Step 1: Use the parametric form of the ellipse instead of solving for $y$ explicitly:
Parametrize with $x=a\cos\theta$, $y=b\sin\theta$, $\theta$ from $\pi/2$ down to $0$ traces the first-quadrant arc.
Step 2: Set up the area as $4\int y\,dx$ in terms of $\theta$:
$dx=-a\sin\theta\,d\theta$, so $\text{Area}=4\displaystyle\int_{\pi/2}^{0} b\sin\theta\,(-a\sin\theta)\,d\theta = 4ab\int_0^{\pi/2}\sin^2\theta\,d\theta$ (flipping the limits removes the minus sign).
Step 3: Evaluate $\int_0^{\pi/2}\sin^2\theta\,d\theta$ using the half-angle identity:
$\sin^2\theta=\dfrac{1-\cos2\theta}{2}$, so $\displaystyle\int_0^{\pi/2}\sin^2\theta\,d\theta = \left[\dfrac{\theta}{2}-\dfrac{\sin2\theta}{4}\right]_0^{\pi/2} = \dfrac{\pi}{4}-0 = \dfrac{\pi}{4}$.
Step 4: Substitute back:
$\text{Area}=4ab\cdot\dfrac{\pi}{4}=\pi ab$.
Final Answer:
The area of the ellipse is $\pi ab$ square units.
\[ \boxed{\pi ab} \]