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 [itex]x= \rho cos(\theta) sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], and [itex]z= \rho cos(\phi)[/itex] with [itex]\theta[/itex] going from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] form 0 to [itex]\pi[/itex].
On a sphere of radius R, we have [itex]\rho= R[/itex] and so can write the surface in terms of the two parameters, [itex]\theta[/itex] and [itex]\phi[/itex] as a vector equation:
[tex]\vec{p}(\theta, \phi)= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}[/tex]
The derivatives of that with respect to [itex]\theta[/itex] and [itex]\phi[/itex] are tangent vectors to that surface along the respective coordinate axes:
[tex]\vec{p}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
and
[tex]\vec{p}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{i}[/tex]
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.
[tex]\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|[/tex]
[tex]= -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}[/tex]
The length of that vector,
[tex]\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", [itex]R^2 sin^2(\phi)d\theta d\phi[/itex].<br />
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The surface area of the sphere is <br />
[tex]\int_{\theta=0}^{2\pi}\int_{\phi= 0}^\pi R^2 sin^2(\phi)d\phi d\theta[/tex]<br />
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To integrate that, use the standard identity [itex]sin^2(\phi)= (1/2)(1- cos(\phi))[/itex]:<br />
[tex]R^2/2 \int_{\theta= 0}^{2\pi} [1- cos(2\phi)]d\phi d\theta[/itex][/tex][/tex]
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