Surface integrals to derive area of sphere

[bawbag]

Given a sphere $$x^2 + y^2 + z^2 = a^2$$ how would I derive the surface area by using surface integrals?


The method I've tried is as follows: $$dA = sec\ \gamma \ dxdy$$ where gamma is the angle between the tangent plane at dA and the xy plane. $$sec \gamma = \frac{|\nabla \varphi|}{\partial \varphi /\partial z}$$$$= \frac{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}{2z}$$$$= \frac{a}{\sqrt{a^2 - x^2 - y^2}}$$

converting to polar coordinates and integrating the expression for $sec \gamma$$$\int^{a}_{0}\int^{2 \pi}_{0} \frac{a}{\sqrt{a^2 - r^2}} r \ drd \theta$$ and using substitution $u = a^2 - r^2$ yields $$-a \pi [2 \sqrt{a^2 - a^2} - 2 \sqrt{a^2 - 0}]$$$$= 2 \pi a^2$$ Obviously the area of a sphere is $4 \pi a^2$ so did I do the calculation wrong, or does this method only find the area of one hemisphere, in which case I should multiply the answer by 2?

Thanks in advance

 

[LCKurtz]

Yes, you only did the top half of the sphere. It would be much easier using spherical coordinates where$$
dS = a^2\sin\phi~ d\phi d\theta$$

 

[bawbag]

Thanks, I thought as much. I know how to do it using the spherical polars, but I was just curious as to whether the surface integral method worked as well.

Every day's a school day. Thanks again!