The curve given by $$ r = 4\theta^2, \theta \in [0, \infty) $$ intercepts the x-axis an infinite amount of times. Find the arc length of the curve from the first x-intercept and the third x-intercept. Count $\theta = 0$ as the first x-intercept.
Im not able to find the x-intercepts and therefore not the integration limits.
On
This is a spiral and will intercept the $x$-axis for every $\theta=n\pi$. Therefore, the third intercept will be at $\theta=2\pi$. See the figure below shown for $\theta\in[0,2\pi]$.
The arc length in polar coordinates is given by
$$ \begin{align} s &=\int \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}~d\theta\\ &=\int_0^{2\pi} \sqrt{16\theta^4+64\theta^2}~d\theta\\ &=\int_0^{2\pi} 4\theta\sqrt{\theta^2+4}~d\theta\\ &=\frac43(\theta^2+4)^{3/2}\biggr|_0^{2\pi}\\ &=\frac{32}{3}\left[(1+\pi^2)^{3/2}-1 \right]\approx 371.58 \end{align} $$
I have verified this solution numerically.
$$x(t)=r\cos (t)=4t^2\cos (t) $$ $$y(t)=4t^2\sin (t) $$
$$x'(t)=t(8\cos (t)-4t\sin (t)) $$ $$y'(t)=t(8\sin (t)+4t\cos (t)) $$
$${x'}^2+{y'}^2=t^2 (64+16t^2) $$
the length is
$$4\int_0^{2\pi}t\sqrt {4+t^2}dt $$ The curve intercepts the x-axis when $t=0 \;,\; t=\pi \;,\; t=2\pi $.