In this video we will discuss how to derive the area
formula for a circle by integrating the formula for the
circumference of the circle.
Remember in an earlier video we discovered how the
derivative of the area formula for a circle is equal
to 2 pi r, which is the circumference of a circle.
And therefore, it should make sense if we integrate
the circumference we should be able to recover the
formula for the area.
To begin, we know the circumference of a circle equals
2 pi r, and the area equals pi r squared.
If we differentiate A with respect to r or find A
prime of r we apply the power rule of differentiation
by multiplying by two and then subtracting one from
the exponent which gives us dA, dr equals 2 pi r,
which again is the circumference.
If we write this in differential form by multiplying
both sides by differential r we have differential A
which represents the change in the area, is equal to
two pi r times differential r, where differential r
represents the change in the radius.
Looking at the circle on the right, if we focus on
this green ring, the thickness would be from a small
change in r and therefore the thickness is differential r.
The area of this green ring would represent the change
in the area with a small change in the radius, and the
change in the area is represented by differential A.
To find the area of the green ring, we find the
circumference and then multiply by differential r
shown here.
2 pi r times differential r represents the area of the
green ring which is also the change in the area,
represented by differential A.
So it should make sense if we wanted to determine the
area of the circle, for each small change in the radius
we could determine the area of the resulting ring and if we
sum the areas of the rings as the radius increases,
it will give us the total area.
Let's look at our animation to better understand this.
Again if our goal is to determine the area
of this red circle,
we can determine the area by summing the areas of the
rings as the radius increases.
So if we start with this very small ring here, with a
small change in the radius we can determine this area
and then as the radius increases we continue determining
the area of the rings and then summing these areas which
will result in the total area of the circle.
And now let's focus on the circle graphed on the coordinate
plane.
We will now call the thickness of each ring differential
x or delta x.
The sum of the areas of the rings with thickness delta x
or differential x from x equals zero to x equals r
gives the area of the circle.
And now let's set this up using calculus notation.
Let's first focus on this product here.
We have 2 pi times c sub i times delta x.
Each of these products represents the area of a ring
with a small change in the radius which now we denote
using differential x, or delta x, again because we
are now on the coordinate plane.
And c sub i is just some x value in each sub interval
along the x axis.
So 2 pi times c sub i is a circumference times delta x
which is the thickness of the ring gives us the area
of the ring and as we sum the areas of the rings,
it will give us the area of the circle.
As the number of rings approach infinity, we have the
limit as n approaches infinity of the sum from i equals
1 to n of 2 pi times c sub i times delta x which is
equal to the def integral from 0 to r of 2 pi x
differential x.
Notice how the integrating function is a circumference
given by 2 pi x.
The next step, let's factor out 2 pi so we have the area
equals 2 pi times integral from 0 to r of x differential
x and now we integrate.
Remember x is x to the first so we add one of the exponent
and then divide by two.
And now we find big f of b minus big f of a.
We first substitute r for x which gives us r squared
divided by 2, and then we substitute 0 for x, which
gives us 0 squared divided by 2 and now we find the
difference but this is 0, giving us 2 pi times r squared
divided by 2, simplifying 2 divided by 2, simplifies
to 1 and we do get the area equals pi r squared which
is the area formula for a circle.
I hope you found this helpful.