General method for finding bisector of area between two curves

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 $f, g$?

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

• Can you elaborate a bit on where that comes from?
– Ben
Commented Jun 16, 2015 at 18:36
• You have the area between the curve and the line $y=4$ as $\frac{32}{3}$ and the area between the curve and the horizontal line $y=b$ is half this and is given by the integral. The upper limit is $\sqrt b$ and the lower limit is 0 because the function is even (that's why we multiply by 2). Drawing a picture makes it clearer. Commented Jun 16, 2015 at 18:42
• I'm still not quite following. I don't exactly see why the upper limit is $\sqrt{b}$. I would think that $\int_0^b (b-x^2)dx$ might represent the area under $y=b$ and above $y=x^2$ and by setting that = $\frac{16}{3}$ would get me the answer... obviously that doesn't work, but I don't quite get how you're getting to the answer you have.
– Ben
Commented Jun 16, 2015 at 19:02
• The upper limit on the integral is where the curve meets the line $y=b$ Commented Jun 16, 2015 at 19:44
• Ok, I see how this problem is solved. I will not go so far as to say that I have a great sense of what the general method for this sort of problem is.
– Ben
Commented Jun 16, 2015 at 20:03

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 $$y=x$$.