Area of the region [(x, y): x^2+(y-2)^2 4, x^2 2 y] is

Question

Area of the region $$\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}$$ is

Accepted Answer

We have,

$$
x^2+(y-2)^2 \leq 2^2 \text { and } x^2 \geq 2 y
$$

On solving the given equation of parabola and circle, we get

$$
\begin{aligned}
2 y+(y-2)^2 & =4 \\
\Rightarrow y =0 \text { or } 2
\end{aligned}
$$

If $y=0$, then $x=0$ and

If $y=2$, then $x= \pm 2$

$\therefore$ The intersecting point of circle and parabola are $(0,0),(-2,2),(2,2)$

Area of the shaded region = $$
=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi
$$

$$
\begin{aligned}
\text { Required area } & =2\left[\int\limits_0^2 \frac{x^2}{2} d x-(4-\pi)\right] \\
& =2\left[\left.\frac{x^3}{6}\right|_0 ^2-4+\pi\right] \\
& =2\left[\frac{4}{3}+\pi-4\right] \\
& =2\left[\pi-\frac{8}{3}\right] \\
& =2 \pi-\frac{16}{6}
\end{aligned}
$$

Solutions · 1