# Line integral evaluation

Find line integral:

$$I = \oint\limits_{C} (y-z)\mathrm{d}x + (x^2-y)\mathrm{d}y + (z-x)\mathrm{d}z$$

where curve C is given with: $$\begin{array}\\ x = a\cos{t}\\ y = a\sin{t}\\ z = a^2\cos{2t}\\ \end{array}$$

and where $t\in \mathbb{R}, 0<t<2\pi$, and direction of $C$ is the same as the direction of growth of variable $t$.

What I've tried so far:

1) If we go standardly: $\mathrm{d}x=-a\sin{t}\mathrm{d}t$, $\mathrm{d}y=a\cos(t)\mathrm{d}t$, $\mathrm{d}z=-4a^2\sin{t}\cos{t}\mathrm{d}t$, and substitute all into integral, we get: $$\int\limits_{0}^{2\pi}-a^2\sin^2{t} + 3a^3\cos^2{t}\sin{t} - a^3\sin^3{t} + a^3\cos^3{t} - a^2\sin{t}\cos{t}-4a^4\cos^3{t}\sin{t} + 4a^4\sin^3{t}\cos{t})\mathrm{d}t$$

but this is very ugly and I have no idea how to proceed, apart from trying random trig manipulations, which I've tried to no success.

2) If we realize that $z = a^2\cos^2{t} - a^2\sin^2{t} = x^2 - y^2$, and that $\mathrm{d}x=-y\mathrm{d}t$, $\mathrm{d}y=x\mathrm{d}t$, and $\mathrm{d}z=-4xy\mathrm{d}t$, we can substitute that into the integral, and get: $$\int\limits_0^{2\pi}\left(y^2+3x^2y-y^3+x^3-xy-4x^3y+4xy^3\right)\mathrm{d}t,$$ but this is also kind of hopeless :)

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Hi Vidak, welcome to Math SE! It'll help us give you more useful answers if you fill out your questions with such details as how much you know about the topic and what you've tried so far on the problem. If this is homework, please tag it "homework" as well. –  Kevin Carlson Sep 20 '12 at 13:01
@KevinCarlson Hi, edited with my progress so far :) –  Vidak Sep 20 '12 at 13:50
It seems to me that the first approach leads to rather easy integrals. Two of them are slightly more involved, see for example wolframalpha.com/input/?i=Integrate[%28Sin[t]%29^3%2Ct] –  Siminore Sep 20 '12 at 13:55

Since sin and cos are periodic, shift the domain of $\theta$ from $[0,2 \pi]$ to $[- \pi, \pi]$. Since sin is odd, $\int \limits_{-\pi}^\pi \cos^2 t \sin t dt$ , $\int \limits_{-\pi}^\pi \sin^3 t dt$ , $\int \limits_{-\pi}^\pi \cos t \sin t dt$ , $\int \limits_{-\pi}^\pi \cos^3 t \sin t dt$ and $\int \limits_{-\pi}^\pi \cos t \sin^3 t dt$ all vanish, so only $\int \limits_0^{2\pi} \cos^3 t dt$ and $\int \limits_0^{2\pi} \cos^2 t dt$ remain.

Shifting the domain from $[0,2 \pi]$ to $[- {\pi \over 2}, {3\pi \over 2}]$, $$\int \limits_0^{2\pi} \cos^3 t dt = \int \limits_{-\pi \over 2}^{3\pi \over 2} \sin^3 ({\pi \over 2} - t) dt = \int \limits_{-\pi}^\pi \sin^3 u du = 0$$

Using the same trick, $\int \limits_0^{2\pi} \cos^2 t dt = \int \limits_0^{2\pi} \sin^2 t dt$. Hence $$\int \limits_0^{2\pi} \cos^2 t dt = {1 \over 2} \int \limits_0^{2\pi} (\sin^2 t + \cos^2 t) dt = \pi$$ Therefore, answer is $-a^2 \pi$.

Another way to simplify the integral is to find a potential for the field. Let $F = (y-z,x^2-y,z-x)$, $F_1 = (y,x^2,0)$ and $f = xz-{y^2 \over 2} -{z^2 \over 2}$. Then $F = \nabla f + F_1$. Therefore, $$\oint \limits_C \langle F,T \rangle ds = \oint \limits_C \langle F_1,T \rangle ds + 0 = \oint \limits_C \langle F_1,T \rangle ds$$ which gives the same answer.

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Why "this is ugly"?

Following your first step, you only need to find:

$$\int \sin^2 x, \int \cos^2x\sin x, \int \cos^3x-\sin^3x, \int \sin x\cos x, \int \cos^3x\sin x, \int \sin^3x\cos x$$ one by one.

Can you use $\int \cos^2 xd(\cos x)$ for one of them? You might also want to change $\sin^2x$ into a function of $\cos 2x$.

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oh lol, forgot integrals are additive :D thanks! –  Vidak Sep 20 '12 at 14:29

I would continue (1):

$a$ is a constant, doesn't bother, and there are lots to know about $\cos$ and $\sin$. First of all, $\cos 2t$ is fine as it is. $\sin^3t$ will be related to $\cos 3t$ and $\sin 3t$...

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This is probably a comment... –  Siminore Sep 20 '12 at 14:04