Vector calculus - line integral computation

  • #1
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Compute the line integral [tex]\int_{C} F\cdot dr[/tex] where F = -y i + x j. The directed path C in the xy-plane consists of two parts: i) a left semicircle from (0, -1) to (0, 1) with center at the origin, and ii) a straight line segment from (0,1) to (2,1).

i) r(t) = cos t i + sin t j [pi/2 <=t<= 3pi/2]
ii) r(t) = 2t i + j [0<=t<=1]

for i):
F(r (t) ) = - sin t i + cos t j
r'(t) = - sin t i + cos t j

integ F(r(t)) dot r'(t) dt
= integ 1 dt [pi/2<=t<= 3pi/2]
so integral is
3pi/2 - pi/2 = pi ???
is this correct so far.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Compute the line integral [tex]\int_{C} F\cdot dr[/tex] where F = -y i + x j. The directed path C in the xy-plane consists of two parts: i) a left semicircle from (0, -1) to (0, 1) with center at the origin, and ii) a straight line segment from (0,1) to (2,1).

i) r(t) = cos t i + sin t j [pi/2 <=t<= 3pi/2]
ii) r(t) = 2t i + j [0<=t<=1]

for i):
F(r (t) ) = - sin t i + cos t j
r'(t) = - sin t i + cos t j

integ F(r(t)) dot r'(t) dt
= integ 1 dt [pi/2<=t<= 3pi/2]
so integral is
3pi/2 - pi/2 = pi ???
is this correct so far.
Almost- you have the "orientation" backwards. The path does not go "from pi/2 to 3pi/2" (i.e. from (0,1) to (0,-1)), it goes "from 3pi/2 to pi/2" (from (0,-1) to (0,1)). That reverses the sign on the integral.
 

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