Let the area of the bounded region {(x, y): 0 | vedclass

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(A) The region is bounded by the parabola $y^2 = 9x$ and the line $y = 3x - 6$.First,find the points of intersection by substituting $x = \frac{y^2}{9

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$15$

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(A) The region is bounded by the parabola $y^2 = 9x$ and the line $y = 3x - 6$.First,find the points of intersection by substituting $x = \frac{y^2}{9}$ into the line equation:$y = 3(\frac{y^2}{9}) - 6 \implies y = \frac{y^2}{3} - 6 \implies y^2 - 3y - 18 = 0$.Factoring gives $(y - 6)(y + 3) = 0$,so $y = 6$ or $y = -3$.For $y = -3$,$x = \frac{(-3)^2}{9} = 1$. For $y = 6$,$x = \frac{6^2}{9} = 4$.The region is bounded between $y = -3$ and $y = 6$. The line is to the right of the parabola in this region.$A = \int_{-3}^{6} [(\frac{y+6}{3}) - (\frac{y^2}{9})] dy$$A = \frac{1}{3} \int_{-3}^{6} (y + 6 - \frac{y^2}{3}) dy = \frac{1}{3} [\frac{y^2}{2} + 6y - \frac{y^3}{9}]_{-3}^{6}$$A = \frac{1}{3} [(\frac{36}{2} + 36 - \frac{216}{9}) - (\frac{9}{2} - 18 - \frac{-27}{9})]$$A = \frac{1}{3} [(18 + 36 - 24) - (4.5 - 18 + 3)] = \frac{1}{3} [30 - (-10.5)] = \frac{40.5}{3} = 13.5$.Wait,re-evaluating the region $0 \leq 9x \leq y^2$ and $y \geq 3x-6$ specifically for the shaded part in the image:The shaded region is bounded by $x=0$,$y^2=9x$ (lower branch $y = -3\sqrt{x}$) and $y=3x-6$.Intersection of $y = -3\sqrt{x}$ and $y = 3x-6$: $3x-6 = -3\sqrt{x} \implies x-2 = -\sqrt{x}$. Let $\sqrt{x} = t$,$t^2+t-2=0 \implies (t+2)(t-1)=0 \implies t=1 \implies x=1$.$A = \int_{0}^{1} [(-3\sqrt{x}) - (3x-6)] dx = \int_{0}^{1} (-3x^{1/2} - 3x + 6) dx$$A = [-3 \cdot \frac{2}{3} x^{3/2} - \frac{3x^2}{2} + 6x]_0^1 = [-2(1) - 1.5 + 6] = 2.5$.$6A = 6 	imes 2.5 = 15$.

Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

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