# Surface integral, spherical coordinates, earth

1. Feb 10, 2014

### HmBe

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. Feb 10, 2014

### pasmith

In the northern hemisphere, the spherical polar coordinate $\theta$ is equal to $90^{\circ}$ minus the latitude. Thus the correct limits of the $\theta$ integral are 0 and $40^{\circ}$.

3. Feb 10, 2014

### HmBe

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