What is the area of the region in the first quadrant bounded by the $x$-axis, the line $3x = 4y$ and the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$?

Explanation

1. 01

   The line $y = \frac{3x}{4}$ meets the ellipse where $\frac{x^{2}}{16} + \frac{x^{2}}{16} = 1$, at $x = 2\sqrt{2}$.

2. 02

   Under the line up to $x = 2\sqrt{2}$, then under the arc: $\int_0^{2\sqrt{2}} \frac{3x}{4}\,dx + \frac{3}{4}\int_{2\sqrt{2}}^{4} \sqrt{16 - x^{2}}\,dx = 3 + \frac{3}{4}(2\pi - 4)$.

3. 03

   $= \frac{3\pi}{2}$ square units, an eighth of the ellipse's $12\pi$.