You're on a roll — Step 1: Set up the integral for the area between the curves. The problem provides the equations for the two curves, $y_1 = \sqrt{ax}$ and $y_2 = \frac{x^2}{a}$, and the integral setup: $$ A = \int_{0}^{a} (y_1 - y_2) \,dx $$ Substitute the expressions for $y_1$ and $y_2$: $$ A = \int_{0}^{a} \left(\sqrt{ax} - \frac{x^2}{a}\right) \,dx $$ Step 2: Rewrite the terms for easier integration. $$ \sqrt{ax} = \sqrt{a} \cdot \sqrt{x} = \sqrt{a} x^{1/2} $$ $$ \frac{x^2}{a} = \frac{1}{a} x^2 $$ So the integral becomes: $$ A = \int_{0}^{a} \left(\sqrt{a} x^{1/2} - \frac{1}{a} x^2\right) \,dx $$ Step 3: Integrate each term. Recall the power rule for integration: $\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$. $$ \int \sqrt{a} x^{1/2} \,dx = \sqrt{a} \frac{x^{1/2+1}}{1/2+1} = \sqrt{a} \frac{x^{3/2}}{3/2} = \frac{2}{3} \sqrt{a} x^{3/2} $$ $$ \int \frac{1}{a} x^2 \,dx = \frac{1}{a} \frac{x^{2+1}}{2+1} = \frac{1}{a} \frac{x^3}{3} = \frac{x^3}{3a} $$ Step 4: Evaluate the definite integral from $0$ to $a$. $$ A = \left[ \frac{2}{3} \sqrt{a} x^{3/2} - \frac{x^3}{3a} \right]_{0}^{a} $$ Substitute the upper limit $x=a$: $$ \left( \frac{2}{3} \sqrt{a} (a)^{3/2} - \frac{(a)^3}{3a} \right) $$ Simplify the terms: $$ \frac{2}{3} a^{1/2} a^{3/2} - \frac{a^3}{3a} = \frac{2}{3} a^{(1/2 + 3/2)} - \frac{a^2}{3} = \frac{2}{3} a^{4/2} - \frac{a^2}{3} = \frac{2}{3} a^2 - \frac{a^2}{3} $$ Substitute the lower limit $x=0$: $$ \left( \frac{2}{3} \sqrt{a} (0)^{3/2} - \frac{(0)^3}{3a} \right) = 0 - 0 = 0 $$ Subtract the lower limit value from the upper limit value: $$ A = \left( \frac{2}{3} a^2 - \frac{a^2}{3} \right) - 0 $$ $$ A = \frac{2a^2 - a^2}{3} = \frac{a^2}{3} $$ The final answer is $\boxed{\frac{a^2}{3}}$. What's next?