Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = 16x^2$ and $y = 25$ into two regions with equal area. Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = 16x^2$ and $y = 25$ into two regions with equal area. (Round your answer to two decimal places.)
My answer is $16.67$
Am I correct? 
 A: Draw a picture. The line $y=25$ meets our parabola at $x=\pm 5/4$. So the area of the region below the line, and above the parabola, is
$$\int_{-5/4}^{5/4} (25-16x^2)\,dx.$$
I think this  is $\frac{125}{3}$, but am error-prone. 
We want to choose $b$ so that the area below the line $y=b$ and above the parabola is $\frac{125}{6}$.
The line meets the parabola at $\pm \sqrt{b}/4$. So the area is
$$\int_{-\sqrt{b}/4}^{\sqrt{b}/4} (b-16x^2)\,dx.$$
I think this is $\frac{b\sqrt{b}}{3}$. Set this equal to $\frac{125}{6}$ and solve for $b$. We get $b=\frac{25}{2^{2/3}}$. This is some distance from $16.67$.
Remark: We calculated, but one can do better. By considerations of scaling, it is (after a while) clear that the area above a parabola $y=cx^2$ and below the line $y=a$ is a constant (involving $c$) times $a^{3/2}$. Let us not worry about the constant. The point is that to bisect the area below $y=w$ we use the line $y=\frac{w}{2^{2/3}}$. So we could have found the answer with essentially no computation, and we could similarly find the line $y=b$ that divides the full area in any desired ratio whatsoever.  
