- #1

Chenkel

- 482

- 108

Hello everyone,

I found a good proof for the area of a circle being ##{\pi}r^2## but I was recently working on my own proof and I used a change of variables and was wondering if I did it correctly and why a change of variables seems to work.

I start with the equation of a circle ##r^2 = x^2 + y^2## then I set up the integral ##2\int_{-r}^{r} {\sqrt{(r^2 - x^2)}}{dx}##

Now I change variables with ##x=r\cos(\theta)## and ##dx=-r\sin(\theta)d{\theta}##

And I update the integral

##-2\int_{\pi}^{0} {\sqrt{(r^2 - (r\cos(\theta))^2)}}r\sin(\theta){d\theta}##

Which is the same as

##2\int_{0}^{\pi} \sqrt{(r^2 - (r\cos(\theta))^2)}r\sin(\theta){d\theta}##

And using trig identities this is the same as

##2\int_{0}^{\pi} {{r^2}{(\sin(\theta))^2}}{d\theta}##

Finally using the half angle trig identity this is the same as ##{r^2}\int_{0}^{\pi} {(1 - \cos(2\theta))}{d\theta}##

The half angle identity is ##(sin(\theta))^2 = (\frac 1 2)(1 - \cos(2\theta))##

The antiderivatie is ##\theta - \frac {\sin(2\theta)}{2}## the evaluation of the full integral gives ##{\pi}r^2##

I'm just curious if I did these steps right and why the new integral in terms of theta gives the answer to the integral in terms of x appropriately.

Any help is appreciated,

Thank you!

I found a good proof for the area of a circle being ##{\pi}r^2## but I was recently working on my own proof and I used a change of variables and was wondering if I did it correctly and why a change of variables seems to work.

I start with the equation of a circle ##r^2 = x^2 + y^2## then I set up the integral ##2\int_{-r}^{r} {\sqrt{(r^2 - x^2)}}{dx}##

Now I change variables with ##x=r\cos(\theta)## and ##dx=-r\sin(\theta)d{\theta}##

And I update the integral

##-2\int_{\pi}^{0} {\sqrt{(r^2 - (r\cos(\theta))^2)}}r\sin(\theta){d\theta}##

Which is the same as

##2\int_{0}^{\pi} \sqrt{(r^2 - (r\cos(\theta))^2)}r\sin(\theta){d\theta}##

And using trig identities this is the same as

##2\int_{0}^{\pi} {{r^2}{(\sin(\theta))^2}}{d\theta}##

Finally using the half angle trig identity this is the same as ##{r^2}\int_{0}^{\pi} {(1 - \cos(2\theta))}{d\theta}##

The half angle identity is ##(sin(\theta))^2 = (\frac 1 2)(1 - \cos(2\theta))##

The antiderivatie is ##\theta - \frac {\sin(2\theta)}{2}## the evaluation of the full integral gives ##{\pi}r^2##

I'm just curious if I did these steps right and why the new integral in terms of theta gives the answer to the integral in terms of x appropriately.

Any help is appreciated,

Thank you!

Last edited: