# Surface integral, spherical coordinates, earth

## 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 Helper

## 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 $40^{\circ}$.

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