1. Problem: Find the area bounded by the curves $y^2 = 9x$ and $y = 3x$. 2. Formula and rules: The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx.$$ We first find the points of intersection to determine $a$ and $b$. 3. Find intersection points: From $y^2 = 9x$, we have $x = \frac{y^2}{9}$. Substitute $y=3x$ into $y^2=9x$: $$ (3x)^2 = 9x \Rightarrow 9x^2 = 9x \Rightarrow x^2 = x \Rightarrow x(x-1)=0.$$ So $x=0$ or $x=1$. Corresponding $y$ values: $y=3(0)=0$ and $y=3(1)=3$. 4. Express curves in terms of $y$ to integrate with respect to $y$: From $y^2=9x$, $x=\frac{y^2}{9}$. From $y=3x$, $x=\frac{y}{3}$. 5. The area between the curves from $y=0$ to $y=3$ is: $$\int_0^3 \left( \frac{y}{3} - \frac{y^2}{9} \right) dy = \int_0^3 \left( \frac{y}{3} - \frac{y^2}{9} \right) dy.$$ 6. Calculate the integral: $$\int_0^3 \frac{y}{3} dy - \int_0^3 \frac{y^2}{9} dy = \left[ \frac{y^2}{6} \right]_0^3 - \left[ \frac{y^3}{27} \right]_0^3 = \frac{9}{6} - \frac{27}{27} = 1.5 - 1 = 0.5.$$ 7. Final answer: The area bounded by the curves is $\boxed{0.5}$.