Question

Evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results. \(\int_{C}(y+2 z) d x+(x-3 z) d y+(2 x-3 y) d z\) (a) \(C\) : line segment from (0,0,0) to (1,1,1) (b) \(C:\) line segments from (0,0,0) to (0,0,1) to (1,1,1) (c) \(C:\) line segments from (0,0,0) to (1,0,0) to (1,1,0) to (1,1,1)

Short answer

For all paths C, the line integral is 0

Step-by-step solution

Identify the Potential Function

The potential function F can be identified by integrating each component of the given vector field: \[ F(x,y,z) = \int (y+2z)dx + \int (x-3z)dy + \int (2x-3y)dz = xy + x^2z + 2xz - (3/2)y^2 - 3/2*z^2 \]

Evaluate for Path C (a)

Using the Fundamental Theorem of Line Integrals, evaluate F at the endpoints of the path C from (0,0,0) to (1,1,1): \[ \Delta F = F(1,1,1) - F(0,0,0) = 1 + 1 + 2 - 3/2 - 3/2 = 0 \]

Evaluate for Path C (b)

Evaluate F at the endpoints of the path C going from (0,0,0) to (0,0,1) to (1,1,1): \[ \Delta F = F(1,1,1) - F(0,0,1) + F(0,0,1) - F(0,0,0) = 0 \]

Evaluate for Path C (c)

Evaluate F at the endpoints of the path C going from (0,0,0) to (1,0,0) to (1,1,0) to (1,1,1): \[ \Delta F = F(1,1,1) - F(1,1,0) + F(1,1,0) - F(1,0,0) + F(1,0,0) - F(0,0,0) = 0 \]