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  • The area (in sq. units) of the part of the circle which is outside the parabola <img alt="y^2=9x," src="https://entrancecorner.co

The area (in sq. units) of the part of the circle x^{2}+y^{2}=36, which is outside the parabola y^2=9x,is:
Option: 1 12\pi+3\sqrt{3}
Option: 2 12\pi-3\sqrt{3}
Option: 3 24\pi-3\sqrt{3}  
Option: 4 24\pi+3\sqrt{3}

Answers (1)

best_answer

 

The curves intersect at points (3, \pm 3 \sqrt{3})

\begin{aligned} &\text { Required area }\\\\ &\begin{array}{l} =\pi r^{2}-2\left[\int_{0}^{3} \sqrt{9 x} d x+\int_{3}^{6} \sqrt{36-x^{2}} d x\right] \\\\ =36 \pi-12 \sqrt{3}-2\left|\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1}\left(\frac{x}{6}\right)\right|_{3}^{6} \\\\ =36 p-12 \sqrt{3}-2\left(9 \pi-\left(\frac{9 \sqrt{3}}{2}+3 \pi\right)\right)=24 \pi-3 \sqrt{3} \end{array} \end{aligned}

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Suraj Bhandari

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