Epicycloid (Hypercylcoid)

The parametric equation of the epicycloid is $x \; = \; 15\cos t - 3\cos5t \hspace{1cm} y \; = \; 15\sin t - 3\sin5t \hspace{2cm} 0 \le t \le 2\pi$ and $\tfrac12\big(x(t)y'(t) - x'(t)y(t)\big) \; = \; 270\sin^22t$ after much simplification. The enclosed area is $A \; =\; \int_0^{2\pi} \tfrac12\big(x(t)y'(t) - x'(t)y(t)\big)\,dt \; = \; 270\pi$ making the answer $\boxed{270}$ .

Total area under epicycloid = (1+n) (2+n) b^2* $\pi$ .
n=ratio of base circle and generating circle radii=12/3=4.
b=gen.circle radius = 3
$So (1+n)*(2+n)*b^2*= 5*6*3^2=\Large \color{#D61F06}{270}$

The first part of the problem is to find parametric equation of the epicycloid.

Let $P$ be a point where the small green circle with radius $r$ touches big red circle with radius $R$ and let $M$ be a point which traces the epicycloid. Introduce two coordinate systems: $S$ with origin $O$ at center of the big red circles and coordinate axes - $x$ and $y$ - in horizontal and vertical direction, respectively; and $S'$ with origin $O'$ at the center of the small green circle and one axis always pointing to $O$ - $x'$ axis - and the other - $y'$ - rotated $90^{\circ}$ in anticlockwise direction with respect to $x'$ (we could say that $S'$ rotates around the $S$ ). Next, angle between positive direction of $x$ -axis and vector $\overrightarrow{OP}$ denote with $\phi$ and angle between positive direction of $x'$ -axis and vector $\overrightarrow{O'M}$ denote with $\theta$ (see the picture). Since distance traveled by the small circle along the big circle equals the length of the arc subtended by the angle $\phi$ , we have $r\theta = R\phi$ . For better presentation of the figure, magnitude of unit vectors $i'$ and $j'$ in the picture is greater than 1. It follows that the coordinates of origin of $S'$ with respect to $S$ are: $\begin{aligned}x_{O'} &= (R+r)\cos{\phi} = (R+r)\cos{\frac{r\theta}{R}} \\ y_{O'} &= (R+r)\sin{\phi} = (R+r)\sin{\frac{r\theta}{R}},\end{aligned}$ or equivalently: $\overrightarrow{OO'} = (R+r)\cos{\left ( \frac{r\theta}{R} \right )}i + (R+r)\sin{\left ( \frac{r\theta}{R} \right )}j\, ,$ where $i$ and $j$ are unit vectors in $x$ and $y$ direction, respectively. On the other hand, coordinates of point $M$ with respect to $S'$ are: $\begin{aligned} x'_{M} &= r\cos{\theta} \\ y'_{M} &= r\sin{\theta}, \end{aligned}$ or equivalently: $\overrightarrow{O'M} = r\cos{\theta}i' + r\sin{\theta}j'\, ,$ where $i'$ and $j'$ are unit vectors in $x'$ and $y'$ direction, respectively. There are, of course, equations which relate two sets of unit vectors: $\begin{aligned}i' &= \cos{\left ( 180^{\circ} + \phi \right )}i + \sin{\left ( 180^{\circ} + \phi \right )}j = -\cos{\phi}i - \sin{\phi}j \\ j' &= \cos{\left ( 270^{\circ} + \phi \right )}i + \sin{\left ( 270^{\circ} + \phi \right )}j = \sin{\phi}i - \cos{\phi}j \end{aligned}$ Now, one can easily determine the coordinates of point $M$ with respect to $S$ by adding vectors $\overrightarrow{OO'}$ and $\overrightarrow{O'M}$ : $\begin{aligned} \overrightarrow{OM} &= \overrightarrow{OO'} + \overrightarrow{O'M} \\ &= (R+r)\cos{\left ( \frac{r\theta}{R} \right )}i + (R+r)\sin{\left ( \frac{r\theta}{R} \right )}j + r\cos{\theta}{\color{#BBBBBB} i'} + r\sin{\theta}{\color{#BBBBBB} j'} \\ &= (R+r)\cos{\left ( \frac{r\theta}{R} \right )}i + (R+r)\sin{\left ( \frac{r\theta}{R} \right )}j + r\cos{\theta}{\color{#BBBBBB} (-\cos{\phi}i - \sin{\phi}j)} + r\sin{\theta}{\color{#BBBBBB} (\sin{\phi}i - \cos{\phi}j)} \\\vec{x}(\theta) &= \left ( (R+r)\cos{\left ( \frac{r\theta}{R} \right )}-r\cos{\left (\frac{R+r}{R} \theta\right )} \right )i + \left ( (R+r)\sin{\left ( \frac{r\theta}{R} \right )}-r\sin{\left (\frac{R+r}{R} \theta\right )} \right )j\end{aligned}$ Thus, we derived the parametric equation for the epicycloid. Notice that we introduced new, more convenient, notation to represent each point as a vector-valued function $\vec{x}(\theta)$ , and we assume further on to work only with respect to $S$ .

To calculate the enclosed area, we are going to divide it to many infinitesimal triangles with vectors $\vec{x}(\theta)$ , $\vec{x}(\theta + d\theta)$ and $\vec{x}(\theta + d\theta)-\vec{x}(\theta)$ as sides. The area of one infinitesimal triangle is equal to: $\begin{aligned} dA &= \frac{1}{2}\left \| \vec{x}(\theta) \times (\vec{x}(\theta + d\theta)-\vec{x}(\theta)) \right \| \\ &= \frac{1}{2}\left \| \vec{x}(\theta) \times \vec{x}\,'(\theta)\right \| d\theta \\ &= \frac{1}{2}\left | x(\theta)y'(\theta)-x'(\theta)y(\theta) \right |d\theta \\ &= (R+r)\frac{r}{2R}(R+2r)\left |1 - \cos{\theta} \right |d\theta. \end{aligned}$

Finally, the total area is: $\begin{aligned} A &= \frac{r(R+r)(R+2r)}{2R}\int_{0}^{8\pi}\left |1 - \cos{\theta} \right |d\theta \\ &= \frac{r(R+r)(R+2r)}{2R}\int_{0}^{8\pi}(1 - \cos{\theta})d\theta \\ &= \frac{4\pi r(R+r)(R+2r)}{R} \end{aligned}$

Replacing $R = 12$ and $r = 3$ , we get $A = 270$ .