This problem is to find the line integral of a semi-circle, from (3,0) to (1,0), centered at (2,0) above the x-axis, y > 0. The vector field is (x i + y j)

I have calculated that the range for t is: π < t < 0,

Parameterizing of a unit circle:

r(t) = x(t)i + y(t)j = cos(t)i + sin(t)j

x=cos(t)

y=sin(t)

This is what I’ve done, but the answer is obviously incorrect:

The circle’s equation is (x-2)^2 + (y-0)^2 = 1

x-2=cos(t), so that:

x=cos(t)+2

y=sin(t)

F = [(cos(t)+2] i + [sin(t)] j

r’(t) = dr/dt = x’(t) i + y’(t) j = { -sin(t) i + cos(t) j }

dr = { -sin(t) i + cos(t) j } dt

F · dr = ∫(from 0 to π) { [(cos(t)+2] i + [sin(t)] j · -sin(t) i + cos(t) j } dt

= ∫(from 0 to π) { -sin(t)cos(t)-2sin(t)] +cos(t)sin(t) } dt

= { -sin(π)cos(π) - 2sin(π) + cos(π)sin(π) } - { -sin(0)cos(0) - 2sin(0) + cos(0)sin(0) }

= { 0 - 0 + 0 } - { -0 - 0 + 0 } …(all the sin’s equal zero)

= 0

(As a side note, how does π < t < 0 differ from π ≤ t ≤ 0 in calculating the answer?)

Thanks in advance.