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QJEEM23Ma028c: JEE Main 2023 Mathematics - Class 12 JEEM23Ma028c Solution: Area of the region is

Area of the region {(x,y):x2+(y−2)2≤4,x2≥2y}\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\} is

Multiple Choice Options

  • A 2π+1632 \pi+\frac{16}{3}
  • B π−83\pi-\frac{8}{3}
  • C π+83\pi+\frac{8}{3}
  • D 2π−1632 \pi-\frac{16}{3}

Step-by-Step Verified Solution

We have,

x2+(y−2)2≤22 and x2≥2yx^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

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

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

If y=2y=2 , then x=±2x= \pm 2

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

JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Area Under The Curves Question 39 English Explanation 1

JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Area Under The Curves Question 39 English Explanation 2

Area of the shaded region = =2×2−14⋅π⋅22=4−π=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi

 Required area amp;=2[∫02x22dx−(4−π)]amp;=2[x36∣02−4+π]amp;=2[43+π−4]amp;=2[π−83]amp;=2π−166\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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