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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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It seems to me that the first approach leads to rather easy integrals. Two of them are slightly more involved, see for example[%28Sin[t]%29^3%2Ct] – Siminore Sep 20 '12 at 13:55
up vote 1 down vote accepted

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

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