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$\begingroup$

enter image description here

The blue region above is called an asymmetric lens.

Both small circles have radius 1, both large circles have radius 2, and the (four) circles are at center separation $r$.

The area of the blue region is given by an integral of arc lengths: $$ \int_{r-1}^{2} 2 x \arccos\left(\frac{x^2 + r^2 - 1}{2 r x}\right) \, dx $$ moving from left to right. Integration by parts gives this (blue area) as $$ \frac{1}{2} \left(8 \sec ^{-1}\left(\frac{4 r}{r^2+3}\right)-\sqrt{-r^4+10 r^2-9}-2 \tan ^{-1}\left(\frac{r^2-3}{\sqrt{-r^4+10 r^2-9}}\right)+\pi \right) $$ but is there a more elegant way?

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    Split by the line between the corners and use the formula for the circular segment. – 2017-01-12
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    The area is what it is, and integration by parts _is_ elegant. Do you want another formula? – 2017-01-12
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    This formula may simplify? – 2017-01-12
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    @Yves What is the "line between the corners"? – 2017-01-12

2 Answers 2

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You can add the area of two segments. enter image description here

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    So this red line is the area between the corners I see. – 2017-01-12
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    @AlexanderGiles, yes they are the intersection points. – 2017-01-12
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    OK thanks this is clear. – 2017-01-12
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$$\frac{\theta_l-\sin\theta_l}2r_l^2+\frac{\theta_r-\sin\theta_r}2r_r^2,$$

where

$$\cos\frac{\theta_l}2=\frac{d^2-r_r^2+r_l^2}{2d},\cos\frac{\theta_r}2=\frac{d^2-r_l^2+r_r^2}{2d}.$$

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    Ok, I'm trying to integrate a function over this region, so this helps. – 2017-01-12