A park designer draws a flower bed between two curved edges…

Step 1. Meeting points. Put $y = \dfrac{x^2}{6}$ into $y^2 = 6x$. Then $\dfrac{x^4}{36} = 6x$, so $x(x^3 - 216) = 0$. The solutions are $x = 0$ and $x = 6$.

Step 2. Which is above? At $x = 1$ the first curve gives $\sqrt{6} \approx 2.45$ and the second gives $\dfrac{1}{6}$. So $y = \sqrt{6x}$ is above.

Step 3. The area. $\displaystyle\int_0^6 \left(\sqrt{6x} - \dfrac{x^2}{6}\right) dx = \sqrt{6} \times \dfrac{2}{3} \times 6^{3/2} - \dfrac{216}{18} = 24 - 12 = 12$.