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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.
Half a cirle

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

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