Moon area approximation using Calculus
A few days ago, a friend sent me a math problem she had received from another friend of hers. At a first glance, this mensuration problem looked quite simple but apparently, “it wasn’t as easy as it looked,” she said. She is a math major and so returning without a solution was not an option; as the good friend I am, I stopped whatever I was doing (which was honestly nothing) and hopped on a zoom call with her to see what we could come up with.
The problem is to find the area of the shaded region (which I named moon because of its resemblance to a crescent moon shape)
The Solution
After half an hour of continuously lighting up with ideas and later realising they were dead ends, I decided to try calculating the area of the shaded region using calculus. I graphed the two circles on Desmos online graphing tool to get a clearer picture.
From figure 2, the difference between the area of the smaller sector in Cs(any mention of a sector elsewhere refers to this sector) and the area of the red and yellow regions (calculated using calculus) is the area we want to calculate. Using a calculator, we were able to approximate the area of the blue region to be 14.6 square units (3 sig figs). Later on another friend of ours came up with a purely geometric solution — which was quite elegant — I won’t talk about in this post.
Using my methodology, I decided to devise a formula that would make life easier for whoever comes about this problem in the future.
The Generalisation
Let’s assign r as the radius of the bigger circle. The radius of the smaller is therefore r/2. We’ll begin by finding the area of the red and yellows regions and we can only do this if we have the equations Cb, L1 and L2. The Equation of Cb is x²+y² =r²; let’s work towards finding the equation of the lines.
Finding the coordinate of points M and N
To find the equation of the lines, we’ll need to find the gradient of the lines and at least one point that lies in the lines. Using the coordinates of points M and N, we can calculate the gradients of L1 and L2 respectively.
In figure 3, only the positive values of y are considered when expressing x and r in terms of y. This is valid because in figure 2, both points are above the y-axis and thus have positive y-coordinates.
In figure 2, it can be seen that the x-coordinate of point M is less negative than that of point N and so the greater x value calculated is the x-coordinate of point M.
Now that we have calculated the coordinates of points M and N, we can go ahead and find the equations of the lines.
Finding the equation of lines L1 and L2
In figure 8, it can be seen that radius r gets cancelled out. I was honestly not surprised because the sectors that can be formed given r = (0,ထ) are similar. This means the angle inside the sector will be equal for all of such sectors and so will the gradient of the two lines that form the angle.
The square of a term in the form (a+b√c), where a, b, c are integers and c>0, is of the same form as its square root. This observation prompted me to do the algebraic manipulation in line 10 of figure 8 and line 8 of figure 9, so that I can eliminate the square root and easily simplify the gradients.
But how did I find the square root? Well, it was some sort of reverse technique: you reverse the process of squaring the sum of two numbers.
Now that we have found the equation of the lines, we can go ahead and find the area of the red and yellow regions. We’re going to use calculus to do that. We’ll find the area of the region under Yb from point N to point M and then subtract from that the area under L2 from point N to the point -r/2 and that of L1 from point -r/2 to point M.
Area of red and yellow regions
Let’s ignore the limits and integrate the equations to produce the area equations.
The integration method used here is trig substitution.
Now that we have integrated all the equations, we can go ahead and insert the limits into the area equations.
We have calculated the areas under Yb, L1 and L2. Let’s go ahead and insert them into the area formula in figure 11 to find the area of the red and yellow regions.
Next, we’ll find the area of the whole sector.
Area of sector
The angle, θ, is in radians.
𝜋-θ is the angle inside the sector. This can deduced from figure 2.
Let’s go ahead and find the area of the moon-shaped region (violet region).
Area of moon-shaped region (violet region)
The area of the violet region is the difference between the area of the sector and the areas of the red and yellow regions. I wish there was a way to simplify the constant in the bracket: anyone with an idea can comment it and I’ll gladly check it out.
For the engineers, this is a more friendly form of the general formula. You’re welcome :)
Conclusion
Using calculus we have been able to devise a general formula for finding the area formed outside the bigger circle when a smaller circle whose radius is half that of the bigger circle is inscribed in a quadrant of the bigger circle. To find the area, on has to multiply the square of the radius of the bigger circle (or twice the area of the smaller circle) by the constant 0.14638... I’ll name this constant the Bosome Constant. *“Bosome” is the Akan name for moon*
In a case where the smaller circle is inscribed in another quadrant of the bigger circle, the formula can still be used. This is because the 1st, 3rd and 4th quadrants are just reflections of the 2nd quadrant with y-axis, x-axis and y=x as mirrors respectively.
Ultimately, I enjoyed this tedious work: I learned a number of nice algebraic techniques.
Leave the thanks in the comment section.
