Evaluating $\int_C \frac{x\mathrm{d}y-y\mathrm{d}x}{x^2+y^2}$, where $C$ has equation $x^{48} + y^{76} = 987654321^{123456789}$

Question

I am trying to evaluate the integral of $w = r^{-2}(x\mathrm{d}y-y\mathrm{d}x)$, where $r^2= (x^2+y^2)$ on a very nasty curve $C: x^{48} + y^{76} = 987654321^{123456789}$.

This is what I've attempted so far.

Let $987654321^{123456789} = \alpha$. Parametrize $C$ by $c(t) = \left( \alpha^{\frac1{48}} \cos^{\frac1{24}} t, \alpha^{\frac1{76}} \sin^{\frac1{38}} t \right), 0 \le t \le 2\pi$. Then \begin{align*} \int_C w &= \int_0^{2\pi} c^* w \\ &= \int_0^{2\pi} \frac{x}{r^2} \alpha^{\frac1{76}} \frac{1}{38} \sin^{-\frac{37}{38}} t \cos t dt + \frac{y}{r^2} \alpha^{\frac{1}{48}} \frac{1}{24} \cos^{-\frac{23}{24}} t \sin t dt \\ &= \frac{1}{38} \alpha^{\frac1{48} + \frac1{76}} \int_0^{2\pi} \frac{\sin^{-\frac{37}{38}} t \cos^{\frac{25}{24}} t}{ \alpha^{\frac1{24}} \cos^{\frac1{12}} t+ \alpha^{\frac1{38}} \sin^{\frac1{19}} t} dt \\ &+ \frac1{24} \alpha^{\frac1{48} + \frac1{76}} \int_0^{2\pi} \frac{ \sin^\frac{39}{38} t \cos^{-\frac{23}{24}} t}{ \alpha^{\frac1{24}} \cos^{\frac1{12}} t+ \alpha^{\frac1{38}} \sin^{\frac1{19}} t} dt \end{align*}

But at this point, I have no idea how to proceed. I think there must be some theorem that I can use so I don't avoid evaluating this nasty constant $\alpha$. I can only think of Stoke's theorem, but I don't think $C$ has boundaries. Can someone have me a hint? Is there a way to simplify my evaluation? Thanks in advance!

Answer 1

The point of this computation was not to parametrize directly, but to use Green's theorem (Stokes' theorem in the plane). Consider the region bounded on the outside by that monstrosity, and bounded on the inside by the unit circle $x^2+y^2=1$ - so the region looks like a donut. Then by Green's theorem we have that

$$\int_C w -\int_{x^2+y^2=1}w = \int dw = 0$$

because $w$ is a closed form. The reason the line integrals are subtracted is to preserve orientation (if the outer one is counter clockwise as viewed from above, the second one should be clockwise). Thus we have

$$\int_C w = \int_{x^2+y^2=1}w = 1^{-2}\int_0^{2\pi} \cos^2t+\sin^2 t\:dt = 2\pi $$

Answer 2

Let $F(x,y)= M(x,y)i+N(x,y)j$ be a vectorial field. Then your are asking for $\int_C{F·ds}$

The given function $F=\frac{-ydx}{x^2+y^2}i+\frac{xdy}{x^2+y^2}j$ verifies $rot(F)=0$ because $\frac{\delta M}{\delta y}=\frac{\delta N}{\delta x}$
Thus, exists an scalar field $U(x,y)$ such that $\nabla f = F$

Solving $\frac{\delta M}{\delta x}=\frac{-y}{x^2+y^2}$ and $\frac{\delta N}{\delta y}=\frac{x}{x^2+y^2}$ gives $U(x,y)= -arctan(\frac {x}{y}) + C$

Because $rot(F)=0$ the field F is conservative.
Thus we don't need to solve the integral, but just plug points in the field $U$

$I= U(b)-U(a)$ where $a,b$ are the start and end points over the curve (not currently given in your question).
For a closed curve that means $I=0$

Notice that $U$ is not defined at $(0,0)$. So, we can not say $I=0$ in this case (that point is inside the curve).

The given monster is closed. But, because F is conservative, we can use any other closed curve. One very easy is a unit circle around the origin.

This circle in parametrics is
$x=cos(t) ==>dx=-sin(t)dt$
$y=sin(t) ==>dy=cos(t)dt$
$ 0 \le t \le 2\pi$

Plugging in the integral:

$I=\int _0 ^{2\pi} dt \;=\;2 \pi$