# Calc1.86

Derive the formula for the area of a circle with arbitrary radius r.

Moving a cirlce does not change its area so let us position this circle so that its center is at the point $(0,0)$. Then the equation for this circle is

$x^{2}+y^{2}=r^{2}\,$

In finding the area of this circle, we will use symmetry. If we find the area of the portion in the first quadrant and multiply this by 4, we have the area of the whole circle. The equation of the upper half of the circle is given by solving for y and taking only the positive square root,

$y={\sqrt {r^{2}-x^{2}}}$

Thus, the area of this circle is given by 4 times the integral of this function while integrating over the x-values for the first quadrant.

$A=4\int _{{0}}^{{r}}{\sqrt {r^{2}-x^{2}}}\,dx\,$

This integral can be done by trig substitution. Let

$x=r\sin \theta \,$

$dx=r\cos \theta \,d\theta \,$

Substituting gives

$A=4\int _{{x=0}}^{{x=r}}{\sqrt {r^{2}-r^{2}\sin ^{2}\theta }}(r\cos \theta )\,d\theta =4\int _{{x=0}}^{{x=r}}r^{2}\cos ^{2}\theta \,d\theta =4r^{2}\int _{{x=0}}^{{x=r}}{\frac {1+\cos(2\theta )}{2}}\,d\theta =2r^{2}{\bigg [}\theta +{\frac {1}{2}}\sin(2\theta ){\bigg ]}_{{x=0}}^{{x=r}}=2r^{2}{\bigg [}\theta +\sin \theta \cos \theta {\bigg ]}_{{x=0}}^{{x=r}}\,$

Now, substituting back in to get the expression in terms of x gives

$A=2r^{2}{\bigg [}\arcsin {\frac {x}{r}}+{\frac {x}{r}}{\frac {{\sqrt {r^{2}-x^{2}}}}{r}}{\bigg ]}_{{0}}^{{r}}=2r^{2}\left[{\frac {\pi }{2}}+0-(0+0)\right]=\pi r^{2}\,$

This, of course is the formula we expected to get. For an arbitrary circle with radius r, the area is given by $\pi r^{2}$. The formula for this shape and many others can be found using calculus.