The area of an ellipse

Setup. By symmetry, the area is $4$ times the area in the first quadrant:
$$\mathcal A=4\int_0^ab\sqrt{1-\frac{x^2}{a^2}}\,dx.$$

Substitution. Let $x=a\sin\theta$ with $\theta\in\left[0,\frac\pi2\right]$, so $dx=a\cos\theta\,d\theta$ and $\sqrt{1-\sin^2\theta}=\cos\theta$ (non-negative on this range):
$$\mathcal A=4ab\int_0^{\pi/2}\cos^2\theta\,d\theta=4ab\cdot\frac\pi4=\pi ab.$$

Remark. For $a=b=r$ this recovers the area $\pi r^2$ of a disc. In general the ellipse is the disc of radius $a$ compressed vertically by the factor $\frac ba$, so its area is multiplied by $\frac ba$.