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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:

  1. $y=x^2$ and $y=c$;
  2. $y=x^2$, $y=b$ and $x=\sqrt c$;
  3. $y=x^2$, $y=4$ and $x=\sqrt b$.

Image showing shaded regions

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

  1. $\int_0^c \mathrm{\sqrt y},\mathrm{d}y$ --> (area bounded by the curve)
  2. $\int_c^b \mathrm{\sqrt y},\mathrm{d}y$ - $\sqrt c(b-c)$ --> (area bounded by the curve less area bounded by adj rectangle)
  3. $\int_b^4 \mathrm{\sqrt y},\mathrm{d}y$ - $\sqrt b(4-b)$ --> (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?

Lalit Tolani
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rwmiller
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  • Have you found any of the areas? Take a line $y = t$. What is the area bound by that line and the curve $y = x^2$ in first quadrant? – Math Lover Sep 05 '21 at 06:54
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    I've update the question with my progress so far. I think i'm able to find the first area, but struggling with the resultant equation when making the three areas equal to each other – rwmiller Sep 05 '21 at 07:58
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    Two equations with two variables. You should get cubic equation at end, taking, for example, $x=c^{1/2},y=b^{1/2}$. – zkutch Sep 05 '21 at 09:12

1 Answers1

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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 $$c\approx0.7922892326$$

David Quinn
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