JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of the part of the circle \(x ^{2}+ y ^{2}=36\) which is outside the parabola \(y ^{2}=9 x ,\) is
- A \(24 \pi+3 \sqrt{3}\)
- B \(12 \pi-3 \sqrt{3}\)
- C \(24 \pi-3 \sqrt{3}\)
- D \(12 \pi+3 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(24 \pi-3 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Required area \(=\pi \times(6)^{2}-2 \int_{0}^{3} \sqrt{9} x d x-\int_{3}^{6} \sqrt{36-x^{2}} d x\) \(=36 \pi-12 \sqrt{3}-2\left(\frac{ x }{2} \sqrt{36- x ^{2}}+18 \sin ^{-1} \frac{ x }{6}\right)_{3}^{6}\) \(=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)\)…
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