JEE Mains · Maths · STD 12 - 8. Application and integration
Area of the region \(\left\{(x, y): x^2+(y-2)^2 \leq 4\right.\), \(\left.x^2 \geq 2 y\right\}\) is
- A \(2 \pi-\frac{16}{3}\)
- B \(\pi-\frac{8}{3}\)
- C \(\pi+\frac{8}{3}\)
- D \(2 \pi+\frac{16}{3}\)
Answer & Solution
Correct Answer
(A) \(2 \pi-\frac{16}{3}\)
Step-by-step Solution
Detailed explanation
\(x^2+(y-2)^2 \leq 2^2 \text { and } x^2 \geq 2 y\) Solving circle and parabola simultaneously : \(2 y+y^2-4 y+4=4\) \(y^2-2 y=0\) \(y=0,2\) Put \(y =2\) in \(x ^2=2 y \rightarrow x = \pm 2\) \(\Rightarrow(2,2)\) and \((-2,2)\)…
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