Key Concepts and Formulas

• Area Between Curves: If $x = f(y)$ and $x = g(y)$ are continuous functions and $f(y) \geq g(y)$ on the interval $[c, d]$, then the area of the region bounded by the curves $x = f(y)$, $x = g(y)$, and the lines $y = c$ and $y = d$ is given by $A = \int_c^d [f(y) - g(y)] \, dy$.
• Finding Intersection Points: To find the intersection points of two curves, set their equations equal to each other and solve for the variable.
• Quadratic Formula/Factoring: To solve a quadratic equation of the form $ax^2 + bx + c = 0$, either factor the quadratic or use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step 1: Express the Equations in Terms of $x$ as Functions of $y$

The given region is defined by $0 \leq 9x \leq y^2$ and $y \geq 3x - 6$. We need to rewrite these inequalities to express $x$ as a function of $y$.

• From $0 \leq 9x \leq y^2$, we get $x \geq 0$ and $x \leq \frac{y^2}{9}$. So, we have $x = \frac{y^2}{9}$ as one bounding curve.

• From $y \geq 3x - 6$, we can rewrite it as $3x \leq y + 6$, which gives us $x \leq \frac{y + 6}{3}$. So, we have $x = \frac{y + 6}{3}$ as the other bounding curve.

We also have $x \geq 0$ as a boundary. Thus, the region is bounded by the parabola $x = \frac{y^2}{9}$, the line $x = \frac{y+6}{3}$, and the $y$-axis ($x=0$).

Step 2: Find the Intersection Points of the Curves

To find the area, we need to determine the points of intersection between the parabola $x = \frac{y^2}{9}$ and the line $x = \frac{y+6}{3}$. We set the two expressions for $x$ equal to each other: $\frac{y^2}{9} = \frac{y+6}{3}$

Multiplying both sides by 9 to eliminate the fractions, we get: $y^2 = 3(y+6)$ $y^2 = 3y + 18$ $y^2 - 3y - 18 = 0$

Factoring the quadratic, we get: $(y-6)(y+3) = 0$

So, the $y$-coordinates of the intersection points are $y = 6$ and $y = -3$. The corresponding $x$-coordinates are:

• When $y = 6$, $x = \frac{6^2}{9} = \frac{36}{9} = 4$.
• When $y = -3$, $x = \frac{(-3)^2}{9} = \frac{9}{9} = 1$.

The intersection points are $(4, 6)$ and $(1, -3)$.

Step 3: Determine the Limits of Integration and the Rightmost and Leftmost Curves

Our limits of integration will be $y = -3$ and $y = 6$. Within this interval, we need to determine which function is "rightmost" (larger $x$-value) and which is "leftmost" (smaller $x$-value).

Consider a $y$ value between -3 and 6, say $y = 0$.

• For the parabola, $x = \frac{0^2}{9} = 0$.
• For the line, $x = \frac{0+6}{3} = 2$.

Since $2 > 0$, the line $x = \frac{y+6}{3}$ is to the right of the parabola $x = \frac{y^2}{9}$ in the region of interest. Also, since $x \geq 0$, we don't need to consider the $y$-axis as a boundary.

Step 4: Set Up and Evaluate the Integral

The area $A$ of the region is given by the integral: $A = \int_{-3}^{6} \left( \frac{y+6}{3} - \frac{y^2}{9} \right) dy$

$A = \int_{-3}^{6} \left( \frac{y}{3} + 2 - \frac{y^2}{9} \right) dy$

Now, we integrate: $A = \left[ \frac{y^2}{6} + 2y - \frac{y^3}{27} \right]_{-3}^{6}$

Evaluate the expression at the limits of integration: $A = \left( \frac{6^2}{6} + 2(6) - \frac{6^3}{27} \right) - \left( \frac{(-3)^2}{6} + 2(-3) - \frac{(-3)^3}{27} \right)$ $A = \left( \frac{36}{6} + 12 - \frac{216}{27} \right) - \left( \frac{9}{6} - 6 - \frac{-27}{27} \right)$ $A = \left( 6 + 12 - 8 \right) - \left( \frac{3}{2} - 6 + 1 \right)$ $A = \left( 10 \right) - \left( \frac{3}{2} - 5 \right)$ $A = 10 - \left( \frac{3}{2} - \frac{10}{2} \right)$ $A = 10 - \left( -\frac{7}{2} \right)$ $A = 10 + \frac{7}{2}$ $A = \frac{20}{2} + \frac{7}{2} = \frac{27}{2}$

Step 5: Calculate $6A$

The question asks for the value of $6A$. $6A = 6 \times \frac{27}{2}$ $6A = 3 \times 27$ $6A = 81$

Common Mistakes & Tips

• Incorrectly identifying the rightmost and leftmost curves: Always test a point within the interval to determine which curve is to the right (larger $x$-value) and which is to the left (smaller $x$-value).
• Integration errors: Double-check your integration and evaluation of the definite integral.
• Not expressing $x$ as a function of $y$: When integrating with respect to $y$, ensure that your equations are in the form $x = f(y)$.

Summary

We found the area of the region bounded by the given curves by first expressing the equations as functions of $y$, finding the intersection points, setting up the definite integral with respect to $y$, and evaluating the integral. Finally, we multiplied the area by 6 to obtain the desired result.

The final answer is $\boxed{81}$.