Find the area common to the curves y = sqrt{9 | vedclass

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(D) The given curves are $x^2 + y^2 = 9$ (a circle with center $(0,0)$ and radius $3$) and $x^2 + y^2 - 6x = 0$ (a circle with center $(3,0)$ and radi

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$3\left(\pi - \frac{\sqrt{3}}{4}\right)$

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(D) The given curves are $x^2 + y^2 = 9$ (a circle with center $(0,0)$ and radius $3$) and $x^2 + y^2 - 6x = 0$ (a circle with center $(3,0)$ and radius $3$).To find the intersection points,substitute $x^2 + y^2 = 9$ into the second equation: $9 - 6x = 0 \Rightarrow x = 3/2$.At $x = 3/2$,$y^2 = 9 - (3/2)^2 = 9 - 9/4 = 27/4$,so $y = \pm \frac{3\sqrt{3}}{2}$.The area common to both curves is symmetric about the $x$-axis. The area is given by $A = 2 \left[ \int_{0}^{3/2} \sqrt{6x - x^2} \, dx + \int_{3/2}^{3} \sqrt{9 - x^2} \, dx \right]$.Evaluating these integrals,we get $A = 2 \left[ \left( \frac{9\pi}{8} - \frac{9\sqrt{3}}{8} \right) + \left( \frac{9\pi}{4} - \frac{9\pi}{8} - \frac{3\sqrt{3}}{8} \right) \right] = 3\left( \pi - \frac{\sqrt{3}}{4} \right)$.

Find the area common to the curves $y = \sqrt{9 - x^2}$ and $x^2 + y^2 = 6x$.

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