# Area between two curves. Split so that it is equal.

There was a question I was doing: Find $y = b$ (find $b$) so that it splits the area between $y=4$ and $y = x^2$ into 2 equal areas.

I found the solution which they integrate it to y axis. And discard the other half since it's symmetrical. That made sense. What I did was find the area between the two curves first, which gave me $32/3$. I halved that which gave me $16/3$. Shouldn't this be half the area? Then I integrated $\int^2_{-2}(4-b)dx$ and set that to $16/3$. So that gave me $(8-2b)-(-8+2b) = 16/3$ I solved that and got $8/3$. The answer is different from the other method. Shouldn't the way I did it give me the correct answer? The correct answer is $4^{(2/3)}$.

• This is a Putnam problem? – Yunus Syed Nov 22 '15 at 6:28
• @YunusSyed, I believe it's from the Briggs calculus text; I don't recall it being marked as a competition problem or anything, though... – dmk Nov 22 '15 at 6:32

So your initial evaluation of the area is correct. $\int_{-2}^2(4-x^2)dx$ is indeed $32/3$. However it looks like you made a mistake setting up your second integral. If you look at the integral $\int_{-2}^2(4-b)dx$, you'll see that the area that it is evaluating is the rectangle of height $4-b$ and length $4$ ($-2$ to $2$). This isn't the area you want to find, since the edges of the area lie on the parabola $x^2$ (thus the area can't be a rectangle). You can solve this issue by evaluating the area bounded below by $y=x^2$ and above by $y=b$. Setting the integral equal to $16/3$ and solving for $b$ will give the correct answer, which is in fact $4^{2/3}$.
$\int_{-\sqrt{b}}^\sqrt{b}(b-x^2)dx = bx - \frac{1}{3}x^3|_{-\sqrt{b}}^\sqrt{b} = \frac{4}{3}b^{3/2}$. Now $\frac{4}{3}b^{3/2} = \frac{16}{3}$ and $b=4^{2/3}$.
You are right when you said you need to integrate it with respect to $y$ axis as the bounded area between $y=4$ and $y=x^2$ will be divided symmetrically by $y$ axis as $y=x^2$ is an even function First, we need to change the given function to a explicit form with respect to $y$ $$y=x^2 \implies x=\sqrt y$$ Since area will be equally bisected by the line, $y=b$, it implies $$\int_0^b \sqrt y dy=\int_b^4 \sqrt y dy$$ $$\frac{2b^{3/2}}{3}=\frac{2(4)^{3/2}-2b^{3/2}}{3}$$ $$4b^{3/2}=2(4^{3/2})$$ $$b^{3/2}=2(4^{1/2})$$ $$b^{3/2}=4$$ $$b=4^{2/3}$$
You actually do not need Calculus to solve this problem. It is well-known (at least from Archimedes' time) that given a parabolic segment with base length $l$ and height $h$ (the base being perpendicular to the axis), its area is $A=\frac{2}{3}hl$. We are cutting segments from the parabola $y=x^2$, hence $l$ and $h$ are connected via $h=\frac{l^2}{4}$ and $$A(h)=\frac{4}{3}h^{3/2} = 2\cdot A\left(\frac{h}{\sqrt{4}}\right).$$ This $\sqrt{4}$ factor clearly gives the position of the splitting line and proves that the given problem cannot be solved by straightedge and compass, since it is equivalent to halving a cube.