Split area defined by curve into equal parts I have a curve $y=x^2$ defined for the region: $0\le x\le2$.
What's the best way to work out the values $b$ and $c$ so that the three areas defined by the shaded regions are equal? I believe this is the regions defined between:

*

*$y=x^2$ and $y=c$;

*$y=x^2$, $y=b$ and $x=\sqrt c$;

*$y=x^2$, $y=4$ and $x=\sqrt b$.


This where I'm at so far at calculating the area's:

*

*$\int_0^c \mathrm{\sqrt y},\mathrm{d}y$ --> (area bounded by the curve)

*$\int_c^b \mathrm{\sqrt y},\mathrm{d}y$ - $\sqrt c(b-c)$ --> (area bounded by the curve less area bounded  by adj rectangle)

*$\int_b^4 \mathrm{\sqrt y},\mathrm{d}y$ - $\sqrt b(4-b)$ --> (area bounded by the curve less area bounded  by adj rectangle)

and if $\int \mathrm{\sqrt y},\mathrm{d}y$ = $\frac{{y^3/_2 }}{3/2}$ making the three areas equal to each other get:
$\displaystyle\frac{{c^{3/2} }}{3/2}  =  \frac{{b^{3/2} }}{3/2} - \frac{{c^{3/2} }}{3/2}- bc^{1/2} + c^{3/2}  =  \frac{{4^{3/2} }}{3/2} - \frac{{b^{3/2} }}{3/2} - 4^{1/2} + b^{3/2}$
But then i'm stuck solving the resultant equation..
Can anybody confirm if this is the right way to approach it? if so what's the best way to solve for b and c?
 A: There is a solution to this question but it involves solving two cubic equations numerically so the values found for $b$ and $c$ are approximate.
Writing $b=p^2$ and $c=q^2$, the first leftmost area is:
$$A_1=\frac23p^3$$
The next middle area is:
$$A_2=q^2(q-p)-\frac13(q^3-p^3)=\frac13(2q^3-3pq^2+p^3)$$
The rightmost area is
$$A_2=4(2-q)-\frac13(2^3-q^3)=\frac13(q^3-12q+16)$$
The three areas are equal, so
$$A_1=A_2\implies\frac23p^3=\frac13(2q^3-3pq^2+p^3)$$
This simplifies to become a cubic in $\frac{p}{q}$, namely
$$\left(\frac{p}{q}\right)^3+3\left(\frac{p}{q}\right)-2=0$$
This has one real solution:
$$\frac{p}{q}=0.596071638...$$
Wolfram Alpha provides the exact form $$\frac{(1+\sqrt{2})^{\frac23}-1}{(1+\sqrt{2})^{\frac13}}$$
Now equating $A_1$ and $A_3$ leads to a cubic equation in $q$, namely
$$0.5764298279q^3-12q+16=0$$
This has three real roots, but only one of these makes sense: $$q\approx1.493287463$$
So the final approximate answers for $b$ and $c$ are:
$$b\approx 2.229907447$$ and $$c\approx0.7922892326$$
