arc lengthintegrationparametric
I'm trying to calculate the arc length of the innerloop see picture of a cardioid-like $r=4a\cos^3\frac\theta3$ but don't know where to start. I know you have formulas like $L=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta$, but don't know how to do this for the innerloop.
thanks in advance
Use $$\sin^2+\cos^2=1\implies \sqrt {1-\cos x}=\dfrac {\sin x}{\sqrt {1+\cos x}} $$ Let $ u=1-\cos x $, so $ du =\sin xdx $. Then, we get $$\int \dfrac {1}{\sqrt {1-u}}\cdot \dfrac {\sin x}{\sin x} du$$ Now, cancel out and do one more $u$ substitution, integrate and then plug back in to get $$2a\sqrt {2 (1-\cos x)}|^{2\pi}_{0}$$
Since $$ \frac{\mathrm{d}}{\mathrm{d}\theta}2(\cos(2\theta))^{1/2} =-2(\cos(2\theta))^{-1/2}\sin(2\theta) $$ we get $$ \begin{align} r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2 &=4\cos(2\theta)+\frac{4\sin^2(2\theta)}{\cos(2\theta)}\\ &=\frac4{\cos(2\theta)} \end{align} $$ Doubling the length of one leaf, we get $$ \begin{align} 2\int_{-\pi/4}^{\pi/4}\frac{2\,\mathrm{d}\theta}{\sqrt{\cos(2\theta)}} &=4\sqrt2\operatorname{K}\left(\frac1{\sqrt2}\right)\\ &=\sqrt{\frac2\pi}\,\Gamma\left(\frac14\right)^2\\ &=\frac{4\pi}{\operatorname{AGM}\left(1,\sqrt2\right)}\\[6pt] &=10.488230217168479242 \end{align} $$ where $\operatorname{K}$ is the Complete Elliptic Integral of the First Kind and, as Jack D'Aurizio comments, $\operatorname{AGM}$ is the Arithmetic-Geometric Mean.
A Derivation of the Integral
I've stated several equivalent forms of the integral above. Here is the derivation of one of them.
$$
\begin{align}
2\int_{-\pi/4}^{\pi/4}\frac{2\,\mathrm{d}\theta}{\sqrt{\cos(2\theta)}}
&=4\int_0^{\pi/4}\frac{\mathrm{d}2\theta}{\sqrt{\cos(2\theta)}}\tag{1}\\
&=4\int_0^1\frac{\mathrm{d}u}{\sqrt{u\vphantom{u^2}}\sqrt{1-u^2}}\tag{2}\\
&=2\int_0^1v^{-3/4}(1-v)^{-1/2}\mathrm{d}v\tag{3}\\
&=2\,\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac34\right)}\tag{4}\\
&=2\frac{\sqrt\pi\,\Gamma\left(\frac14\right)^2}{\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)}\tag{5}\\
&=\sqrt{\frac2\pi}\,\Gamma\left(\frac14\right)^2\tag{6}
\end{align}
$$
Explanation:
$(1)$: exploit the symmetry of $\cos(x)$
$(2)$: substitute $u=\cos(2\theta)$
$(3)$: substitute $v=u^2$
$(4)$: use the Beta Function
$(5)$: $\Gamma\left(\frac12\right)=\sqrt\pi$
$(6)$: $\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\pi\csc\left(\frac\pi4\right)=\pi\sqrt2$ using Euler's Reflection Formula
Best Answer
We can take out constants and multiply them back in at the end, so the curve is $r=\cos^3\frac{\theta}3$.
The required length is twice the length of the curve from $\theta=\pi$ to $\theta=\frac{3\pi}2$. Thus $$\frac L2=\int_\pi^{3\pi/2}\sqrt{\cos^6\frac\theta3+\cos^4\frac\theta3\sin^2\frac\theta3}\,d\theta$$ $$=\int_\pi^{3\pi/2}\sqrt{\cos^4\frac\theta3}\,d\theta=\int_\pi^{3\pi/2}\cos^2\frac\theta3\,d\theta=\frac34\left(\frac\pi3-\frac{\sqrt3}2\right)$$ Therefore the length of the inner loop for $r=4a\cos^3\frac\theta3$ is $a(2\pi-3\sqrt3)$.