# Work (in space)



My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

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*}