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