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 September 12th, 2017, 02:46 AM #1 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 876 Thanks: 60 Math Focus: सामान्य गणित 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.
 September 12th, 2017, 04:19 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807 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 $\displaystyle dz= -cos(t)dt$. $\displaystyle \int (zdx+ xdy+ ydz)= \int_0^{2\pi} (2- sin(t))(-sin(t)dt)+ cos(t)(cos(t)dt)+ sin(t)(-cos(t)dt)= \int_0^{2\pi} (sin^2(t)+ cos^2(t)- sin(t)cos(t)- 2sin(t))dt= \int_0^{2\pi} (1- sin(t)cos(t)- 2sin(t))dt$ Thanks from MATHEMATICIAN and Benit13
 September 12th, 2017, 10:07 PM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 876 Thanks: 60 Math Focus: सामान्य गणित Thank you so much country boy

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