There are already a lot of good answers here, so I'm adding this one
primarily to dazzle people w/ my Mathematica diagram-creating skills.
As noted previously,
$x(t)=a \cos (t)$
$y(t)=b \sin (t)$
does parametrize an ellipse, but t is not the central angle. What
is the relation between t and the central angle?:
Since y is bSin[t] and x is aCos[t], we have:
$\tan (\theta )=\frac{b \sin (t)}{a \cos (t)}$
or
$\tan (\theta )=\frac{b \tan (t)}{a}$
Solving for t, we have:
$t(\theta )=\tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
We now reparametrize using theta:
$x(\theta )=a \cos (t(\theta ))$
$y(\theta )=b \sin (t(\theta ))$
which ultimately simplifies to:
$x(\theta)=\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$
$y(\theta)=\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}}$
Note that, under the new parametrization, $y(\theta)/x(\theta) =
tan(\theta)$ as desired.
To compute area, we need $r^2$ which is $x^2+y^2$, or:
$r(\theta )^2 = (\frac{a}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2+
(\frac{a \tan (\theta )}{\sqrt{\frac{a^2 \tan ^2(\theta )}{b^2}+1}})^2$
(note that we could take the square root to get r, but we don't really need it)
The above ultimately simplifies to:
$r(\theta)^2 =
\frac{1}{\frac{\cos ^2(\theta )}{a^2}+\frac{\sin ^2(\theta )}{b^2}}$
Now, we can integrate $r^2/2$ to find the area:
$A(\theta) = (\int_0^\theta
\frac{1}{\frac{\cos ^2(x )}{a^2}+\frac{\sin ^2(x )}{b^2}} \, dx)/2$
which yields:
$A(\theta) = \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
good for $0\leq \theta <\frac{\pi }{2}$
Interestingly, it doesn't work for $\theta =\frac{\pi }{2}$ so we
can't test the obvious case without using a limit:
$\lim_{\theta \to \frac{\pi }{2}} \, \frac{1}{2} a b \tan
^{-1}\left(\frac{a \tan (\theta )}{b}\right)$
which gives us $a*b*Pi/4$ as expected.
Not exactly an answer to your question, but a bit long for a comment:
Since $r$ denotes the radius of your sphere, it's best to use a different symbol, such as $\rho$, to denote the "radius function" (i.e., the distance to the origin).
Be that as it may, you can avoid the nuisance of integrating the angular variables in your question because $\nabla \cdot F = 3\rho^{2}$ is constant on each spherical shell $S_{\rho}$. Consequently,
$$
\iiint_{V} (\nabla \cdot F)\, dV
= \int_{0}^{r} \left(\iint_{S_{\rho}} 3\rho^{2}\, dA\right) d\rho
= \int_{0}^{r} 3\rho^{2} \left(\iint_{S_{\rho}} dA\right) d\rho
= 3(4\pi)\int_{0}^{r} \rho^{4}\, d\rho.
$$
Best Answer
Under a little bit more general settings, a line integral of a vector field $(p,q)$ along some oriented planar curve $L$ has the following form. $$I=\int_L\left(pdx+qdy\right).$$ Denote $\mathbf{r}=(x,y)$ and let $L$ be parameterized as $t\mapsto\mathbf{r}(t)$, $t\in[0,1]$. We can always choose another parameter $s$, the so called arc length parameter, which satisfies $\frac{d\mathbf{r}}{ds}\equiv 1$, or $ds=|\mathbf{r}'(t)|dt$. Follow your notations, $\frac{ds}{dt}(t)=v(t)$. The unit normal vector field $\mathbf{n}$ can be represented as $$\mathbf{n}=(\frac{dy}{ds},-\frac{dx}{ds})=v(t)^{-1}(\frac{dy}{dt},-\frac{dx}{dt}).$$
If we choose $\mathbf{f}=(q,-p)$ as a vector field, then the line integral above can be rewritten as $$I=\int\mathbf{f}\cdot\mathbf{n}ds=\int\mathbf{f}\cdot(\frac{dy}{dt},-\frac{dx}{dt})dt.$$
Now we know that when $L$ is a simple closed curve, and if we choose $\mathbf{f}=\frac{1}{2}\mathbf{r}$, the line integral above gives the area bounded by $L$, i.e. $$area=\frac{1}{2}\int\mathbf{r}\cdot\mathbf{n}ds=\frac{1}{2}\int(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}\int_L( xdy-ydx).$$