Coached Exercise: Alternative Method for the Area of a Circle

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The derivation of the area of a circle you just covered is interesting and was very ingenious for Archimedes. Let's try from another direction to obtain the same result, namely the area of the circle is πR² .

Begin with any circle of radius R. Consider a concentric (same center) circle of radius r. The idea that will lead to the area of the entire circle is to construct a loop of width dr at radius r and determine the area of the loop and then add up the areas of all the loops from r=0 to r=R to obtain the the area of the entire circle.

Step 1: Draw a diagram. Make sure it that it is large enough to work with, that it has the generic loop at radius r and that the width of the loop is labeled. To see the result click here.

Pretty straight forward step.

Step 2: Write the area of the small loop. Remember that your expression will use dr for the width. To see the answer click here.

If you imagine stretching the loop out into a band, then you can ask what is the area of the band (and consequently the area of the loop). The width of the band is dr and the the length of the band is 2πr.

You may wonder why the length is 2πr while the top of the band appears to be slightly longer than the bottom of the band (ie. the outer circle has a slightly larger circumference than the inner circle). Well if the loop had a width of Δr then the outer circle would have a larger circumference, but as Δr goes to zero the difference between the two circumferences approaches zero.
Observe that the inner radius of the loop is r and the outer radius is r + Δr. So the inner circumference is 2πr and the outer circumference is 2π(r + Δr). Taking the limit of the difference in circumference:
lim 2π(r + Δr) - 2πr = Δr → 0 lim 2πΔr = 0 Δr → 0

Indeed in the limit as Δr goes to zero, the difference is zero (ie. the circumference is 2πr) and Δr becomes dr. This is what is meant by dr being the limit as Δr approaches zero.

Can you see what to do next? We have the area of a single loop at radius r and we can construct the entire circle out of many (and in the limit an infinite number of) loops.

Step 3: Add up the all of the loops from r=0 to r=R to obtain the the area of the entire circle. Drawing a diagram may be helpful. To see the answer click here.

To see what is going on here refer to the diagram on the right.

If you divide the circle into n loops then

          (R - 0)
   Δr  =         
             n

and the inner radius, r_k, of the kth loop is r_k = (k-1)Δr. So the total approximate area will be

A_total =

n ∑ 2π((k-1)Δr)Δr = k=1

n ∑ 2πr_kΔr k=1

which is a Riemann sum. You take the limit of a Riemann sum as n goes to infinity by doing the definite integral. Hence exact total area becomes

A_total =

R r=0

2πr dr

By the way: Can you see how, by unrolling each loop and laying the loops out flat, one on top of the other, this problem becomes equivalent to finding the area of a triangle whose height is r and whose base is 2πr?

Step 4: Solve the integral. The 2π is constant and the function being integrated is polynomial. To see the result click here.

R 0

2πr dr =

r² 2π 2

R 0

Thus the area of the circle is πR² , as expected. Isn't calculus and in particular integration cool? We started off with the circumference of the circle and ended up with the area of the circle. Armed with a knowledge of π to an arbitrary number of digits we can find the area of any circle to an arbitrary number of digits.

As an extension, consider what would happen if the circle had a hole (another concentric circle with a radius of a say). How would the integration change? How would the resulting area change?

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