# Thread: line integral of a shifted semi-circle

1. ## line integral of a shifted semi-circle

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?)

2. ## Re: line integral of a shifted semi-circle

Before I go any further, are you only travelling along the curved surface of the semicircle, or are you also travelling along the x axis in order to close the semicircle?

3. ## Re: line integral of a shifted semi-circle

only along the semicircle, not including the x-axis. also , the area of the field is y>0, so the endpoints of the semicircle are not included.

4. ## Re: line integral of a shifted semi-circle

I did more work on this problem. My answer is -4.
Could you please see if I’ve solved it correctly?

(Previously, I miswrote the range. It should be 0 < t < π )

F · dr = ∫ (from 0 to π) -sin(t)cos(t)dt - 2sin(t)dt + sin(t)cos(t)

Integrating this, I broke it into 3 separate parts:

1: ∫ (from 0 to π) -sin(t)cos(t) dt
2: ∫ (from 0 to π) 2sin(t) dt
3: ∫ (from 0 to π) sin(t)cos(t) dt

1: using substitution:
u = cos(t)
du/dt = 1sin(t)
du = -sin(t)dt
dt = du/-sin(t)

∫ (from 0 to π) -sin(t)cos(t)(du/-sin(t))
= ∫ (from 0 to π) cos(t)(du)
= ∫ (from 0 to π) u du
= (1/2)u^2, evaluated from π to 0
= (1/2)cos^2(t), evaluated from π to 0
= (1/2)cos^2(π) - (1/2)cos^2(0)
using the trig identity cos^2(x) = { [1 + cos(2x)] / 2 }
= (1/2) { ( [1+cos(2π)] / 2) - ( [1+cos(2(0))] / 2) }
= 1/2(1/2 +1/2) - 1/2(1/2 +1/2)
= (1/2)(1) - (1/2)(1)
= 0

2:
- ∫ (from 0 to π) 2sin(t) dt
= -2 ∫ (from 0 to π) sin(t) dt
= -2[-cos(t)] evaluated from π to 0
= 2[cos(t)] evaluated from π to 0
= 2[cos(π) - cos(0)]
= 2[(-1) - (1)]
= -4

3:
u = cos(t)
du/dt = 1sin(t)
du = -sin(t)dt
dt = du/-sin(t)

∫ (from 0 to π) -sin(t)cos(t)(du/-sin(t))
= ∫ (from 0 to π) - u du
= (-1/2)u^2, evaluated from π to 0
= (-1/2)cos^2(t), evaluated from π to 0
= (-1/2) [cos^2(π) - cos^2(0)]
= (-1/2) { [1+cos(2π)] / 2) - ( [1+cos(2(0)] / 2) }
= (-1/2) { [(1/2) + (1/2)] - [(1/2) + (1/2)] }
= 0

(0) + (-4) + (0) = -4