The area (in sq. units) of the part of the ci | vedclass

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(C) First,find the intersection points of $x^{2} + y^{2} = 36$ and $y^{2} = 9x$.Substituting $y^{2} = 9x$ into the circle equation: $x^{2} + 9x - 36 =

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$24 \pi - 3 \sqrt{3}$

Vedclass · Answer

(C) First,find the intersection points of $x^{2} + y^{2} = 36$ and $y^{2} = 9x$.Substituting $y^{2} = 9x$ into the circle equation: $x^{2} + 9x - 36 = 0$.$(x + 12)(x - 3) = 0$,so $x = 3$ (since $x \geq 0$ for the parabola).At $x = 3$,$y^{2} = 27$,so $y = \pm 3 \sqrt{3}$.The area inside the circle and outside the parabola is the total area of the circle minus the area inside both the circle and the parabola.The area inside both is $2 \int_{0}^{3} \sqrt{9x} \, dx + 2 \int_{3}^{6} \sqrt{36 - x^{2}} \, dx$.$= 2 	imes 3 \int_{0}^{3} x^{1/2} \, dx + 2 \left[ \frac{x}{2} \sqrt{36 - x^{2}} + \frac{36}{2} \sin^{-1} \left( \frac{x}{6} ight) ight]_{3}^{6}$.$= 6 \left[ \frac{2}{3} x^{3/2} ight]_{0}^{3} + 2 \left[ (0 + 18 \sin^{-1}(1)) - (\frac{3}{2} \sqrt{27} + 18 \sin^{-1}(1/2)) ight]$.$= 4(3 \sqrt{3}) + 2 \left[ 18(\pi/2) - (\frac{9 \sqrt{3}}{2} + 18(\pi/6)) ight] = 12 \sqrt{3} + 2 [9 \pi - \frac{9 \sqrt{3}}{2} - 3 \pi] = 12 \sqrt{3} + 12 \pi - 9 \sqrt{3} = 12 \pi + 3 \sqrt{3}$.Total area of circle = $36 \pi$.Required area = $36 \pi - (12 \pi + 3 \sqrt{3}) = 24 \pi - 3 \sqrt{3}$.

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:

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