I constructed interesting figure and want to know its function on Cartesian plane and the area enclosed by this figure.
Suppose you have the sine function (explicitly, $m \cdot \sin(kx)$) from 0 to $\frac{\pi}{k}$. Now we rotate this piece of sine function by some angle $\theta$ and linking new rotated sine function with the original one. Do it $n$-times. We will have $n$-star which has sine function "spikes" (these "spikes" are not sharp as real spikes). I want to know what the function will describe this figure and what is the area of this figure in terms of $m$, $n$ and $k$? The difficulty of the problem, as I think, lies in the identification of the function that describes this figure (obviously, this function will be implicit, meaning, it will have form of $f (x, y)$) and following integration of the function (finding the area of the figure $n$-star). I assume it will need a strong multivariable calculus knowledge to do this, however, I am only starting my journey in this field. I will be very happy if you can help me with something of this! Thanks in advance for the answer!
I think the natural way to represent your curve is as a parametric curve $x = X(t)$, $y = Y(t)$, $a \le t \le b$. The area enclosed by a closed parametric curve (not self-intersecting except at the endpoints) is $$ \pm \int_a^b Y(t) X'(t)\; dt $$ $+$ if the curve is traversed clockwise as $t$ increases and $-$ if it is counterclockwise.
[EDIT] For example, consider $y = \sin (3 x)$ and its rotation clockwise by $\pi/4$, which is the parametric curve $x = \cos(\pi/4) t + \sin(\pi/4) \sin(3t)$, $y = -\sin(\pi/4) t + \cos(\pi/4) \sin(3t)$. Here's a picture:
They intersect when $x = \cos(\pi/4) t + \sin(\pi/4) \sin(3t)$ and $\sin(3x) = -\sin(\pi/4) t + \cos(\pi/4) \sin(3t)$. One solution is at $x=0, t=0$ and the next (obtained numerically) at approximately $x = 0.9191977679$, $t = 0.3850664671$. We could combine the two pieces of curves in one parameterization, using $s = 0$ to $1$ for the first piece and $1$ to $2$ for the second:
$$ \eqalign{x &= \cases{ 0.9191977679 s & if $0 \le s \le 1$\cr 0.3850664671 \cos(\pi/4) (2-s) + \sin(\pi/4) \sin(3 \times 0.3850664671 (2-s)) & if $1 \le s \le 2$\cr}\cr y &= \cases{\sin(3 \times 0.9191977679 s) & if $0 \le s \le 1$\cr - 0.3850664671 \sin(\pi/4) (2-s) + \cos(\pi/4) \sin(3 \times 0.3850664671 (2-s)) & if $1 \le s \le 2$\cr}}$$
But it would be more convenient to just use the original parameterizations, taking into account that we want to go "backwards" on the second curve. Thus the area is
$$ \int_0^{0.9191977679} \sin(3x)\; dx - \int_0^{0.3850664671} (-\sin(\pi/4) t + \cos(\pi/4) \sin(3t))(\cos(\pi/4) + 3 \sin(\pi/4) \cos(3t))\; dt$$