Element of surface area in spherical coordinates

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Homework Statement:

In spherical polar coordinates, the element of volume for a body that is symmetrical about the polar axis is,
$$\begin{equation}
dV=2\pi sin{\theta} dr d\theta
\end{equation}$$
Whilst its element of surface area is,
$$\begin{equation}
dS=2\pi rsin{\theta} \sqrt{dr^2+r^2d\theta^2}
\end{equation}$$

Although the homework statement continues, my question is actually about how the expression for dS given in the problem statement was arrived at in the first place.

Relevant Equations:

This question is at the end of a chapter covering multiple integrals, including a change of variables in multiple integrals where, for a change of variables from Cartesian coordinates to spherical polar coordinates, we have,
$$\begin{equation}
dxdydz = Det[J] dr d\phi d\theta = r^2 sin\theta dr d\phi d\theta
\end{equation}$$
Where ## Det[J]## is the determinant of the 3x3 Jacobian Matrix containing all the partial derivatives of ##x,y,z## with respect to ##r,θ,ϕ##
r,θ,ϕ
For integration over the ##x y plane## the area element in polar coordinates is obviously ##r d \phi dr ## I can also easily see ,geometrically, how an area element on a sphere is ##r^2 sin\theta d\phi ## And I can verify these two cases with the Jacobian matrix. So that's where I'm at. That's what I know. However, trying to take the determinant of the Jacobian in this case obviously does not get me the expression given in the problem statement and I'm not sure where to begin. Thank you for your help. I'm not actually in school but looking to go back for Physics and don't have any teacher to turn to.
 

Answers and Replies

  • #2
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Just realized I forgot the ##d\theta## in the area element on a sphere ##(dS=r^2sin\theta d\phi d\theta) ##
 
  • #4
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Doesn't it depend on the surface? If you're integrating over x-y plane its just dxdy
 
  • #5
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I don't see how that helps me though
 
  • #6
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Last edited:
  • #7
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My question is actually about how the expression for dS given in the problem statement (##dS=2πrsinθ\sqrt{dr^2+r^2dθ^2}##) was arrived at in the first place, given the equations stated above (which was all that we were assumed to know as of attempting this problem). The only example problems up to this point involved computing Jacobians to make a change of variables in multiple integrals. A basic example of such a transformation would be how the volume element in cartesian coordinates ##dxdydz## transforms to ##r^2 sin\theta dr d\theta d\phi## where ##r^2 sin\theta## is the determinant of the Jacobian matrix. The area element on a sphere can be found by similar means, but can also be arrived at geometrically: ##dS=r^2 sin\theta d\theta d\phi## But this type of procedure would never lead me to the expression given in the problem statement. I can see how, for a problem with symmetry about the polar axis (like this one) the phi part of the integral would just integrate to ##2\pi ## (the leading factor in the given area element expression). So I expected something of the form ##2 \pi (something) dr d\theta##. Since the surface I'm integrating over may not necessarily be on a sphere, the dr makes an appearance this time. ##2\pi (something) dr d\theta## needs to be an area. An area is a displacement time a displacement. ##dr## is a displacement itself (its scale factor is one). To convert the infinitesimal change in ##\theta## to a displacement, it needs to be multiplied by a scale factor. Well, I thought, multiplying by r would do that. So my conclusion is that an area element (for an arbitrary body with symmetry with respect to ##\phi## which leads to the ##2\pi## factor) should be ##2\pi r d\theta dr ##. On comparison with the area element in the problem statement, this leads me to the conclusion that ##d\theta dr = sinθ \sqrt{dr^2+r^2 dθ^2}## which I cannot reconcile and that's annoying because I know somewhere I'm being stupid and my reasoning is faulty. That's why I made my way to PF and spent way too much time on this question. I really appreciate the time and help. I will be sure to pay it forward at some point. One clue I've picked up on for my question is that the square root factor in the expression is the magnitude of a general displacement vector in spherical coordinates (where the phi component is zero). That's all I've got. Hopefully this essay clarifies my question a little more. Thanks again
 
  • #8
13,797
10,940
My question is actually about how the expression for dS given in the problem statement (##dS=2πrsinθ\sqrt{dr^2+r^2dθ^2}##) was arrived at
From where, given what? What is ##S## in this specific case?
... in the first place, given the equations stated above ...
Which equations? ##dV## or ##dx\wedge dy\wedge dz##?
...
Just answer these few questions: formulas, no essays.
 

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