Answer :
Presumably you should be doing this using calculus methods, namely computing the surface integral along [tex]\mathbf r(u,v)[/tex].
But since [tex]\mathbf r(u,v)[/tex] describes a sphere, we can simply recall that the surface area of a sphere of radius [tex]a[/tex] is [tex]4\pi a^2[/tex].
In calculus terms, we would first find an expression for the surface element, which is given by
[tex]\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j[/tex]
[tex]\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k[/tex]
[tex]\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u[/tex]
So the area of the surface is
[tex]\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u[/tex]
[tex]=-2\pi a^2(\cos\pi-\cos 0)[/tex]
[tex]=-2\pi a^2(-1-1)[/tex]
[tex]=4\pi a^2[/tex]
as expected.
But since [tex]\mathbf r(u,v)[/tex] describes a sphere, we can simply recall that the surface area of a sphere of radius [tex]a[/tex] is [tex]4\pi a^2[/tex].
In calculus terms, we would first find an expression for the surface element, which is given by
[tex]\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k[/tex]
[tex]\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j[/tex]
[tex]\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k[/tex]
[tex]\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u[/tex]
So the area of the surface is
[tex]\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u[/tex]
[tex]=-2\pi a^2(\cos\pi-\cos 0)[/tex]
[tex]=-2\pi a^2(-1-1)[/tex]
[tex]=4\pi a^2[/tex]
as expected.