# Arc length of cardioid using line integral techniques

#### romsek

MHF Helper
you seem to be doing it the hard way

$r=r(\theta)$

$ds=\sqrt{r^2 + r_{\theta}^2} ~d\theta$

$L=\int ds$

$r(\theta)=\cos(\theta)+1$

$r_{\theta}=-\sin(\theta)$

$ds=\sqrt{(1+\cos(\theta))^2+\sin^2(\theta)}=$

$\sqrt{1+2\cos(\theta)+\cos^2(\theta)+\sin^2( \theta)}=$

$\sqrt{2+2\cos(\theta)}$

so

$L = \displaystyle{\int_0^{2\pi}}\sqrt{2+2\cos(\theta)}~d\theta=8$

#### Archie

Yes, you are on the right track. Your curve is parametrised by $$\displaystyle \theta$$, so you just need to build the integral $$\displaystyle \int \limits_C \mathrm d s = \int \limits_C \big| \big|\mathbf{ r}'(\theta) \big| \big| \, \mathrm d \theta$$.

MHF Helper

#### Archie

You have $$\displaystyle cos^4 x + 2\cos^2 x \sin^2 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2$$ and $$\displaystyle 2\cos x\sin^2 x + 2 \cos ^3 x = 2\cos x(\sin^2 x + \cos^2 x)$$.

• MechanicalPencil and (deleted member)

#### MechanicalPencil

Thanks everyone! I finally got through it, I appreciate your help!