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In the diagram,the curved paths are arcs of circles centered at vertices $A$ and $B$ of a square of side $6$. Find the area of the shaded section $BCD$.

I've been stuck on this problem for days. I have tried to get the shaded section by puzzling with the figures but failed. I've tried to add more lines to get a set of equations but that failed. But I still want to be able to solve it by myself so if you guys can give me only hints, that would be best. enter image description here

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  • $\begingroup$ One possible way should be splitting the area with the normal on BC through A. $\endgroup$
    – coproc
    Aug 18, 2015 at 13:51
  • $\begingroup$ Are you familiar with integrals? $\endgroup$
    – anak
    Aug 18, 2015 at 14:14
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    $\begingroup$ Since you can calculate the area of BCA, it suffices to compute BDA and subtract. Hint: BDA is almost a sector of the circle centered on A, but DA is curved. Similarly, it's almost a sector of the circle centered on B. If you computed the area of the triangle ΔBDA, as well as the sectors, how would that help you? $\endgroup$
    – Akiva Weinberger
    Aug 18, 2015 at 14:54
  • $\begingroup$ Here’s another hint. imgur.com/5NHedVO The pink area and the green area in this picture are “segments” of a circle or radius 6. You can find their areas if you can figure out the angle of the arc they span. Then the area you want is a triangle (one side is the radius of a circle, and you can probably figure out the angles) plus the green segment plus half the pink segment. $\endgroup$
    – Steve Kass
    Aug 18, 2015 at 15:00

1 Answer 1

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Here is a hint with double integrals. Split the area in two:

enter image description here

Then

\begin{align} S_{CBED}&= S_{CED}+S_{CBE} = \int_{x_1}^{x_2}\int_{f_2(x)}^{f_1(x)} dy dx + \int_{x_2}^{x_3}\int_{f_3(x)}^{f_1(x)} dy dx \end{align} where

\begin{align} f_1(x)&=\sqrt{6^2-(x-6)^2}, \\ f_2(x)&=6-\sqrt{6^2-(x-6)^2}, \\ f_3(x)&=x, \\ x_1&=6-3\sqrt{3}, \\ x_2&=6-3\sqrt{2}, \\ x_3&=6. \end{align}

$S_{CBED}=\tfrac{15}{2}\pi-9\sqrt{3}\approx 7.97348763$.

Another way to split the area:

enter image description here

suggests a geometric solution as a sum of the sector $BCD$ and a difference between the sector $ABD$ and $\triangle ABD$.

$\triangle ABD$ is equilateral (why?), so

\begin{align} S_1&=\tfrac12 \cdot 36(\tfrac\pi3-\tfrac\pi4)=\tfrac32 \pi \\ S_2&=\tfrac12 \cdot 36\tfrac\pi3=6\pi \\ S_3&=9\sqrt3 \end{align}

And the answer is $\tfrac{15}2\pi-9\sqrt3$, as above.

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  • $\begingroup$ thank you for showing both methods. $\endgroup$
    – Nameless
    Aug 18, 2015 at 16:19

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