You don't say whether you are required to Cartesian coordinates but if not, here is how I would find the surface area.
Spherical coordinates are given by x= \rho cos(\theta) sin(\phi), y= \rho sin(\theta)sin(\phi), and z= \rho cos(\phi) with \theta going from 0 to 2\pi and \phi form 0 to \pi.
On a sphere of radius R, we have \rho= R and so can write the surface in terms of the two parameters, \theta and \phi as a vector equation:
\vec{p}(\theta, \phi)= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}
The derivatives of that with respect to \theta and \phi are tangent vectors to that surface along the respective coordinate axes:
\vec{p}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}
and
\vec{p}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{i}
The cross product of those tangent vectors (called the "fundamental vector product" for the surface) is perpendicular to the surface and incorporates all metric information- it gives the "vector differential of surface area" for the surface.
\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\-Rsin(\theta)sin(\phi) & Rcos(\theta)sin(\phi) & 0 \\ Rcos(\theta)cos(\phi) & Rsin(\theta)cos(\phi) & -Rsin(\phi)\end{array}\right|
= -R^2cps)\theta sin^2\phi\vec{i}+ R^2 sin(\theta)sin^2(\phi)\vec{j}- R^2 sin(\phi)cos(\phi)\vec{k}
The length of that vector,
\sqrt{R^4(sin^4(\phi)+ sin^2(\phi)cos^2(\phi))}= \sqrt{R^4sin^2(\phi)}= R^2 sin^2(\phi)[/itex] <br />
gives the "differential of surface area", R^2 sin^2(\phi)d\theta d\phi.<br />
<br />
The surface area of the sphere is <br />
\int_{\theta=0}^{2\pi}\int_{\phi= 0}^\pi R^2 sin^2(\phi)d\phi d\theta<br />
<br />
To integrate that, use the standard identity sin^2(\phi)= (1/2)(1- cos(\phi)):<br />
R^2/2 \int_{\theta= 0}^{2\pi} [1- cos(2\phi)]d\phi d\theta[/itex]
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