# Why is the area of a circle  πr^2?

I wondered if I could prove why the area of a circle is πr2.
This requires something called trigonometric substitution. Also this involves substituting numerous parts in order to get an answer.
First of all, get an equation for drawing half a circle:
y = √(1-x2)
Now I wanted to put a radius element in to it which I call r:
y = r√(1-x2/r2) = √(r2-x2) but I will use the longer version.

So if I just integrated this from -r to r and multiplied it by two, I will get the area of a circle.
 Area of a circle = 2r ∫ r √( 1 - x2 ) dx -r r2
I moved 'r' outside the integral symbol because it is a constant.
To try to solve this, I need something called trigonometric substitution. If an integrand contains a2 - x2 then let x = a•sin θ and use the identity:
1 - sin2(θ) = cos2(θ)
So I try to get the bit under the square root to be cos2(θ) which equals cos (θ) once the square root is taken.

First of all, I'm going to just find the antiderivative.
I am going to start with the bit in the square root. I will replace x:
x = r • sin θ
Now when I replace x with this, I can get rid of the r2part:
 1 - x2 = 1 - (r • sin θ)2 = 1 - r2•sin2θ = 1 - sin2θ = cos2θ r2 r2 r2
I can replace the part under the square root with cos2θ but I also need to replace dx with something. If I take the substitution for x above and differentiate this, I can use that to find something to replace dx:
 x = r • sin θ, dx = r • cos θ, dx = r • cos θ dθ dθ
Now I can take my equation for the area and replace the part under the square root and dx which gives me:
2r ∫√(cos2θ) • r • cos θ dθ = 2r ∫ cosθ • r • cos θ dθ = 2r ∫ r • cosθ • cos θ dθ
= 2r2 ∫ cos2θ dθ (again, I took 'r' out of the integral symbol and just multiplied it with the other 'r')

Now I can integrate this which gives me:
 2r2 ( θ + sin(2θ) ) + C 2 4
 = r2 ( θ + sin(2θ) ) + C 2
Now I need to replace θ and the sine part to get things back in terms of x. The θ part is easy because I have already x = r • sin θ so θ = arcsin(x / r).

I will find an equivalent equation to replace sin(2θ) / 2. In situations like this, I sometimes try to find this out with my graphical calculator, graphing two things until they are equal or you can look up trigonometric identities. It turns out that:
sin(2θ) / 2 = sin θ • cos θ
This gives me a further problem. Now I have to substitute the sine and the cosine with something in terms of x. The sine is easy because I used x = r • sin θ so sin θ = x / r. I stated an equation earlier which I can use for cos θ:
 1 - x2 = 1 - (r • sin θ)2 = 1 - r2•sin2θ = 1 - sin2θ = cos2θ r2 r2 r2
This means that:
 1 - x2 = cos2θ r2
Taking the square root of both sides gives:
 cos θ = √( 1 - x2 ) r2

We have:
 θ = arcsin ( x ) r
 sin θ = x r
 cos θ = √( 1 - x2 ) r2
I can now put these back in to the equation:
 r2 ( θ + sin(2θ) ) + C 2
= r2(θ + sin (θ) • cos(θ)) + C
= r2(arcsin(x / r) + (x / r)√(1- x2/r2) + C
To calculate the area under the circle from -r to r, I replace the terms for x as normal. This makes the part under the square root equal to zero and the square root of zero is zero so I can ignore that. I am left with:
Area = (r2•arcsin(r / r)) - (r2•arcsin(-r / r))
= (r2• π / 2) - (r2• -π / 2)
= r2• π / 2 + r2• π / 2
= 2(r2• π / 2)

= πr2