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How can we calculate the line integral $$\int_r (xy^2z^2-1, x^2yz^2, x^2y^2z)\cdot ds$$ where $$r=r_1\cup r_2$$ where the parametric representation of $r_1$ is $$\sigma_1 (t)=(\sin t, \cos t, t-\frac{\pi}{2}), t\in [0,\frac{\pi}{2}] $$ and the parametric representation of $r_2$ is $$\sigma_2 (t)=(\cos^3 \phi, \sin^3 \phi, 0), \phi \in [0,\pi] $$ ?

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I have done the following:

$$\int_{r} (xy^2z^2-1, x^2yz^2, x^2y^2z)\cdot ds=\int_{\sigma_1} \left (\left (\sin t \cos^2 t\left (t-\frac{\pi}{2}\right )^2-1\right )\cos t-\sin^2 t\cos t \left (t-\frac{\pi}{2}\right )^2\sin t+\sin^2 t\cos^2 t\left (t-\frac{\pi}{2}\right )\right )dt+ \\ \int_{\sigma_2} 3\cos^2\phi \sin \phi d\phi \\ =-1+2=1 $$

Is this correct?

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    $\begingroup$ Is it supposed to be $\cos^4 t$ and $\sin^3 t$ in $\sigma_2(t)$? $\endgroup$ – Arthur Jan 28 '16 at 12:09
  • $\begingroup$ I edited it and added also that what I have done... Could you take a look at it? @Arthur $\endgroup$ – Mary Star Jan 28 '16 at 12:25
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Since the integrand is a total differential, there is no need to carry out any explicit integration!

Once one figure out $r$ start at $\left[0,1,-\frac{\pi}{2}\right]$ and end at $[-1,0,0]$, one can evaluate the integral by computing the differences of the "primitive" at the endpoints.

$$\int_r (xy^2z^2-1, x^2yz^2, x^2y^2z)\cdot ds = \int_r d\left(\frac{(xyz)^2}{2} - x\right) = \left[\frac{(xyz)^2}{2} - x \right]_{[0,1,-\frac{\pi}{2}]}^{[-1,0,0]}\\ = \left( \frac{0^2}{2} - (-1) \right) - \left( \frac{0^2}{2} - 0 \right) = 1$$

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  • $\begingroup$ The integral over $r_2$ is $$\int_0^{\pi}3\cos^2\phi\sin\phi d\phi=\int_0^{\pi}d(-\cos^3\phi)=-(-1-1)=2$$ or not? $\endgroup$ – Mary Star Jan 28 '16 at 13:15
  • $\begingroup$ I edited it now... It should be $\sigma_2(\phi) = ({(\cos\phi)^3},(\sin\phi)^3,0)$. $\endgroup$ – Mary Star Jan 28 '16 at 13:19
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    $\begingroup$ @MaryStar I updated the answer (and fix some errors), the final answer is $1$. $\endgroup$ – achille hui Jan 28 '16 at 13:24
  • $\begingroup$ Great!! Thanks a lot!! :-) $\endgroup$ – Mary Star Jan 28 '16 at 13:26

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