<s>[INST] Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating 
$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$
So we should be able to find the area of the ellipse by 
$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$
$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$
$$=\pi\frac{a^2+b^2}{2}.$$
First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.
What am I doing wrong?
 [/INST] 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. 
</s>