In this video,
we will derive the area of a circle
using integration in the equation of a circle
x squared plus y squared equals r squared.
To begin, here is a graph of the circle
centered at the origin with a radius of r.
Let's use the symmetry of the circle
on the coordinate plane when integrating.
For example, let's determine the area in the first quadrant,
which is 1/4 of the circle.
Then if we multiply this area by four,
we will have the total area of the circle.
Let's decide to integrate with respect to x,
and therefore we will need to solve the equation
x squared plus y squared equals r squared
for y so that we have y as a function of x.
So beginning with the equation x squared
plus y squared equals r squared,
let's subtract x squared on both sides,
which gives us y squared equals
r squared minus x squared.
Now to solve for y, we take the square root
of both sides of the equation.
But when doing this,
we will have a plus or minus on the right.
Simplifying, we have y equals plus or minus the square root
of the quantity r squared minus x squared,
where the positive square root gives the top of the circle
and the negative square root gives the bottom of the circle.
But because we are using the symmetry
and going to determine the area in the first quadrant
and then multiply it by four,
we will use the function y equals the positive square root
of the quantity r squared minus x squared.
And therefore, when we set up the integral,
notice how integrand function is the square root
of the quantity r squared minus x squared.
Then we have dx because
we're integrating with respect to x along the x-axis.
Limits of integration are going to be from zero to r
because we already said the circle is r.
And then again, we are multiplying by four
because the integral gives us the area in the first quadrant
or 1/4 of the area.
Multiplying by four gives us the area of the entire circle.
And now to evaluate the integral,
we'll perform a change of variables,
where we let x equal r sine theta,
and therefore differential x, or dx,
equals r cosine theta d theta.
Remember dx is equal to the derivative of r sine theta
with respect to theta times d theta.
Performing this substitution gives us A
equals four times the integral.
Now these limits of integration are for x.
We will need to determine new limits of integration
in terms of theta after performing the substitution.
Performing the substitution, we have the square root
of r squared minus,
x squared is going to be the square of r sine theta,
which gives us r squared sine squared theta.
And then dx equals r cosine theta d theta.
And now because we're integrating with respect to theta,
our limits of integration must be in terms of theta.
And because we're integrating in the first quadrant,
theta is going to be from zero radians
to the positive y-axis, which is pi over two radians.
So now the limits of integration are from zero
to pi over two.
We could also determine limits of integration
by substituting zero for x in this equation
and solving for theta,
to determine the lower limit of integration.
Then to determine the upper limit of integration,
we could substitute r for x and again solve for theta.
For the next step,
let's factor r squared minus r squared sine squared theta
by factoring out r squared.
This gives us the square root
of r squared times the quantity
one minus sine squared theta,
and we still have r cosine theta d theta.
And now using the identity
sine squared theta plus cosine square theta equals one,
if we subtract sine squared theta on both sides,
we have one minus sine squared theta
equals cosine squared theta.
So let's substitute cosine squared theta here.
And the square root of r squared is equal to r.
Because we have another factor of r here,
we are going to have an r squared outside the square root.
So this is equal to four
times the integral from zero to pi over two.
Let's show some extra work here.
We have r from the square root of r square,
and then we still have the square root of,
one minus sine squared theta is equal
to cosine squared theta.
And then we still have r cosine theta d theta.
Well, r times r is equal to r squared,
and the square root cosine squared is equal
to one factor of cosine times this factor of cosine theta,
which gives us cosine squared theta.
So this gives us four times the integral
from zero to pi over two of r squared
cosine squared theta d theta.
Let's continue on the next slide.
For the next step,
let's perform a substitution for cosine squared theta.
Looking at the identities here below,
cosine squared theta equals 1/2
times the quantity one plus cosine two theta.
We can also factor out r squared
because we are integrating with respect to theta not r.
So now we have the area is equal to four
r squared times the integral from zero to pi over two.
And then again, we're performing a substitution
for cosine squared theta,
which again is equal to 1/2
times the quantity one plus cosine two theta.
Let's factor out the 1/2, so we can divide this by two.
Integrand function is now one plus cosine two theta.
And now let's break this up into two separate integrals
because to integrate cosine two theta,
we will need to perform u-substitution.
Of course, we can also simplify
four r squared divided by two, which is two r squared.
So we have the area equals two r squared
times
the integral from zero to pi over two of one d theta
plus the integral from zero to pi over two
of cosine two theta d theta.
The integral of one respect to theta is theta.
Then we have plus,
and now we need to integrate cosine two theta,
which again requires u-substitution,
where U is equal to two theta.
So if u is equal to two theta,
du is equal to two d theta.
Dividing both sides by two,
we have 1/2 du equals d theta.
So in terms of u,
we can think of this as the integral of cosine u.
Then again, d theta is 1/2 du.
So the entire derivative is 1/2 sine u,
which is 1/2 sine two theta.
And now need to find big F of B minus big F of A.
When theta is equal to pi over two, we have pi over two
plus 1/2 sine of,
two times pi over two is pi.
Then minus, when theta is zero, we have zero
plus 1/2 sine, two times zero is zero.
Well, sine pi is equal to zero, so this is zero.
This is zero.
And sine zero is also zero.
So this gives us the area is equal to two r squared
times pi over two.
Simplifying, two divided by two simplifies to one,
giving us the area
equals pi r squared,
which is the area of a circle,
which does give us the area formula for a circle.
The area of a circle
is equal to pi r squared.
I hope you found this helpful.