Consider the vector field F(x, y, z) = (3 + 2xe^y − z^2sin(x))i + (x^2 e^y − 6 y^2 z)j + (2z cos(x) − 2 y^3)k. Find the work done by the force field F on an object that moves from (0,−1,−1) to (π, 1, 1).

The work done by a force field on an object moving along a path is given by the line integral of the force field along the path. In this case, the force field is F and the path is a straight line from (0,-1,-1) to (π,1,1).

The line integral of a vector field F along a path C from point A to point B is given by:

∫_C F • dr

where dr is a differential displacement vector along the path, and "•" denotes the dot product.

First, we parameterize the path from (0,-1,-1) to (π,1,1). A possible parameterization is:

r(t) = (πt, t-1, t-1) for 0 ≤ t ≤ 2

Then, dr/dt = (π, 1, 1)

Next, we substitute r(t) into F to get F(r(t)):

F(r(t)) = (3 + 2πte^(t-1) - (t-1)^2sin(πt))i + ((πt)^2 e^(t-1) - 6(t-1)^2(t-1))j + (2(t-1)cos(πt) - 2(t-1)^3)k

Then, we take the dot product of F(r(t)) and dr/dt:

F(r(t)) • dr/dt = (3 + 2πte^(t-1) - (t-1)^2sin(πt))π + ((πt)^2 e^(t-1) - 6(t-1)^2(t-1)) + (2(t-1)cos(πt) - 2(t-1)^3)

Finally, we integrate this from t=0 to t=2 to get the work done:

W = ∫ from 0 to 2 of F(r(t)) • dr/dt dt

This integral can be computed numerically to get the final answer.