JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 12)
Area of the region $$\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}$$ is
$$2 \pi+\frac{16}{3}$$
$$\pi-\frac{8}{3}$$
$$\pi+\frac{8}{3}$$
$$2 \pi-\frac{16}{3}$$
Explanation
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 $(0,0),(-2,2),(2,2)$
_11th_April_Morning_Shift_en_12_1.png)
_11th_April_Morning_Shift_en_12_2.png)
Area of the shaded region = $$ =2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi $$
$$ \begin{aligned} \text { Required area } & =2\left[\int\limits_0^2 \frac{x^2}{2} d x-(4-\pi)\right] \\\\ & =2\left[\left.\frac{x^3}{6}\right|_0 ^2-4+\pi\right] \\\\ & =2\left[\frac{4}{3}+\pi-4\right] \\\\ & =2\left[\pi-\frac{8}{3}\right] \\\\ & =2 \pi-\frac{16}{6} \end{aligned} $$
$$ x^2+(y-2)^2 \leq 2^2 \text { and } x^2 \geq 2 y $$
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 = $$ =2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi $$
$$ \begin{aligned} \text { Required area } & =2\left[\int\limits_0^2 \frac{x^2}{2} d x-(4-\pi)\right] \\\\ & =2\left[\left.\frac{x^3}{6}\right|_0 ^2-4+\pi\right] \\\\ & =2\left[\frac{4}{3}+\pi-4\right] \\\\ & =2\left[\pi-\frac{8}{3}\right] \\\\ & =2 \pi-\frac{16}{6} \end{aligned} $$
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