Fundamental Theorem for Line Integrals

Question

Find the work done in moving a particle along a curve from point A(1, 0, −1) to B(2, 2, −3) via the conservative force field ~F (x, y, z) = 〈2y³ − 6xz, 6xy² − 4y, 4 − 3x²〉.

(a) using the Fundamental Theorem for Line Integrals;

(b) by explicitly evaluating a line integral along the curve consisting of the line segment from A to P (1, 2, −1) followed by the line segment from P to B.

Answer 1

a) dM/dx = 2y³- 6xz; M = ∫(2y³ - 6xz)dx = 2y³x - 3x²z + h(y,z);

dM/dy = 6y²x + dh/dy = 6xy² - 4y; dh/dy = -4y; h = - 2y² + f(z)

M = 2y³x - 3x²z - 2y² + f(z); dM/dz = - 3x² + f'(z) = 4 - 3x²; f = 4z + C

M(x,y,z)= 2y³x - 3x²z - 2y² + 4z + C.

So M(x, y, z) is potential of this vector field.

W = ∫_A^BF•dr = M(B) - M(A) = (32 + 36 - 8 - 12 + C) - (0 + 3 - 0 - 4 + C) = 45

b) A(1,0,-1) and P(1,2,-1); AP = <0,2,0>; (x - 1)/0 = y/2 = (z + 1)/0 = t; x = 1; y = 2t; z = - 1; 0 ≤ t ≤1

W₁ = ∫_A^PF•dr = ∫₀¹(6·4t²- 8t)·2dt = (8t³ - 4t²)₀¹ = 4

P(1,2,-1) and B(2,2,-3): PB = <1, 0, -2>; (x - 1)/1 = (y - 2)/0 = (z + 1)/-2 = s; x = s + 1; y = 2; z = -2s - 1

0 ≤ s ≤1

W₂ = ∫_P^BF•dr = ∫₀¹[16 - 6(s + 1)(-2s -1)]ds + [4 - 3(s + 1)²](-2)ds = ∫₀¹(16 + 12s² + 18s + 6 - 8 + 6s² + 12s + 6)ds = ∫₀¹(18s²+ 30s + 20)ds = (6s³ + 15s² + 20s)₀¹ = 6 + 15 + 20 = 41

W = W₁ + W₂ = 45

Answer 2

(a)

Let me restate the Fundamental Theorem for Line Integrals:

If G(x,y,z) is such that F = ∇G then for any curve from A to B, ∫F.ds = G(B) - G(A).

So the constraints on G are

∂G/∂x = 2y³ − 6xz

∂G/∂y = 6xy² − 4y

∂G/∂z = 4 − 3x²

By integration

G = ∫(2y³ − 6xz)dx = 2xy³ - 3x²z + C1(y, z)

G = ∫(6xy² − 4y)dy = 2xy³ - 2y² + C2(x, z)

G = ∫(4 − 3x²)dz = 4z - 3x²z + C3(x, y)

where C1, C2, C3 are constants of integration -- functions of the other two variables,

excluding the variable of integration.

If we just collect all the unique terms from the three integrals, we get

G(x, y, z) = 2xy³ - 3x²z - 2y² + 4z

whose gradient is indeed F.

Then

G(B) = 2×2×2³ - 3×2²×(-3)- 2×2² + 4×(-3) = 48

G(A) = <0 - 3×1²×(-1) - 0 + 4×(-1) = -1

∫F.ds = G(B) - G(A) = 49

(b)

From (1, 0, -1) to (1, 2, -1)

the variables x=1 and z=-1 remain constant, while y goes from 0 to 2.

So we can write ds = <0, dy, 0>, and

F.ds = (6xy² - 4y)dy = (6y² - 4y)dy

And the line integral becomes the definite integral from 0 to 2

∫(6y² - 4y)dy = 2y³ - 2y² from 0 to 2 = 8

From (1, 2, -1) to (2, 2, -3) the variable y=2 remains constant while x and z vary.

We can express that by making them functions of a variable t ranging from 1 to 2:

x(t) = t

y(t) = 2

z(t) = 1-2t

Then ds = <dx, dy, dz> = <dt, 0, -2dt>

The line integral from P to B becomes the definite integral as t ranges from 1 to 2:

∫<2y³ − 6xz, 6xy² − 4y, 4 − 3x²>.<dt, 0, -2dt> =

∫(2y³ − 6xz)dt - 2(4 − 3x²)dt =

∫(2×2³ − 6t(1-2t))dt - 2(4 − 3t²)dt =

∫(8 - 6t + 18t²)dt =

8t - 3t² + 6t³

When evaluated from 1 to 2:

(8×2 - 3×2² + 6×2³) - (8×1 - 3×1² + 6×1³) = 41

So the two integrals from A to P and to P to B add up to 49, which is the same as the result from (a).