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.
The area of a circular sector of radius $r$ and angle $d\theta$ is $\pi r^2 \frac{d\theta}{2\pi} = \frac{1}{2} r^2 d\theta$.
The right side of the square ($0 < \theta < \pi/4$) is the line $x = 5$, which in polar coordinates is $r = \frac{5}{\cos \theta}$ (not $\sin$).
Putting this together, the integrand should be $\frac{1}{2}(\frac{5}{\cos \theta})^2 d\theta = \frac{25}{2 \cos^2 \theta} d\theta$.
This works out to $12.5$. Since it only covers half of the square, double it to get the $25$ you expect.
Best Answer
Notice, since the ellipse: $x=a\cos\theta$ & $y=b\sin\theta$ is equally divided into four symmetrical regions hence, the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$ Now changing in Polar-coordinates by setting $y=b\sin\theta$ & $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$ $$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$ $$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$ $$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$
$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$