The area of the region (in sq. units) bounded | vedclass

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(C) The given curves are the circle $x^2+y^2=16$ (center $(0,0)$,radius $r=4$) and the parabola $y^2=6x$ (vertex $(0,0)$,opening right).To find the in

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$\frac{4}{3}(4 \pi+\sqrt{3})$

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(C) The given curves are the circle $x^2+y^2=16$ (center $(0,0)$,radius $r=4$) and the parabola $y^2=6x$ (vertex $(0,0)$,opening right).To find the intersection points,substitute $y^2=6x$ into $x^2+y^2=16$:$x^2+6x-16=0 \implies (x+8)(x-2)=0$.Since $x \ge 0$ for the parabola,we have $x=2$.Then $y^2=6(2)=12 \implies y = \pm 2\sqrt{3}$.The area is symmetric about the $x$-axis,so Area $= 2 \int_{0}^{2} \sqrt{6x} \, dx + 2 \int_{2}^{4} \sqrt{16-x^2} \, dx$.First part: $2 \sqrt{6} \int_{0}^{2} x^{1/2} \, dx = 2 \sqrt{6} [\frac{2}{3} x^{3/2}]_{0}^{2} = 2 \sqrt{6} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{8}{3} \sqrt{12} = \frac{16\sqrt{3}}{3}$.Second part: $2 [\frac{x}{2} \sqrt{16-x^2} + \frac{16}{2} \sin^{-1}(\frac{x}{4})]_{2}^{4} = 2 [ (0 + 8 \sin^{-1}(1)) - (\sqrt{12} + 8 \sin^{-1}(\frac{1}{2})) ]$.$= 2 [ 8(\frac{\pi}{2}) - 2\sqrt{3} - 8(\frac{\pi}{6}) ] = 2 [ 4\pi - 2\sqrt{3} - \frac{4\pi}{3} ] = 2 [ \frac{8\pi}{3} - 2\sqrt{3} ] = \frac{16\pi}{3} - 4\sqrt{3}$.Total Area $= \frac{16\sqrt{3}}{3} + \frac{16\pi}{3} - 4\sqrt{3} = \frac{16\pi}{3} + \frac{4\sqrt{3}}{3} = \frac{4}{3}(4\pi+\sqrt{3})$.

The area of the region (in sq. units) bounded by the curves $x^2+y^2=16$ and $y^2=6x$ is

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