Surface integral, spherical coordinates, earth

  • Thread starter HmBe
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Homework Statement



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




Homework Equations



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


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.
 

Answers and Replies

  • #2
pasmith
Homework Helper
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Homework Statement



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




Homework Equations



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


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.
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].
 
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You're a hero, I have mental problems. Thank you so much.
 

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