Surface integral, spherical coordinates, earth

[HmBe]

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.

 

[pasmith]

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 \theta is equal to 90^{\circ} minus the latitude. Thus the correct limits of the \theta integral are 0 and 40^{\circ}.

 

[HmBe]

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