For her next show, Frederica wants to mix things up. Instead of placing the light on the circumference of the wheel, she will pick a random point inside the circle. (Before you ask, let me clarify what “random” means here: Any two regions with the same area are equally likely to contain the point.)
As before, she will roll the wheel for one revolution along the ground and capture the motion with a single long exposure on her camera. On average, what can she expect the length of the path to be?
As before we will compute for any given starting point $$(x_0, y_0) \in \{ (x,y) \mid \sqrt{x^2 + (y-1)^2} \leq 1 \}$$ and calculate the path that it traces out as the wheel makes a full rotation. If the point $(x_0,y_0)$ is on the circumference of the circle, the path is known as a cycloid; however, if $(x_0,y_0)$ is in the interior of the wheel it is known as a curtate trochoid. Anywhoozle, let's again define our parametric equations ...
Let's assume that the point of light is at $x(0) = x_0, y(0) = y_0$, where without loss of generality let's assume that $x_0 \gt 0,$ and of course $r = \sqrt{x_0^2 + (y_0-1)^2} \leq 1.$ Then we see that we can write $x(0) = r \sin \theta, y(0) = 1 + r \cos \theta,$ for $\theta = \tan^{-1} \frac{x_0}{y_0 - 1} \in [0, \pi].$ Since after a rotation of $t \in [0,2\pi)$ we will have the wheel centered at $(t,1),$ while the angle that the point of light now makes as measured with respect to the positive $y$-axis, will now be $\theta + t.$ So at time $t$ the point of light would be at \begin{align*} x(t) &= t + r \sin (\theta + t) \\ y(t) &= 1 + r \cos (\theta + t).\end{align*} Differentiating with respect to time we get \begin{align*} \frac{dx}{dt} &= 1 + r \cos (\theta + t) \\ \frac{dy}{dt} &= -r \sin (\theta + t), \end{align*} so we see that the arclength of the path the light takes is \begin{align*}\ell &= \int_0^{2\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 }\,dt \\ &= \int_0^{2\pi} \sqrt{ \left( 1 + r \cos (\theta + t)\right)^2 + \left( - r \sin (\theta + t) \right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos (\theta + t) } \,dt.\end{align*} We can easily set $u = \theta + t$ and then through periodicity recognize that we have \begin{align*}\ell = \ell(r) &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos(\theta+t) } \,dt\\ &= \int_\theta^{\theta + 2\pi} \sqrt{ 1 + r^2 + 2r \cos u} \,du\\& = \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos u } \,du,\end{align*} regardless of the the value of $\theta.$ Now at this point, if we wanted to we could do some more algebra and appeal to the complete elliptic integrals of the second kind to find that $$\ell(r) = 4(1+r)E\left(\frac{2 \sqrt{r}}{1+r}\right),$$ where $$E(k) = \int_0^{\pi/2} \sqrt{1-k^2\sin^2t}\,dt;$$ however, since we would still need to plug this into another integral, this doesn't really do us any good.
However, we can us the integral definition of $\ell(r)$ and the definition of the randomly chosen starting point to get an expression for the expected length of the path. If we want to get the probability that some starting point $(x_0,y_0)$ is in the annulus centered at $(0,1)$ between radii $r$ and $r + dr,$ we get that the area of the annulus is $2\pi r \,dr,$ while the total probability of the unit disk centered at $(0,1)$ is $\pi,$ then we see that the conditional probability of the starting point being in the annulus between radii $r$ and $r + dr$ is $2r\, dr.$ Therefore, we get from the law of total expectation that $$\mathcal{L} = \mathbb{E} \left[ \ell \right] = \int_0^1 \ell(r) 2r \,dr.$$ Plugging back in the integral form of $\ell(r),$ we get $$\mathcal{L} = \int_0^1 \ell(r) 2r\,dr = \int_0^1 \int_0^{2\pi} 2r \sqrt{ 1 + r^2 + 2r \cos u } \,du \,dr,$$ which we can also recognize as the polar area integral of the function $$f(x,y) = 2\sqrt{ (1-x)^2 + y^2 }$$ over the unit disk. That is, $$\mathcal{L} = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{(1-x)^2 + y^2} \,dy \,dx.$$
