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How to use Mathematica to compute a circle integration?

Posted 10 years ago

Hi, I am a newbie in this software. I want to compute the circle integration.

Given:

A = {3y, -xz, yz^2}
S: x^2+y^2 = 2z
C: z=2
Circle Integration A belongs to S and C.

I used those command:

ss[x_^2+y_^2] := 2*z
cc[z_]:=2
A={3*y, -x*z, y*z^2}
Integrate[A,{x,y,z} \[Element] ss, {x,y,z} \[Element] cc]

But I got a error: integrate:ilim: invalid integartion variable or limit(s) in {x,y,z} [Element] ss, How to compute this, thanks a lot!

POSTED BY: MOP MOP
6 Replies

Your region reg is a circle, not a surface. Your vector n is normal to the circle, but not to the surface. What you are calculating is not a surface integral.

I would do the surface integral like this:

surf = ImplicitRegion[{x^2 + y^2 == 2*z, z <= 2}, {x, y, z}];
n = Normalize[D[x^2 + y^2 - 2*z, {{x, y, z}}]] /. Abs -> Identity;
Integrate[
 Dot[Curl[{3*y, -x*z, y*z^2}, {x, y, z}], n], {x, y, z} \[Element] 
  surf]

The sign of the final result depends on the choice of the orientation of the surface, i.e., on the choice between n and -n.

POSTED BY: Gianluca Gorni
Posted 10 years ago

enter image description here

The left side of the equation, I could use Curl to compute now, as following:

n = {0, 0, 1}
s = x^2 + y^2 == 2*z
c = z == 2
a = {3*y, -x*z, y*z^2}
reg = ImplicitRegion[{s, c}, {x, y, z}]
Integrate[Dot[Curl[a, {x, y, z}], n], {x, y, z} \[Element] reg]

enter image description here

Give the right answer, -20*pi, but how about the right side of the equation

POSTED BY: MOP MOP
POSTED BY: Gianluca Gorni
POSTED BY: Daniel Lichtblau
Posted 10 years ago

A is a Vector field S is a surface C is a circle intersection on the surface S

It use stroke's theorem to compute the flux, so the answer is -20*pi, But how can I use Mathematica to compute?

POSTED BY: MOP MOP

You can use ImplicitRegion:

reg = ImplicitRegion[{x^2 + y^2 == 2 z, z == 2}, {x, y, z}]
Integrate[{3 y, -xz, yz^2}, {x, y, z} \[Element] reg]

There are different kinds of integrals along lines. The code above gives the integral with respect to the (absolute) arc length, I think. This may not be what you meant.

POSTED BY: Gianluca Gorni
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