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JEE Mains · Maths · STD 12 - 8. Application and integration

The area of the region, inside the circle \((x-2 \sqrt{3})^2+y^2=12\) and outside the parabola \(y^2=2 \sqrt{3} x\) is :

  1. A \(3 \pi+8\)
  2. B \(6 \pi-16\)
  3. C \(3 \pi-8\)
  4. D \(6 \pi-8\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6 \pi-16\)

Step-by-step Solution

Detailed explanation

Required area \(=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}\right) d x \) \( =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}\right) d x \)…
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