer being \sqrt{3}\pi a^2. The normal vector and curl of the given function are calculated, and the surface integral is evaluated using the appropriate parameters and coordinate system. The final answer is confirmed to be correct.
  • #1
linearfish
25
0

Homework Statement


Use Stokes Theorem to compute
[tex]\int_{L}^{} y dx + z dy + x dx[/tex]
where L is the circle x2 + y2 + z2 = a2, x + y + z = 0

The Attempt at a Solution


I feel like this problem shouldn't be that hard but I can't get the right answer: (pi)a2/3.

I calculated the curl of F as: -(i + j + k)
and the normal vector as:
[tex]\frac{i + j + k}{\sqrt{3}}[/tex]

So:
[tex]\int_{L}^{} y dx + z dy + x dx = \int \int -(i + j + k) \cdot (\frac{i + j + k}{\sqrt{3}}) ds = -\frac{3}{\sqrt{3}} \int \int ds[/tex]

Here's where I'm stuck. I think the integral should just be the area of the circle (pi*a2) but maybe I'm thinking about it wrong. Thanks.
 
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  • #2
Is this integral supposed to be



[tex]
\int_{L}^{} y dx + z dy + x dz
[/tex] ?

[last variable of integration being z, not x?]

I think you'll find that, although the plane goes through the origin, it does not intersect the sphere at an equator, so the area of the circle is not [tex]\pi \cdot a^2[/tex].

EDIT: Here's a way to see that. Write out the equation
x + y + z = 0 using spherical coordinates. You can divide through by r, since we know that isn't zero on the circle of intersection. You now have

[tex]sin \theta cos \phi + sin \theta sin \phi + cos \theta = 0 [/tex]

Make a convenient choice of the azimuthal angle [tex]\phi[/tex], say, 0º. You now have an equation in [tex]\theta[/tex] only, and can see immediately that the polar angle (or co-latitude) can't be 90º, so the intersection circle is not an equator of the sphere. Where must it be and what is the circle's area there?
 
Last edited:
  • #3
Is this integral supposed to be
[tex]\int_{L}^{} y dx + z dy + x dz[/tex]

[last variable of integration being z, not x?]

Yes, thanks.

I'm still confused, though. I thought that if a plane cut through the center of a sphere then it was a great circle, hence its area was a quarter of the surface area of the sphere.
 
  • #4
linearfish said:
Yes, thanks.

I'm still confused, though. I thought that if a plane cut through the center of a sphere then it was a great circle, hence its area was a quarter of the surface area of the sphere.

You're right linearfish. Because the sphere's center is at the origin and the plane intersects the origin, the intersection must, necessarily, be a great circle.

EDIT: Of course, it is not the "equator" of the sphere (in the xy-plane).
 
  • #5
Okay, I get that. So is my calculation off somewhere else?
 
  • #6
Consider: what is the surface? Is it the interior of the circle? Or is it the half-sphere? Both surfaces have the same boundary.
 
  • #7
cellotim said:
You're right linearfish. Because the sphere's center is at the origin and the plane intersects the origin, the intersection must, necessarily, be a great circle.

EDIT: Of course, it is not the "equator" of the sphere (in the xy-plane).

Sorry, yes, I really should have checked my development of that thought with more points on the sphere. So ignore post #2 as the ravings of a sleep-deprived person...

The problem is with n ds , which is normal to the surface of the hemisphere bounded by the inclined great circle, and so isn't [tex]
\frac{i + j + k}{\sqrt{3}}[/tex].
 
  • #8
When one sleep deprived person is trying to converse with another, chaos is bound to ensue. I appreciate the help from both of you. I'll admit though that I'm still at a loss. I know what the normal should be so that the problem works out but I don't know how to get that.

(I will also admit that I took vector calculus 6 years ago and I'm not doing such a great job of reteaching myself).
 
  • #9
Sorry, Stokes theorem will give the same answer irrespective of the surface. One problem is that the normal is probably negative. Besides that you've the done it correctly. To see why, rotate the space so that [tex]\nabla\times \vec{F} = -\sqrt{3}\vec{k}[/tex]. Now the circle lies in the xy plane. Parameterize the surface in the new coordinate system: [tex]\vec{\phi} = (rcos\theta,rsin\theta,0)[/tex]. If you carry out the surface integral explicitly, you get the same answer, [tex]\sqrt{3}\pi a^2[/tex].
 
  • #10
That makes sense. It's not the answer in my notes but maybe I wrote it down wrong.
 

1. What is Stokes Theorem?

Stokes Theorem is a fundamental theorem in vector calculus that relates the line integral of a vector field over a closed curve to the double integral of the curl of the vector field over the surface bounded by the curve. It is used to evaluate integrals that are difficult to solve using traditional methods.

2. How is Stokes Theorem used to solve integrals?

Stokes Theorem allows us to convert a difficult line integral into a simpler double integral over a surface. This makes it easier to evaluate integrals that involve vector fields, such as the one given in the question.

3. What is the significance of the vector field in Stokes Theorem?

The vector field represents a physical quantity, such as velocity or force, that can vary at different points in space. In Stokes Theorem, the vector field is used to calculate the work done along a closed curve or the flux through a closed surface.

4. Can Stokes Theorem be applied to any type of curve or surface?

Stokes Theorem is applicable to any smooth curve and surface, as long as the vector field is differentiable along the curve and the surface is orientable. It is a generalization of the more commonly known Green's Theorem, which only applies to plane regions.

5. What other theorems are related to Stokes Theorem?

Stokes Theorem is closely related to other fundamental theorems in vector calculus, such as the Divergence Theorem and the Fundamental Theorem of Calculus. It is also related to other theorems in mathematics, such as the Cauchy-Riemann equations in complex analysis and the Poincaré-Hopf theorem in topology.

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