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 _____

Step 1: Identify the curves and intersection points

The region is bounded by two curves: a parabola $x = \frac{y^2}{9}$ and a straight line $x = \frac{y+6}{3}$. To find the area of the region, we first need to determine the points where these two curves intersect. We can do this by setting their $x$ values equal to each other.

Step 2: Find the intersection points

To find the intersection points, we equate the expressions for $x$. This gives us a quadratic equation in $y$. Solving this quadratic equation by factoring, we find two values for $y$: $y=6$ and $y=-3$. These are the $y$ -coordinates of the intersection points.

Step 3: Determine the x-coordinates of intersection

Now we substitute the $y$ -values back into either of the original equations to find the corresponding $x$ -coordinates. For $y=6$, $x=4$, giving the point $(4{,}6)$. For $y=-3$, $x=1$, giving the point $(1,-3)$. These are the two points where the line and the parabola intersect.

Step 4: Set up the integral for the area

Since the line $x = \frac{y+6}{3}$ is to the right of the parabola $x = \frac{y^2}{9}$ in the region of interest, we integrate the difference of the $x$ -values with respect to $y$. The limits of integration are the $y$ -coordinates of the intersection points, from $-3$ to $6$.

Step 5: Evaluate the integral

We now evaluate the definite integral. First, we find the antiderivative of each term. Then, we substitute the upper limit $y=6$ and the lower limit $y=-3$ into the antiderivative and subtract the results. This calculation yields the area $A = \frac{27}{2}$.

Step 6: Calculate 6A

The problem asks for the value of $6A$. We multiply the calculated area $A = \frac{27}{2}$ by $6$ to get the final answer.

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