Q.Find the area of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$, using integration.

Comparing $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ with the standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ gives $a^2=9\Rightarrow a=3$ and $b^2=4\Rightarrow b=2$. The upper half of the ellipse is $y=\dfrac{b}{a}\sqrt{a^2-x^2}=\dfrac{2}{3}\sqrt{9-x^2}$. The first-quadrant quarter-area is

The remaining integral is the quarter-circle integral of radius $3$, equal to $\dfrac{\pi(3)^2}{4}=\dfrac{9\pi}{4}$. So …

Read off $a$ and $b$ from the standard equation, and either apply $\pi ab$ directly or derive it via the quarter-ell …

Swapping which denominator is $a^2$ and which is $b^2$; forgetting the final $\times 4$ scaling from t …