Error evaluating the area inside an epicycloid, spot my mistake?

Question

Hi! this is my first post on mathstackexchange, so I hope I can navigate successfully, including entering the LaTeX.

And, this will be embarrassing. But never mind, I really want to know what I'm doing wrong mathematically, and I'm hoping someone can point out my error.

Trying to evaluate the area inside an epicycloid: small circle rolling around the outside of a fixed bigger circle, a point $P$ on the circumference of the smaller circle traces this curve.

Somewhat general case: fixed circle radius $R$ centered on the origin $(0,0)$, small rolling circle radius $r$ centered initially at $(R+r,0)$, with $P$ at initial position $(R,0)$. Assume ratio $k=R/r$ is a positive integer. To keep the details to a minimum, consider a particular example. $R=3,\;r=1,\;$ so $k=3$, $P$ starts at $(3,0)$, this epicycloid has three lobes and it is given by $$x(\theta)=4\cos\theta-\cos(4\theta),\quad y(\theta)=4\sin\theta-\sin(4\theta).$$

The area inside this example can be calculated (as it can for the area of any such simple curve enclosing the origin) as $$A=\dfrac{1}{2}\int_0^{2\pi}(x^2+y^2)\,d\theta.$$ And in this particular case, I am confident that the area should work out to $20\pi$. Everyone says so, - even including my own accurate drawing of this rather nice trefoil (counting the little squares on a large sheet of graph paper). But when I do the calculation, below, I get that the area is $17\pi$? Where is my mistake, where am I missing something?

In this example, substituting for $x$ and $y$ : $$A=\dfrac{1}{2}\int_0^{2\pi}\left((4\cos\theta - \cos 4\theta)^2 + \left(4\sin\theta - \sin 4\theta\right)^2\right) \,d\theta$$

Expanding and rearranging, $$A=\dfrac{1}{2}\int_0^{2\pi}\bigl(\;16\cos^2\theta+16\sin^2\theta\;\;+\;\cos^2 4\theta+\sin^2 4\theta\;\; -\; 8(\cos 4\theta\cos\theta + \sin 4\theta\sin\theta )\;\bigr)\,d\theta$$

Simplifying,  $$A=\dfrac{1}{2}\int_0^{2\pi}\bigl(\;16\;\;+\;1\;\; -\; 8\cos( 4\theta -\theta) \;\bigr)\,d\theta=\dfrac{1}{2}\int_0^{2\pi}(17-8\cos 3\theta)\,d\theta$$

This integrates to $$A=\dfrac{1}{2}\;\left|\;17\theta-\dfrac{8}{3}\sin 3\theta\;\right|_0^{2\pi}\;= 17\pi.$$

Presumably the mistake is that the "(+1)" in the integrand should somehow be, "(+4)"? But that does not look right.

Please, what is wrong with the above? I suppose the mistake is probably glaringly obvious, but I just can't see it. Grateful in advance for any assistance. Abashedly stumped, Terry

Answer 1

You wrongly assumed that your $\theta$ would be the polar angle (i.e. to the surface point P of the smaller circle). Instead it is just a parameter (which in fact describes the angle to the center of the smaller circle). The polar angle $\varphi$ here would be given by

$$\tan(\varphi)=\frac{y(\theta)}{x(\theta)}$$

--- rk

Edit: A better Ansatz here would be $$A = \int_0^{2\pi}y(\theta)\cdot x'(\theta)\ d\theta$$

Answer 2

The mistake is subtle. The correct formula for the area enclosed by the epicycloid in polar coordinates is $$ A=\frac{1}{2}\int_0^{2\pi}\rho^2\,d\phi, \tag{1} $$ where $\phi$ is the angle between the radius vector $\vec{\rho}=(x,y)$ and the $x$-axis. This angle $\phi$ is not the same as the angle $\theta$ used in the parameterization of the epicycloid; instead, it is given by $$ \phi=\arctan\!\left(\frac{y}{x}\right). \tag{2} $$ In terms of $\theta$, Eq. $(1)$ can be rewritten as $$ A=\frac{1}{2}\int_0^{2\pi}[x^2(\theta)+y^2(\theta)]\frac{d\phi}{d\theta}\,d\theta, \tag{3} $$ where $$ \frac{d\phi}{d\theta}=\frac{d}{d\theta}\arctan\!\left(\frac{y}{x}\right) =\frac{x(\theta)y'(\theta)-x'(\theta)y(\theta)}{x^2(\theta)+y^2(\theta)}. \tag{4} $$ Therefore, \begin{align} A&=\frac{1}{2}\int_0^{2\pi}[x(\theta)y'(\theta)-x'(\theta)y(\theta)]\,d\theta \\ &=\underbrace{-\frac{1}{2}x(\theta)y(\theta)|_0^{2\pi}}_{=\,0}+\int_0^{2\pi}x(\theta)y'(\theta)\,d\theta \\ &=\int_0^{2\pi}[4\cos(\theta)-\cos(4\theta)][4\cos(\theta)-4\cos(4\theta)]\,d\theta \\ &=20\pi.\phantom{\int_0^{2\pi}} \tag{5} \end{align}

Answer 3

I also agree that it's a mistake here to use the area formula for a polar curve as $\theta$ here is not the polar angle (this was already noted out in the other answers). Instead, we can use parametric curve area formula: $$A=\int_0^{2\pi}(4\cos\theta - \cos 4\theta)(4\sin\theta - \sin 4\theta )'d\theta= \\ \int_0^{2\pi}(4\cos\theta -4 \cos 4\theta)(4\cos\theta -4 \cos 4\theta )d\theta = \\ \int_0^{2\pi}(16\cos^2\theta +4 \cos^2 4\theta)d\theta = 20 \pi $$