Find the area of the region $\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\}$

(D) The area bounded by the curves $\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\}$ is determined by finding the intersection points of the parabola $y^{2}=4 x$ and the circle $x^{2}+y^{2}=\frac{9}{4}$.
Substituting $y^{2}=4 x$ into the circle equation: $4 x^{2}+4(4 x)=9 \Rightarrow 4 x^{2}+16 x-9=0$.
Solving the quadratic equation: $(2 x-1)(2 x+9)=0$,we get $x=\frac{1}{2}$ (since $x \geq 0$ for the parabola).
For $x=\frac{1}{2}$,$y^{2}=4(\frac{1}{2})=2 \Rightarrow y=\pm \sqrt{2}$. The intersection points are $(\frac{1}{2}, \sqrt{2})$ and $(\frac{1}{2}, -\sqrt{2})$.
The area is symmetric about the $x$-axis,so the total area is $2 \times$ (Area in the first quadrant).
Area $= 2 \left[ \int_{0}^{1/2} \sqrt{4x} \, dx + \int_{1/2}^{3/2} \sqrt{\frac{9}{4}-x^2} \, dx \right]$.
$= 2 \left[ \int_{0}^{1/2} 2\sqrt{x} \, dx + \frac{1}{2} \int_{1/2}^{3/2} \sqrt{(3/2)^2-x^2} \, dx \right]$.
$= 2 \left[ 2 \cdot \frac{2}{3} x^{3/2} \Big|_0^{1/2} + \frac{1}{2} \left( \frac{x}{2} \sqrt{\frac{9}{4}-x^2} + \frac{9}{8} \sin^{-1}(\frac{2x}{3}) \right) \Big|_{1/2}^{3/2} \right]$.
$= 2 \left[ \frac{4}{3} (\frac{1}{2\sqrt{2}}) + \frac{1}{2} \left( (0 + \frac{9}{8} \sin^{-1}(1)) - (\frac{1}{4} \sqrt{2} + \frac{9}{8} \sin^{-1}(\frac{1}{3})) \right) \right]$.
$= 2 \left[ \frac{\sqrt{2}}{3} + \frac{9\pi}{16} - \frac{\sqrt{2}}{8} - \frac{9}{16} \sin^{-1}(\frac{1}{3}) \right] = 2 \left[ \frac{5\sqrt{2}}{24} + \frac{9\pi}{16} - \frac{9}{16} \sin^{-1}(\frac{1}{3}) \right] = \frac{5\sqrt{2}}{12} + \frac{9\pi}{8} - \frac{9}{8} \sin^{-1}(\frac{1}{3}) \text{ sq. units.}$