Question

Calculate the work done by a force F = 3x2ˆi + (xz − y)ˆj + 2z kˆ
r
in moving a particle
along the curve x = 2t2; y = t; z = 4t2 − t from t = 0 to t = 1. Is the force conservative?

Answer 1

Denote the curve by the letter C, and its parameterization by

$$x(t) = 2t^2, \; y(t) = t, \; z(t) = 4t^2-t.$$

Then

$$x'(t) = 4t, \; y'(t) = 1, \; z'(t) = 8t-1.$$

The work done by a force is calculated using a curvilinear integral of the second kind, which reduces to a definite integral [https://en.wikipedia.org/wiki/Work_(physics)#Path_dependence]:

$$\int\limits_C 3x^2dx+(xz-y)dy+2zdz = \\ \int\limits_0^1 \bigl(3x^2(t) x'(t) + (x(t)z(t)-y(t)) y'(t) + 2z(t)z'(t)\bigr) dt = \\ \int\limits_0^1 (48 t^5+8 t^4+62 t^3-24 t^2+t) dt = \\ 8 t^6+\frac{8 t^5}{5}+\frac{31 t^4}{2}-8 t^3+\frac{t^2}{2} \biggl\rvert_0^1 = \frac{88}{5}.$$

Check the conservativity of the force by calculating the curl [https://en.wikipedia.org/wiki/Conservative_force#Mathematical_description]:

$$\nabla \times (3x^2, xz-y,2z) = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3x^2 & xz-y & 2z \end{vmatrix} = (-x,0,z).$$

It is nonzero, which means that the force is non-conservative.