General method for finding bisector of area between two curves

Question

I am looking for the general method for solving a problem like this:

Find a number, $b$, s.t. the line $y=b$ divides the region between the curves $y=x^2$ and $y=4$ into two regions with equal area.

Unfortunately, I don't have the solution available to check my thinking, but here's what I've tried.

Finding the area between the curves is pretty clear to me: $A(x)= \int_{-2}^2 4-x^2 dx $ since $x^2 = 4 \Leftrightarrow x=\pm2$, then $A(x)=\frac{32}{3}$ .

So now I initially think I want to find $y=b$ s.t. $\int_{-2}^{2} b-(4-x^2)dx = \frac{16}{3}$. But also, this doesn't seem right--the limits of integration seem out of place.

Next, I try the same strategy for determining the limits of integration above, $b-(4-x^2)=0 \Leftrightarrow x = \pm \sqrt{4-b}$, which gives me $\int_{-\sqrt{4-b}}^{\sqrt{4-b}} b-(4-x^2)dx = \frac{16}{3}$ and this leads to $\frac{3x}{3}+bx -4x \rvert_{-\sqrt{4-b}}^{\sqrt{4-b}} \rightarrow 2b\sqrt{4-b}$ so I have $2b \sqrt{4-b}=\frac{16}{3}$; however, this seems bad... Particularly, it appears there are no real roots to this equation. It's possible that I integrated wrong, but I haven't yet identified an error there.

So I step back, and try to turn the problem on its side, as it were. I note that $f(x) = 4-x^2 \Leftrightarrow f(y) = \pm\sqrt{4-y}$ so maybe I can say $\int_0^b \sqrt{4-y}dy = \frac{16}{3}$ and taking that integral gives me, after a bit of u-substitution, $\frac{2}{3}u^{3/2} \rvert_{4-b}^{4}$, and by setting that equal to the half the area as above, if my algebra is correct, gives me $b=16^{2/3} -4 \approx 2.35$

Now none of this quite gives me the sense that I know what I'm about. Am I even on the right track? Am I missing some more obvious approach?

My loose definition of the general problem: Given two curves on some interval $I$, $f(x), g(x)$ find a line $y=b$ s.t. the line divides the region into two regions of equal area, on the interval $I$. How do you tackle this problem, step by step, for any $f, g$?

Answer 1

You need to solve for $b$ the equation $$\frac{16}{3}=2\int^\sqrt{b}_0(b-x^2)dx$$ Which leads to $b=4^{\frac23}$

Answer 2

Suppose you are presented with a slightly different type of problem: you are given a function $f$, and you have to find $a$ such that the line $x=a$ divides the area under the curve $y=f(x)$ between $x=0$ and $x=4$ into two equal areas.

This is much easier! You compute $I=\int_0^4 f(x)\;dx$, and you look for $a$ such that $\int_0^a f(x)\;dx=I/2$.

But this is exactly the same as your problem, except that $x$ and $y$ have been swapped. So now try to solve the following problem:

Find a number, $b$, such that the line $x=b$ divides the region between the curves $y^2=x$ and $x=4$ into two regions with equal area.

This is in fact just what you ended up doing, but I hope it gives you a better sense of what you are about :-) In effect, you are reflecting the whole problem about the line $y=x$.