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

Write the two inequalities separately.
  • From $$0 \le 9x \le y^{2}$$ we get $$x \ge 0$$ and $$y^{2} \ge 9x$$.
  • From $$y \ge 3x-6$$ we have the straight line $$y = 3x-6$$ (the region is above this line).

The parabola $$y^{2}=9x$$ can be written as two branches:
  Upper branch  $$y = 3\sqrt{x}$$,
  Lower branch  $$y = -3\sqrt{x}$$.

The condition $$y^{2}\ge 9x$$ means the point must lie outside the parabola, that is,
  either $$y \ge 3\sqrt{x}$$ or $$y \le -3\sqrt{x}$$.

The extra condition $$y \ge 3x-6$$ cuts away much of this unbounded region.
Let us find where the lower branch $$y=-3\sqrt{x}$$ meets the line $$y=3x-6$$:

Set them equal:
$$-3\sqrt{x}=3x-6$$   $$\Longrightarrow$$    divide by 3: $$-\sqrt{x}=x-2$$.
Put $$t=\sqrt{x}\; (t\ge 0)$$,
$$-t = t^{2}-2 \quad\Longrightarrow\quad t^{2}+t-2=0$$.
Solving, $$t=\dfrac{-1+\sqrt{1+8}}{2}=1 \;(\text{positive root})$$,
so $$\sqrt{x}=1 \Longrightarrow x=1$$ and the common point is $$(1,-3).$$

Next, compare the two curves for $$0\le x\le 1$$.
Take any $$x$$ in this interval (say $$x=0.25$$):
  • $$y=-3\sqrt{0.25}=-3\times 0.5=-1.5,$$
  • $$y=3(0.25)-6=-5.25.$$
Thus $$-3\sqrt{x} > 3x-6$$ in $$0\le x\le 1$$, so for these $$x$$ the region that satisfies
$$y\le -3\sqrt{x}\quad\text{and}\quad y\ge 3x-6$$ lies between the two curves.

For $$x\gt 1$$ the inequality reverses, giving no overlap between $$y\le -3\sqrt{x}$$ and $$y\ge 3x-6$$.
Hence the only bounded part of the region is the strip
$$0\le x\le 1,\quad 3x-6\le y\le -3\sqrt{x}.$$

Now compute its area $$A$$ by integrating with respect to $$x$$:

$$ \begin{aligned} A &= \int_{0}^{1}\Bigl[\;y_{\text{top}}-y_{\text{bottom}}\Bigr]\,dx \\ &= \int_{0}^{1}\Bigl[\,-3\sqrt{x}\;-\;(3x-6)\Bigr]\,dx \\ &= \int_{0}^{1}\!\bigl(-3x^{1/2}-3x+6\bigr)\,dx. \end{aligned} $$

Integrate term by term:

$$ \begin{aligned} A &= \left[-2x^{3/2}-\dfrac{3}{2}x^{2}+6x\right]_{0}^{1} \\ &= \Bigl(-2(1)^{3/2}-\dfrac{3}{2}(1)^{2}+6(1)\Bigr) \;-\;\Bigl(0+0+0\Bigr) \\ &= \bigl(-2 -1.5 +6\bigr)=2.5=\dfrac{5}{2}. \end{aligned} $$

Finally, the question asks for $$6A$$:

$$6A = 6 \times \dfrac{5}{2} = 15.$$

Therefore, $$6A = 15.$$