Area of polar curve

Question

So I have the polar curve

$r=\sqrt{|\sin(n\theta)|}$

Which I am trying to evaluate between $0$ and $2\pi$. By smashing it into wolfram it returns a constant value 4 for any $n$.

I tried calculating it manually (I suspect my calculation might be wrong), but I arrived at

$$\textrm{I}=\frac{1}{2}\int_{0}^{2\pi}|\sin(n\theta)|= \left[\text{sgn}(\sin(nx))\frac{\cos(nx)}{n}\right]_{0}^{2\pi} $$ Which when looking at it can't be evaluated at $0$ or $2\pi$ with taking the limits. I suspect that whenever $n$ increases, the curve becomes "tighter", which could explain why in the integral stays permanently at 4, but I can't come up with a sound argument for it, so if someone could give me a pointer, that would be greatly appreciated.

Answer 1

The constant value of $4$ is normal $$\int_{0}^{2\pi}|\sin(t)|\,dt=\int_{0}^{\pi}\sin(t)\,dt-\int_{\pi}^{2\pi}\sin(t)\,dt=2+2$$ $$\int_{0}^{2\pi}|\sin(2t)|\,dt=\int_{0}^{\frac\pi 2}\sin(2t)\,dt-\int_{\frac\pi 2}^{\pi}\sin(2t)\,dt+\int_{\pi}^{\frac{3\pi} 2}\sin(2t)\,dt-\int_{\frac{3\pi} 2}^{2\pi}\sin(2t)\,dt=1+1+1+1$$ Just continue.

Answer 2

Use the variable change $t=n\theta$ to rewrite the integral as

$$\textrm{I}=\frac{1}{2}\int_{0}^{2\pi}|\sin(n\theta)|d\theta = \frac1{2n} \int_{0}^{2\pi n}|\sin t|dt$$

Due to the periodicity of the sine function

$$\textrm{I} = \frac1{2n} \cdot n\int_{0}^{2\pi}|\sin t|dt =\frac1{2}\int_{0}^{2\pi}|\sin t|dt =\frac1{2}\cdot 4 \int_{0}^{\pi/2}\sin tdt = 2$$