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Surface integral, spherical coordinates, earth

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the surface area of the earth (as a fraction of the total surface of the earth) that lies above 50 degrees latitude North.




    2. Relevant equations

    $$A = \int_R\sqrt{|\det(g)|}d\theta d\phi$$


    3. The attempt at a solution

    Hence I get

    $$\int_0^{360} \int_{50}^{90} r^2 \sin(\theta) d\theta d\phi$$

    Which gives 0.321 as the answer, which just isn't right. The actual answer I know should be 0.117, or

    $$\frac{1-\sin(50)}{2}$$

    I'm sure something must be wrong with my integral, but I can't figure out what.
     
  2. jcsd
  3. Feb 10, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    In the northern hemisphere, the spherical polar coordinate [itex]\theta[/itex] is equal to [itex]90^{\circ}[/itex] minus the latitude. Thus the correct limits of the [itex]\theta[/itex] integral are 0 and [itex]40^{\circ}[/itex].
     
  4. Feb 10, 2014 #3
    You're a hero, I have mental problems. Thank you so much.
     
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