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Find the area of the portion of the circle $x^2+y^2=64$ which is exterior to the parabola $y^2 =12x$. |
$\frac{16}{3}(8π-\sqrt{3})$ $\frac{4}{3}(8π-\sqrt{3})$ $\frac{5}{3}(8π-\sqrt{3})$ $\frac{2}{3}(8π-\sqrt{3})$ |
$\frac{16}{3}(8π-\sqrt{3})$ |
Required Area = $π8^2-2\int\limits_0^A[\sqrt{64-y^2}-\frac{y^2}{12}]dy$
$⇒\frac{16}{3}(8π-\sqrt{3})$ |
