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Surface integral problem - don't need to use Jacobian for polar?

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral.
    ∫∫S x^2*z^2 dS
    S is the part of the cone z^2 = x^2 + y^2 that lies between the planes z = 1 and z = 3.


    2. Relevant equations

    [itex]\int \int _{S}F dS = \int \int _D F(r(u,v))|r_u\times r_v|dA[/itex]
    [itex]x=rcos(\theta)[/itex]
    [itex]y=rsin(\theta)[/itex]

    3. The attempt at a solution
    First, I parametrized the cone.
    [itex]z^2=x^2+y^2\Rightarrow z=\sqrt{x^2+y^2} \Rightarrow z=r[/itex]
    Therefore, the cone's vector equation should be
    [itex]{\bf R}(r,\theta)=rcos(\theta){\bf i}+rsin(\theta){\bf j}+r{\bf k}[/itex]
    [itex]{\bf R}_r=cos(\theta){\bf i}+sin(\theta){\bf j}+{\bf k}[/itex]
    [itex]{\bf R}_\theta=-rsin(\theta){\bf i}+rcos(\theta){\bf j}[/itex]
    [itex]|{\bf R}_r \times {\bf R}_\theta| = r\sqrt{2}[/itex]
    [itex]\int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}[/itex]

    Now, this is the right answer as per the book. My question is, when we go from [itex]dA[/itex] to [itex]drd\theta[/itex], why don't we use the Jacobian for polar coordinates, r? Had we included the Jacobian, the degree of r in the final double integral would have been six instead of five, giving us a completely different answer. My confusion here: Why is [itex]dA[/itex] not equal to [itex]rdrd\theta[/itex]?
     
    Last edited by a moderator: Dec 7, 2011
  2. jcsd
  3. Dec 6, 2011 #2
    Sorry, the final step of the integration process didn't come out right. Here it is:

    [itex]\int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}[/itex]
     
  4. Dec 6, 2011 #3

    Dick

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    The |R_r x R_theta| factor already includes the r in dA. If the surface were just the x-y plane and they wanted you to find area by integrating 1. Think about what that factor would be.
     
    Last edited: Dec 6, 2011
  5. Dec 7, 2011 #4
    I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating [itex]dA[/itex] to [itex]rdrd\theta[/itex] in all scenarios) or because of my own negligence. Does the [itex]|{\bf R}_r\times{\bf R}_\theta|[/itex] factor always take care of the Jacobian for surface integrals, then? In something akin to spherical coordinate integrals (e.g., if my vector [itex]{/bf R}[/itex] wer
     
  6. Dec 7, 2011 #5
    God, I'm bad at this whole Latex + forum thing. I'm so sorry about the double posts... I think I accidentally hit submit or something.

    I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating [itex]dA[/itex] to [itex]rdrd\theta[/itex] in all scenarios) or because of my own negligence. Does the [itex]|{\bf R}_r\times{\bf R}_\theta|[/itex] factor always take care of the Jacobian for surface integrals, then? In something akin to spherical coordinate integrals, would that factor take care of the whole [itex]\rho^2sin(\phi)[/itex] factor for me?
     
  7. Dec 7, 2011 #6

    Dick

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    r and theta here are really just a convenient parametrization of the surface. [itex]dS=|r_u\times r_v| du dv[/itex]. I wouldn't substitute dA for du dv, if it's going to confuse you.
     
  8. Dec 7, 2011 #7
    I see, that makes sense. Are there any situations in which I would ever have to actually worry about a Jacobian factor when doing a surface integral similar to the one I've described above? Or is it something I should only worry about when explicitly executing coordinate transforms?
     
  9. Dec 7, 2011 #8

    Dick

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    You don't have to worry about it if you use that formula. Like I said, the 'jacobian' part is the [itex]|r_u\times r_v|[/itex] factor.
     
  10. Dec 7, 2011 #9
    Got it. Thanks very much, Dick.
     
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