Question

A particle moves along line segments from the origin to the points (3, 0, 0), (3, 5, 1), (0, 5, 1), and back to the origin under the influence of the force field F(x, y, z) = z2i + 5xyj + 2y2k. Find the work done.

Answer 1

Parameterize each line segment from $(x_0,y_0,z_0)$ to $(x_1,y_1,z_1)$ by

$\vec r(t) = (1-t) (x_0\,\vec\imath + y_0\,\vec\jmath + z_0\,\vec k) + t (x_1\,\vec\imath + y_1\,\vec\jmath + z_1\,\vec k$

with $0\le t\le1$. The work done by $\vec F$ on the particle along each segment is given the line integral of $\vec F$ with respect to that segment,

$\displaystyle \int_{C_i} \vec F \cdot d\vec r = \int_0^1 \vec F(\vec r_i(t)) \cdot \dfrac{d\vec r_i(t)}{dt} \, dt$

• (3, 0, 0) to (3, 5, 1)

$\vec r_1(t) = 3\,\vec\imath + 5t\,\vec\jmath + t\,\vec k$

$W_1 = \displaystyle \int_0^1 \left(t^2\,\vec\imath + 75t\,\vec\jmath + 50t^2\,\vec k\right) \cdot \left(5\,\vec\jmath + \vec k\right) \, dt \\\\ ~~~~~~~~ = \int_0^1 (375t + 50t^2) \, dt = \frac{1225}6$

• (3, 5, 1) to (0, 5, 1)

$\vec r_2(t) = 3(1-t)\,\vec\imath + 5(1-t)\,\vec\jmath + \vec k$

$W_2 = \displaystyle \int_0^1 \left(\vec\imath + 75(1-t)\,\vec\jmath + 50 \,\vec k\right) \cdot \left(-3\,\vec\imath - 5\,\vec\jmath\right) \, dt \\\\ ~~~~~~~~ = -3 \int_0^1 \,dt = -3$

• (0, 5, 1) to (0, 0, 0)

$\vec r_3(t) = 5(1-t)\,\vec\jmath + (1-t)\,\vec k$

$W_3 = \displaystyle \int_0^1 \left((1-t)^2\,\vec\imath + 50(1-t)^2\,\vec k\right) \cdot \left(-5\,\vec\jmath - \vec k\right) \, dt \\\\ ~~~~~~~~ = \int_0^1 (-50 + 100t - 50t^2) \, dt = -\frac{50}3$

Then the total work done by $\vec F$ on the particle is

$W = W_1 + W_2 + W_3 = \boxed{\dfrac{369}2}$