A window is shaped like the ellipse x²/9 + y²/4 = 1, with x and y in decimetres (dm)…

Step 1. Meeting points. The bar meets the $x$-axis at $(3, 0)$ and the $y$-axis at $(0, 2)$. Both points lie on the ellipse. So the limits are $x = 0$ and $x = 3$.

Step 2. Which is above? At $x = 1$ the ellipse gives $\dfrac{2}{3}\sqrt{8} \approx 1.886$ and the bar gives $\dfrac{4}{3}$. The ellipse is above.

Step 3. The area. $\displaystyle\int_0^3 \left(\dfrac{2}{3}\sqrt{9 - x^2} - 2 + \dfrac{2x}{3}\right) dx$. The first part is a quarter of the ellipse, $\dfrac{\pi \times 3 \times 2}{4} = \dfrac{3\pi}{2}$. The other part is the triangle, $\dfrac{3 \times 2}{2} = 3$.