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, 6xy³ − 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

Note: the middle component of the force should be 6xy^2 - 4y in order to be conservative

a) ∇f = 〈∂f/∂x, ∂f/∂y, ∂f/∂z〉= F

∂f/∂x = 2y³ − 6xz

f(x,y,z) = 2xy³ - 3x²z + C(y,z)

∂f/∂y = 6xy²+ ∂C/∂y

Thus, ∂C/∂y = -4y, C(y,z) = -2y²+D(z)

f(x,y,z) = 2xy³ - 3x²z - 2y²+D(z)

∂f/∂z = - 3x² + ∂D/∂z

Thus, ∂D/∂z = 4, D(z) = 4z

Finally, f(x,y,z) = 2xy³ - 3x²z - 2y² + 4z

Work is equal to f(B) - f(A) = 49

b) We must write an equation for a line from A to P

A(1, 0, -1) and P(1(1, 2, -1) so r₁(t) = 〈1, 2t, -1〉which travels from A to P from t = 0 to t = 1 seconds

We now write F in terms of time since Work = ∫F•dr

F (x, y, z) = 〈2y³ − 6xz, 6xy² − 4y, 4 − 3x²〉

r₁(t) = 〈x(t), y(t), z(t)〉= 〈1, 2t, -1〉

F(t) = 〈2*(2t)³ − 6(1)(-1), 6(1)(2t)² − 4(2t), 4 − 3(1)²〉= 〈16t³+6, 24t² − 8t, 1〉

dr = 〈0, 2, 0〉* dt

W = ∫₀¹F(t) • dr = ∫₀¹2*(24t² − 8t) * dt = 8

Follow a similar procedure to find the work from P to B. Work = 41

Then add the work to find the total work from A to B.

Work from A to B = Work from A to P + Work from P to B = 49, same as part a)