Cutting a Parabolic Region in Half

The line and curve $y = b$ and $y = ax^2$ intersect at $(\pm \sqrt{\frac{b}{a}}, b)$ , and since the area of a parabola is $\frac{2}{3}$ the area of its inscribed rectangle, the area of the large parabola is $A_L = \frac{2}{3} \cdot 2\sqrt{\frac{b}{a}} \cdot b = \frac{4b\sqrt{b}}{3\sqrt{a}}$ .

Similarly, the area of the small parabola bounded by $y = c$ and $y = ax^2$ is $A_S = \frac{4c\sqrt{c}}{3\sqrt{a}}$ .

To cut the region in half, $2A_S = A_L$ , or $2 \frac{4c\sqrt{c}}{3\sqrt{a}} = \frac{4b\sqrt{b}}{3\sqrt{a}}$ , and this solves to $c = \boxed{2^{-\frac{2}{3}}b}$ .

From the given conditions of the problem we get $\displaystyle \int_0^b \sqrt y dy=2\displaystyle \int_0^c \sqrt y dy\implies b^{\frac{3}{2}}=2c^{\frac{3}{2}}\implies c=\boxed {2^{-\frac{2}{3}}b}$ .