Evaluate the Line integral ∫c (2x + y² - z) ds where C is the curve counterclockwise around a triangle with vertices (0,0,0), (0,1,0), (0,1,1).
Hint: Parameterize the piece-wise smooth curve first. Be sure and figure out the range (limits) for your parameter correctly for each piece of the path. Give exact answer only.
Hint: Parameterize the piece-wise smooth curve first. Be sure and figure out the range (limits) for your parameter correctly for each piece of the path. Give exact answer only.
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Note that we want positive orientation. In order for this to occur, we need to go from (0, 0, 0) to (0, 1, 0) to (0, 1, 1).
The vector equation of the line from (0, 0, 0) to (0, 1, 0) is:
a(t) = <0, 0, 0> + t<0, 1, 0> = <0, t, 0>,
so we can parametrize this line as:
x = 0, y = t, z = 0 (0 <= t <= 1).
Then, the line segment from (0, 1, 0) to (0, 1, 1) is parametrized as:
x = 0, y = 1, z = t (0 <= t <= 1),
and the line segment from (0, 1, 1) to (0, 0, 0) is parametrized as:
x = 0, y = t, z = t (0 <= t <= 1).
If we call the line from (0, 0, 0) to (0, 1, 0) C₁, the line from (0, 1, 0) to (0, 1, 1) C₂, and the line from (0, 1, 1) to (0, 0, 0) C₃:
∫C (2x + y^2 - z) ds, with C = C₁ U C₂ U C₃
= ∫C₁ (2x + y^2 - z) ds + ∫C₂ (2x + y^2 - z) ds + ∫C₃ (2x + y^2 - z) ds
= ∫ [(0 + t^2 - 0)(1) + (0 + 1 - t)(1) + (0 + t^2 - t)(1)] dt (from t=0 to 1)
= ∫ (t^2 - t + 1 + t^2 - t) dt (from t=0 to 1)
= ∫ (2t^2 - 2t + 1) dt (from t=0 to 1)
= [(2/3)t^3 - t^2 + t] (evaluated from t=0 to 1)
= (2/3 - 1 + 1) - (0 - 0 + 0)
= 2/3.
I hope this helps!
The vector equation of the line from (0, 0, 0) to (0, 1, 0) is:
a(t) = <0, 0, 0> + t<0, 1, 0> = <0, t, 0>,
so we can parametrize this line as:
x = 0, y = t, z = 0 (0 <= t <= 1).
Then, the line segment from (0, 1, 0) to (0, 1, 1) is parametrized as:
x = 0, y = 1, z = t (0 <= t <= 1),
and the line segment from (0, 1, 1) to (0, 0, 0) is parametrized as:
x = 0, y = t, z = t (0 <= t <= 1).
If we call the line from (0, 0, 0) to (0, 1, 0) C₁, the line from (0, 1, 0) to (0, 1, 1) C₂, and the line from (0, 1, 1) to (0, 0, 0) C₃:
∫C (2x + y^2 - z) ds, with C = C₁ U C₂ U C₃
= ∫C₁ (2x + y^2 - z) ds + ∫C₂ (2x + y^2 - z) ds + ∫C₃ (2x + y^2 - z) ds
= ∫ [(0 + t^2 - 0)(1) + (0 + 1 - t)(1) + (0 + t^2 - t)(1)] dt (from t=0 to 1)
= ∫ (t^2 - t + 1 + t^2 - t) dt (from t=0 to 1)
= ∫ (2t^2 - 2t + 1) dt (from t=0 to 1)
= [(2/3)t^3 - t^2 + t] (evaluated from t=0 to 1)
= (2/3 - 1 + 1) - (0 - 0 + 0)
= 2/3.
I hope this helps!
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From (0,0,0) to (0,1,0):
0
x(t) = 0, x'(t) = 0
y(t) = t, y'(t) = 1
z(t) = 0, z'(t) = 0
ds/dt = Sqrt(0^2 + 1^2 + 0^2) = 1 ---> ds=dt
Our integral becomes:
(2*0 + t^2 - 0)dt = (t^2)dt --> (1/3)t^3 = (1/3)(1 - 0) = 1/3
From (0,1,0) to (0,1,1)
0
x(t) = 0, x'(t) = 0
y(t) = 1, y'(t) = 0
z(t) = t, z'(t) = 1
ds/dt = sqrt(0^2 + 0^2 + 1^2) = 1 ---> ds = dt
Our integral becomes:
(2*0 + 1^2 - t)dt = (1 - t)dt ---> (t - (t^2)/2) = (1 - 1/2) - (0 - 0) = 1/2
From (0,1,1) to (0,0,0)
0
x(t) = 0, x'(t) = 0
y(t) = 1 - t, y'(t) = -1
z(t) = 1 - t, z'(t) = -1
ds/dt = sqrt(0^2 + (-1)^2 + (-1)^2) = sqrt(2) --> ds = Sqrt(2)dt
Our integral becomes:
(2*0 + (1 - t)^2 - (1 - t))dt = (t^2 - 2t + 1 - 1 + t)dt = (t^2 - t)dt --> (1/3)t^3 - (1/2)t^2 = (1/3) - (1/2) = -1/6
The sum of these integrals is:
1/3 + 1/2 - 1/6 = 2/3
0
x(t) = 0, x'(t) = 0
y(t) = t, y'(t) = 1
z(t) = 0, z'(t) = 0
ds/dt = Sqrt(0^2 + 1^2 + 0^2) = 1 ---> ds=dt
Our integral becomes:
(2*0 + t^2 - 0)dt = (t^2)dt --> (1/3)t^3 = (1/3)(1 - 0) = 1/3
From (0,1,0) to (0,1,1)
0
x(t) = 0, x'(t) = 0
y(t) = 1, y'(t) = 0
z(t) = t, z'(t) = 1
ds/dt = sqrt(0^2 + 0^2 + 1^2) = 1 ---> ds = dt
Our integral becomes:
(2*0 + 1^2 - t)dt = (1 - t)dt ---> (t - (t^2)/2) = (1 - 1/2) - (0 - 0) = 1/2
From (0,1,1) to (0,0,0)
0
x(t) = 0, x'(t) = 0
y(t) = 1 - t, y'(t) = -1
z(t) = 1 - t, z'(t) = -1
ds/dt = sqrt(0^2 + (-1)^2 + (-1)^2) = sqrt(2) --> ds = Sqrt(2)dt
Our integral becomes:
(2*0 + (1 - t)^2 - (1 - t))dt = (t^2 - 2t + 1 - 1 + t)dt = (t^2 - t)dt --> (1/3)t^3 - (1/2)t^2 = (1/3) - (1/2) = -1/6
The sum of these integrals is:
1/3 + 1/2 - 1/6 = 2/3