\(\)

Let \(\vec F\) be a vector field representing a force $$ \vec F = P\hat\imath + Q\hat\jmath + R\hat k = \left\langle P, Q, R \right\rangle \nonumber $$

Let \(C\) be a curve in space, and space vector \(d\vec r\) with components \(d\vec r=\left\langle dx,dy,dz\right\rangle\). Then, the work done by the field $$ \shaded{ \rm{Work} = \int_C\vec F\cdot d\vec r = \int_C P\,dx+Q\,dy+R\,dz } \nonumber $$

The method is the same as in the plane: evaluate by parameterizing \(C\), express \(x,y,z,dx,dy,dz\) in terms of the parameter.

## Examples

Let \(\vec F\) be a force field. What work does the force exert on the particle in two different trajectories? $$ \vec F=\left\langle yz,xz,xy\right\rangle $$

### Trajectory A

Let a particle move over curve \(C\), in force field \(\vec F\). What work does the force exert on the particle? $$ C: x=t^2, y=t^2, z=t, 0 \le t \le 1 $$

Express \(x,y,z\) in terms of parametric variable \(t\) $$ \begin{align*} x=t^3 &\Rightarrow dx = 3t^2\;dt \\ y=t^2 & \Rightarrow dy = 2t\;dt \\ z=t & \Rightarrow dz = dt \end{align*} $$

Substituting \(x,y,z,dx,dy,dz\) in the integral expression $$ \begin{align} \int_C\vec F\cdot d\vec r &=\int_C yz\;dx+xz\;dy+xy\;dz \label{eq:example3dwork} \\ &=\int_0^1 t^3.3t^2dt+t^4.2t\;dt+t^5dt \nonumber \\ &= \int_0^1 6t^5dt = \left[t^6\right]_0^1=1 \nonumber \end{align} $$

### Trajectory B

Let a particle move over curve \(C\) consisting of three line segments, in the same force field \(\vec F=\left\langle yz,xz,xy\right\rangle\). What work does the force exert on the particle? $$ \begin{align*} C_1&: \mathrm{from\ } (0,0,0) \mathrm{\ to\ }(1,0,0) \\ C_2&: \mathrm{from\ } (1,0,0) \mathrm{\ to\ }(1,1,0) \\ C_3&: \mathrm{from\ } (1,1,0) \mathrm{\ to\ }(1,1,1) \end{align*} $$

Curves \(C_1,C_2\) are in the \(xy\)-plane, so \(z=0\) and therefore \(dz=0\).

Substituting \(z\) and \(dz\) in equation \(\eqref{eq:example3dwork}\), only leaves $$ \begin{align*} \int_{C_{1,2}}\vec F\cdot d\vec r &=\int_C y\bcancel{z}_0\;dx+x\bcancel{z}_0\;dy+xy\;\bcancel{dz}_0 \\ &=\int_C xz\;dy = 0 \end{align*} $$

For \(C_3\): $$ \begin{array}{l} x=1 \Rightarrow dx=0 \\ y=1 \Rightarrow dy=0 \\ z \mathrm{\ from\ } 0 \mathrm{\ to\ } 1 \end{array} \nonumber $$

Substituting \(x,y,dx,dy\) in the line integral along \(C_3\) $$ \begin{align*} \int_{C_3}\vec F\;d\vec r &=\int_C yz\;dx+xz\;dy+xy\;dz \\ &=\int_C z.0+1.z.0+1.1\;dz \\ &=\int_0^1 dz = \left[z\right]_0^1=1 \end{align*} $$

Adding the 3 terms together $$ \int_C\vec F\;d\vec r= \int_{C_1}\vec F\;d\vec r + \int_{C_2}\vec F\;d\vec r + \int_{C_3}\vec F\;d\vec r = 1 \nonumber $$

### Compare

Trajectory A and B have the same answer, because the vector field is a (conservative) gradient field. Both paths go from the origin to \((1,1,1)\).

Find a function, what’s gradient is this vector field. It might be \(xyz\) $$ \newcommand{pdv}[1]{\tfrac{\partial}{\partial #1}} \begin{align*} \vec F &= \nabla(xyz) \\ &= \left\langle \pdv{x}xyz, \pdv{y}xyz, \pdv{z}xyz \right\rangle \\ &= \left\langle yz, xz, xy \right\rangle \end{align*} $$

Knowing this, we could have used the fundamental theorem of calculus for line integrals (written using the symbolic \(\nabla\)-operator) $$ \shaded{ \int_C\nabla f\cdot d\vec r = f(P_1)-f(P_o) } \nonumber $$

Substituting \(f(x,y,z)=xyz\) $$ \begin{align*} \int_C\nabla f\cdot d\vec r &= f(1,1,1) – f(0,0,0) \\ &= 1 – 0 = 1 \end{align*} $$