# Solving a curve integral around part of an elipse

I'm having trouble calculating a curve integral in a vector field:

$\int_C y (18x + 1)\ dx + 2y^2\ dy$

where $C$ is the curve along the ellipse $9x^2 + y^2 = 64$
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3}$ , $4$) to the point (-$\frac{4}{3}$ , $4\sqrt{3}$)
Thats almost one "lap" around the ellipse..

When making a parametrisation I come up with:

x = $\sqrt{ \frac{64}{9}\ } \cos t$

y = $\sqrt{ 64 } \sin t$

$- \frac{\Pi}{3} \le t < \arctan(3 \sqrt{3})$

But the integral created from this parametrization give an answer involving arctan. Any ideas to get a rational answer?

I had an idea to split the curve into multiple curves and integrating them piecewise, but that gets really messy aswell.

This is my calculations when making a variable change: First the variable change:

$u= 3x$
$du = 3dx$

$\ u^2 + y^2 =64$

$\int y(18x + 1)dx + 2y^2dy = \int y(6u + 1)\frac{du}{3} + 2y^2dy$

$u = \sqrt{64}\cos t$
$y = \sqrt{64}\sin t$
$du = -\sqrt{64}\sin tdt$
$dy = \sqrt{64}\cos tdt$

The startpoint after making variable change: $( -4\sqrt{3},4)$

$arctan( \frac{4}{-4\sqrt{3}}) = -\Pi/6$
Startangle: $\frac{5\Pi}{6}$

The endpoint after making variable change: $( -4,4\sqrt{3})$

$arctan( \frac{-4\sqrt{3}}{4}) = -\Pi/3$

End angle: $\frac{2\Pi}{3}$

$\int y(6u + 1)\frac{du}{3} + 2y^2dy = \int -\frac{1}{3}\sqrt{64}\sin t(6\sqrt{64}\cos t + 1)\sqrt{64}\sin tdt + 128\sin^2t\sqrt{64}\cos tdt =$

$\int -\frac{64}{3}\sin^2tdt$

$\frac{5\Pi}{6} \le t \le \frac{2\Pi}{3}$

The answer becomes: $\frac{16}{9}\Pi$

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The first thing that occurs to me is to make the substitution $u = 3x$ (so that $C$ is $u^2 + y^2 = 64$) and then integrate around the arc of a circle instead of around the arc of an ellipse. – dfan Feb 28 '14 at 15:55
I've tried that and arrived at the answer $\frac{16}{9}\Pi$ but it seems incorrect.. – DJJQ Feb 28 '14 at 17:10
@DJJQ I completed my answer, arriving at $-16\pi/9$, please countercheck :) – flonk Mar 3 '14 at 10:51
Because neither $+-16\pi/9$ is the answer in the book :) – DJJQ Mar 3 '14 at 12:04
@DJJQ I think I found and solved it, see my edited answer below :) – flonk Mar 3 '14 at 13:21

As @dfan suggested, substitute $u=3x$ leading to a circular curve $C'$ defined by $$u^2+y^2=r^2$$ of radius $r=8$. Your integral becomes $$J:=\int_C dx\,y(18x+1)+\int_C dy\,2y^2=\frac{1}{3}\int_{C'}du\,y(6u+1)+2\int_{C'}dy\,y^2.$$ Taking the total differential of the circle equation, we have $udu=-ydy$, so the second integral cancels against the $6u$-term from the first one and we are left with

$$J=\frac{1}{3}\int_{C'}du\,y.$$

Next, introduce polar coordinates $$u=r\cos\phi,\\y=r\sin\phi,$$parametrizing the curve with $\phi$. Your integral limits transform to $$(x,y)_0=(-4/\sqrt{3},4)\to u_0=-4\sqrt{3}\quad \to\phi_0=\arccos\frac{-\sqrt{3}}{2}=\frac{5\pi}{6},$$ $$(x,y)_1=(-4/3,4\sqrt{3})\to u_1=-4\quad \to\phi_1=\arccos\frac{-1}{2}+2\pi=\frac{8\pi}{3}.$$ Here it is essential to add $2\pi$ to the upper limit $\phi_1$, such that $\phi_0<\phi_1$ and the integral passes the curve in the correct rotation sense. Finally, with the line element

$$du=-d\phi\,r\sin\phi,$$ the integral becomes

$$J=-\frac{r^2}{3}\int_{\phi_0}^{\phi_1} d\phi\,\sin^2\phi=-\frac{r^2}{6}\left( \phi-\frac{1}{2}\sin2\phi \right)_{\phi_0}^{\phi_1}$$ and using the given values for $r,\phi_0,\phi_1$ we obtain $$J=-\frac{176\pi}{9}.$$

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Thanks for the answer. I think you mixed up the end and startpoints, hence J should equal = $\frac{16\Pi}{9}$ – DJJQ Mar 3 '14 at 11:51
@DJJQ Apart from that sign, why you think $\pm\frac{16\pi}{9}$ should be incorrect? – flonk Mar 3 '14 at 12:06
It's indeed the right answer, the same answer would also be given by -7Pi/6 to 2Pi/3 I guess – DJJQ Mar 3 '14 at 14:20
@DJJQ yepp, in principal you can go from $\frac{5\pi}{6}+2\pi k$ to $\frac{2\pi}{3}+2\pi(k+1)$. – flonk Mar 3 '14 at 14:27

Draw a figure! The parametrization of the ellipse is $$t\mapsto \bigl(x(t),y(t)\bigr):=\left({8\over3}\cos t,8\sin t\right)\qquad(t\in{\mathbb R})\ .\tag{1}$$ Now we have to find the boundary values which are relevant for the integral $J$ in question. Both endpoints $p$ and $q$ of the arc lie in the second quadrant. At the starting point we have $8\sin t=4$, or $\sin t={1\over2}$, which amounts to $t=-\pi-{\pi\over6}$, and at the endpoint we have $8\sin t=4\sqrt{3}$, which amounts to $t={2\pi\over3}$.

Therefore we obtain $$J=\int_{-7\pi/6}^{2\pi/3} \left(y(t)(18 x(t)+1)\dot x(t) + 2y^2(t)\dot y(t)\right)\ dt\ ,$$ where you now have to plug in the parametrization $(1)$. The computation can be simplified somewhat by observing that $$\int\nolimits_\gamma 2y^2\ dy={2\over3}y^3\biggr|_p^q\quad .$$

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In my reasoning this is also what I have done, except my startpoint is defined as $\frac{5\Pi}{6}$, which is the same as $\frac{-7\Pi}{6}$ right? – DJJQ Mar 3 '14 at 11:52
@DJJQ: You have to choose the $t$-interval $[\alpha,\beta]$ in such a way that the parametrization $(1)$ actually produces the intended arc and not some other arc with the same endpoints. – Christian Blatter Mar 3 '14 at 12:04