The larger of the two areas (in sq. units) in | vedclass

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(A) For the point of intersection, substitute $y^2=6ax$ into $x^2+y^2=16a^2$: $x^2+6ax-16a^2=0$ $(x+8a)(x-2a)=0$ Since $x \ge 0$ for the parabola, we

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$\frac{4a^2}{3}(8\pi-\sqrt{3})$

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(A) For the point of intersection, substitute $y^2=6ax$ into $x^2+y^2=16a^2$: $x^2+6ax-16a^2=0$ $(x+8a)(x-2a)=0$ Since $x \ge 0$ for the parabola, we have $x=2a$. The area of the smaller region is $A_1 = 2 \left[ \int_0^{2a} \sqrt{6ax} \, dx + \int_{2a}^{4a} \sqrt{16a^2-x^2} \, dx \right]$. Calculating the integrals: $2 \int_0^{2a} \sqrt{6a} \sqrt{x} \, dx = 2 \sqrt{6a} \left[ \frac{2}{3} x^{3/2} \right]_0^{2a} = 2 \sqrt{6a} \cdot \frac{2}{3} \cdot 2a \sqrt{2a} = \frac{8a^2 \sqrt{12}}{3} = \frac{16a^2 \sqrt{3}}{3}$. $2 \int_{2a}^{4a} \sqrt{(4a)^2-x^2} \, dx = 2 \left[ \frac{x}{2} \sqrt{16a^2-x^2} + \frac{16a^2}{2} \sin^{-1} \left( \frac{x}{4a} \right) \right]_{2a}^{4a}$ $= 2 \left[ (0 + 8a^2 \cdot \frac{\pi}{2}) - (a \sqrt{12a^2} + 8a^2 \cdot \frac{\pi}{6}) \right] = 2 \left[ 4\pi a^2 - 2a^2 \sqrt{3} - \frac{4\pi a^2}{3} \right] = 2 \left[ \frac{8\pi a^2}{3} - 2a^2 \sqrt{3} \right] = \frac{16\pi a^2}{3} - 4a^2 \sqrt{3}$. Total smaller area $A_1 = \frac{16a^2 \sqrt{3}}{3} + \frac{16\pi a^2}{3} - 4a^2 \sqrt{3} = \frac{16\pi a^2}{3} + \frac{4a^2 \sqrt{3}}{3} = \frac{4a^2}{3}(4\pi + \sqrt{3})$. The larger area is the total area of the circle minus the smaller area: $A_2 = \pi(4a)^2 - A_1 = 16\pi a^2 - \left( \frac{16\pi a^2}{3} + \frac{4a^2 \sqrt{3}}{3} \right) = \frac{32\pi a^2}{3} - \frac{4a^2 \sqrt{3}}{3} = \frac{4a^2}{3}(8\pi - \sqrt{3})$ sq. units.

The larger of the two areas (in sq. units) into which the circle $x^2+y^2=16a^2$ is divided by the parabola $y^2=6ax$, is

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