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September 12th, 2017, 03:46 AM   #1
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Line integration

Evaluate:
$\displaystyle \int_{C} (z~dx +x~dy+y~dz)$

where,
C is the intersection of $\displaystyle x^{2}+y^{2}=1$ and the plane $\displaystyle y+z=2$. Orient C counter clockwise.
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September 12th, 2017, 05:19 AM   #2
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A standard parameterization for $\displaystyle x^2+ y^2= 1$ is $\displaystyle x= cos(t)$, $\displaystyle y= sin(t)$. From $\displaystyle y+ z= 2$, $\displaystyle z= 2- y= 2- sin(t)$. So we can parameterize the curve with $\displaystyle x= cos(t)$, $\displaystyle y= sin(t)$, $\displaystyle z= 2- sin(t)$ with t from 0 to $\displaystyle 2\pi$.

So $\displaystyle dx= -sin(t)dt$, $\displaystyle dy= cos(t)dt$ and $\displaystyle dz= -cos(t)dt$.

$\displaystyle \int (zdx+ xdy+ ydz)= \int_0^{2\pi} (2- sin(t))(-sin(t)dt)+ cos(t)(cos(t)dt)+ sin(t)(-cos(t)dt)= \int_0^{2\pi} (sin^2(t)+ cos^2(t)- sin(t)cos(t)- 2sin(t))dt= \int_0^{2\pi} (1- sin(t)cos(t)- 2sin(t))dt$
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September 12th, 2017, 11:07 PM   #3
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Thank you so much country boy
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