JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region \( A = \{(x, y): 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\} \) is:
- A \(\frac{\pi}{2}+2\)
- B \(\pi+\frac{2}{3}\)
- C \(\pi+4\)
- D \(\frac{\pi}{2}+\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\pi+\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(A=\int_0^2 2 \sqrt{x} d x+2 \int_1^{\sqrt{2}} \sqrt{8-4 x^2} d x\) \(=\frac{8}{3}\left(x^{\frac{3}{2}}\right) \int_0^1+4 \int_1^{\sqrt{2}} \sqrt{2-x^2} d x\)…
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