Since $$\int \sqrt{a^2 + t^2} \,dt = \frac{t}{2} \sqrt{a^2 + t^2} + \frac{a^2}{2} \ln \left| t + \sqrt{a^2 + t^2} \right| + C,$$ we can do this integration directly. We first get that \begin{align*}g(x) &= \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \int_0^{\sqrt{1-x^2}} \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \left[ \frac{y}{2} \sqrt{ (1-x)^2 + y^2 } + \frac{(1-x)^2}{2} \ln \left| y + \sqrt{ (1-x)^2 + y^2 } \right| \right]_0^{\sqrt{1-x^2}} \\ &= 4 \left[ \frac{\sqrt{1-x^2} \sqrt{2 - 2x}}{2} + \frac{(1-x)^2}{2} \ln \left| \sqrt{1-x^2} + \sqrt{2-2x} \right| - \frac{(1-x)^2}{2} \ln |1-x| \right] \\ &= 2\sqrt{2} (1-x) \sqrt{1+x} + 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right).\end{align*} So we now have \begin{align*}\mathcal{L} &= \int_{-1}^1 g(x) \,dx \\ &= \int_{-1}^1 2\sqrt{2} (1-x)\sqrt{1+x} \,dx + \int_{-1}^1 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right) \,dx.\end{align*} Using integration by parts, we have first integral is \begin{align*}I_1 &= \int_{-1}^1 2\sqrt{2}(1-x) \sqrt{1+x} \,dx \\ &= 2 \sqrt{2} \left( \left[ \frac{2}{3} (1-x) (1+x)^{3/2} \right]_{-1}^1 + \frac{2}{3} \int_{-1}^1 (1+x)^{3/2} \,dx \right) \\ &= 2 \sqrt{2} \left.\frac{4}{15} (1+x)^{5/2} \right|_{-1}^1 = 2\sqrt{2} \frac{4}{15} 2^{5/2} = \frac{64}{15}.\end{align*} We can break the second integral further into two integrals, namely $$I_2 = \int_{-1}^1 2(1-x)^2\ln \left( \sqrt{1+x} + \sqrt{2} \right) \,dx$$ and $$I_3 = \int_{-1}^1 2(1-x)^2 \ln \sqrt{1-x} \,dx = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx.$$
If we set $u = \ln \left( \sqrt{1+x} + \sqrt{2} \right)$ then we see that $$du = \frac{dx}{2 \sqrt{1+ x} ( \sqrt{1+x} + \sqrt{2} )} = \frac{ \sqrt{2} - \sqrt{1+x} }{2 \sqrt{1+x} (1-x)} dx,$$ so then if we pursue integration by parts we get to obtain \begin{align*}I_2 &= \int_{-1}^1 2(1-x)^2 \ln \left(\sqrt{1+x} + \sqrt{2}\right) \,dx \\ &= 2 \left( \left[ -\frac{(1-x)^3}{3} \ln \left( \sqrt{1+x} + \sqrt{2} \right) \right]_{-1}^1 + \int_{-1}^1 \frac{(1-x)^3}{3} \frac{\sqrt{2} - \sqrt{1+x}}{2 \sqrt{1+x} (1-x) } \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \int_{-1}^1 \frac{\sqrt{2}}{6} \frac{(1-x)^2}{\sqrt{1+x}} \,dx - \int_{-1}^1 \frac{(1-x)^2}{6} \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \left[ \frac{\sqrt{2}}{3} (1-x)^2 \sqrt{1+x} \right]_{-1}^1 + \frac{2\sqrt{2}}{3} \int_{-1}^1 (1-x) \sqrt{1+x} \,dx + \left[ \frac{(1-x)^3}{18} \right]_{-1}^1 \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \frac{4}{9} + \frac{1}{3} \left( \frac{64}{15} \right) \right) = \frac{56}{15} -\frac{8}{3} \ln 2, \end{align*} where we are using our precalculated knowledge of $I_1$ in the second to last step.
Using integration by parts gives $$\int t^2 \ln t \,dt = \frac{t^3}{3} \ln t - \int \frac{t^2}{3} \,dt = \frac{t^3}{3} \ln t - \frac{t^3}{9} + C.$$ With the u = 1-x substituion we get $$I_3 = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx = \int_{0}^2 u^2 \ln u \,du = \left[\frac{u^3}{3} \ln u - \frac{u^3}{9} \right]_{0}^2 = \frac{8}{3} \ln 2 - \frac{8}{9}.$$ So putting it altogether we get the expected length of the uniformly random point of light to be \begin{align*}\mathcal{L} &= I_1 + I_2 + I_3 \\ &= \frac{64}{15} + \left( \frac{56}{15} - \frac{8}{3} \ln 2 \right) + \left( \frac{8}{3} \ln 2 - \frac{8}{9} \right) \\ &= 8 - \frac{8}{9} = \frac{64}{9} = 7.111111\dots.\end{align*}
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