Question:
The area (in sq. units) of the part of the circle $x^{2}+y^{2}=36$, which is outside the parabola $\mathrm{y}^{2}=9 \mathrm{x}$, is :
Correct Option: , 3
Solution:
Required area
$=\pi \times(6)^{2}-2 \int_{0}^{3} \sqrt{9} x d x-\int_{3}^{6} \sqrt{36-x^{2}} d x$
$=36 \pi-12 \sqrt{3}-2\left(\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1} \frac{x}{6}\right)_{3}^{6}$
$=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)$
$=24 \pi-3 \sqrt{3}$
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