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(A) The equations are $x^2+y^2=16a^2$ and $y^2=6ax$. Substituting $y^2=6ax$ into the circle equation: $x^2+6ax-16a^2=0$. Factoring gives $(x+8a)(x-2a)

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$\frac{4a^2}{3}(4\pi+\sqrt{3})$

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(A) The equations are $x^2+y^2=16a^2$ and $y^2=6ax$. Substituting $y^2=6ax$ into the circle equation: $x^2+6ax-16a^2=0$. Factoring gives $(x+8a)(x-2a)=0$. Since $x \ge 0$ for the parabola,we take $x=2a$. At $x=2a$,$y^2=6a(2a)=12a^2$,so $y=\pm 2a\sqrt{3}$. The area $A$ is $2 \int_{0}^{2a} \sqrt{6ax} \, dx + 2 \int_{2a}^{4a} \sqrt{16a^2-x^2} \, dx$. First integral: $2\sqrt{6a} \int_{0}^{2a} x^{1/2} \, dx = 2\sqrt{6a} [\frac{2}{3}x^{3/2}]_{0}^{2a} = 2\sqrt{6a} \cdot \frac{2}{3} (2a\sqrt{2a}) = \frac{8a^2\sqrt{12}}{3} = \frac{16a^2\sqrt{3}}{3}$. Second integral: $2 [\frac{x}{2}\sqrt{16a^2-x^2} + \frac{16a^2}{2} \sin^{-1}(\frac{x}{4a})]_{2a}^{4a} = 2 [ (0 + 8a^2 \cdot \frac{\pi}{2}) - (a\sqrt{12a^2} + 8a^2 \cdot \frac{\pi}{6}) ] = 2 [ 4\pi a^2 - 2a^2\sqrt{3} - \frac{4\pi a^2}{3} ] = 2 [ \frac{8\pi a^2}{3} - 2a^2\sqrt{3} ] = \frac{16\pi a^2}{3} - 4a^2\sqrt{3}$. Total Area = $\frac{16a^2\sqrt{3}}{3} + \frac{16\pi a^2}{3} - 4a^2\sqrt{3} = \frac{16\pi a^2}{3} + \frac{4a^2\sqrt{3}}{3} = \frac{4a^2}{3}(4\pi+\sqrt{3})$.

The area of the region that is common to the circle $x^2+y^2=16a^2$ and the parabola $y^2=6ax$ is

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