Line integrals

Our last topic!!

Falling and work

Suppose we shoot a 1kg object horizontally off of platform that's 64 feet hight with an initial speed of 10 feet per second.

A parabolic path

What is the work done by gravity on this object?

Defining formulae

In the preceding slide,

  • Path: $\mathbf{r}(t) = \langle 10t, 64-16t^2 \rangle$
  • Gravitational field: $\mathbf{F}(x,y) = \langle 0, -32 \rangle$.

The vector field is constant in this case but need not be in general.

Formula for solution

We can express the work as a line integral:

$$ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^2 \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$
  • The left side is an abstract formulation where $C$ is the path, $\mathbf{F}$ is the vector field, and $\mathbf{r}$ is a parametrization of the path.
  • The right side allows you to evaluate the line integral by translating it to a normal integral.


Using the line integral formulation, the work is

$$\begin{align} \int_C \mathbf{F} \cdot d\mathbf{r} &= \int_0^2 \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt \\ &= \int_0^2 \langle 0, -32 \rangle \cdot \langle 10,-32t \rangle \, dt \\ &= \int_0^2 1024t \, dt = 2048. \end{align}$$


Trajectory with explanation

The work required to move the object from one point to the next is approximately $\mathbf{F}(\mathbf{r}(t_i))\cdot \mathbf {r}'(t_i)\Delta t$. If we add those all up, we get

$$ \sum_{i=1}^n \mathbf{F}(\mathbf{r}(t_i))\cdot \mathbf {r}'(t_i)\Delta t \to \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$

Vector fields

  • A vector field is a function $\mathbf{F}:\mathbb R^n \to \mathbb R^n$.
  • Typically, we visualize a vector field by drawing a collection of vectors each emanating from the point where we evalaute the field.
  • Often, we need to scale the vectors to keep the images fro being too cluttered.

Example 1

The identity field
$$\mathbf{F}(x,y) = \langle x,y \rangle$$

Example 2

A rotational field
$$\mathbf{F}(x,y) = \langle -y,x \rangle$$

Example 3

A 3D field
$$\mathbf{F}(x,y,z) = \langle x,-z,y \rangle$$

Example 4

An electrostatic field

An electrostatic field

A 3D integral

A 3D path

The path of $\textbf{r}(t) = \langle 2 t, t^2, t^3 \rangle$ over the time interval $-1\leq t \leq 1$ through the vector field $\mathbf{F}(x,y,z) = \langle x,-z,y \rangle$.

The corresponding line integral

To find the work done by that vector field on an object moving through that path, we evaluate

$$ \int_C \textbf{F}\cdot d\textbf{r} = \int_{-1}^1 \textbf{F}(\textbf{r}(t))\cdot\textbf{r}(t) \, dt. $$

Since $\mathbf{F}(x,y,z) = \langle x,-z,y \rangle$ and $\textbf{r}(t) = \langle 2 t, t^2, t^3 \rangle$,

$$\textbf{F}(\textbf{r}(t))\cdot \textbf{r}(t) = \langle 2t, -t^3, t^2 \rangle\cdot\langle 2,2t,3t^2\rangle = 4t+t^4.$$

Finishing it

Finishing the integral, we get

$$\begin{align} \int_C \textbf{F}\cdot d\textbf{r} &= \int_{-1}^1 \textbf{F}(\textbf{r}(t))\cdot\textbf{r}(t) \, dt\\ &= \int_{-1}^1 \left(4t+t^4\right) \, dt = 2/5. \end{align}$$

Conservative vector fields

A vector field $\mathbf{F}$ is called conservative if it arises as the gradient of a real valued function.
That is, there is a function $f:\mathbb R^n\to \mathbb R$ such that $\mathbf{F} = \nabla f$.

The function $-f$ is sometimes called the potential of the vector field.


Let $$\mathbf{F}(x,y,z) = - \frac{\langle x,y,z \rangle}{(x^2+y^2+z^2)^{3/2}}.$$ Then $\mathbf{F}$ is conservative because $\mathbf{F} = \nabla f,$ where $$f(x,y,z) = 1/\sqrt{x^2+y^2+z^2},$$ as you can check. This example is is essentially the gravitational field.

Finding the potential

  • Vector fields might or might not be conservative.
  • If $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ is a conservative 2D field, then $$P = \frac{\partial f}{\partial x} \: \text{ and } \: Q = \frac{\partial f}{\partial y},$$ for some $f$. Thus (equating the mixed partials of $f$), $$\frac{\partial P}{\partial y}= \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial Q}{\partial x}.$$ This gives us a test to see if a vector field is conservative.


  • $\mathbf{F}(x,y) = \langle xy, x^2y^2 \rangle$ is not conservative since $$\frac{\partial}{\partial y}xy = x \neq 2xy^2 = \frac{\partial}{\partial x}x^2y^2.$$
  • $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ is conservative since $$\frac{\partial}{\partial y}2xy^3 = 6xy^2 = \frac{\partial}{\partial x}(3x^2y^2+1).$$

Finding the potential

Since $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ is conservative, there must be a function $f$ such that $\mathbf{F}=\nabla f$.

  • Surely, $\frac{\partial f}{\partial x} = 2xy^3$. Thus, $$f(x,y) = x^2y^3 + g(y).$$
  • Note that $g(y)$ is a constant of integration; it is constant as far as $x$ is concerned. To find it more precisely, note that $$\frac{\partial f}{\partial y} = 3x^2y^2 + g'(y) = 3x^2y^2 + 1.$$ Thus, $g'(y)=1$ so $g(y)=y$ and $f(x,y) = x^2y^3+y$.

The fundamental theorem of line integrals

If $\mathbf{F} = \nabla f$ is a conservative vector field and $C$ is a path starting at $A$ and ending at $B$, then

$$\int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A).$$

This makes it easy to compute line integrals!
(for conservative vector fields)


Let $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$, let $\mathbf{r}(t) = \langle t^2,t \rangle$, and $C$ denote the portion of the path parametrized by $\mathbf{r}$ over the time interval $[0,1]$.
Let's evaluate $\displaystyle \int_C \mathbf{F}\cdot d\mathbf{r}.$

Solution: We already know that $\mathbf{F}$ is the gradient of $f(x,y)=x^2y^3+y$. We can also see that $C$ moves from $(0,0)$ to $(1,1)$ over the given time interval. Thus, $$\int_C \mathbf{F}\cdot d\mathbf{r} = f(1,1)-f(0,0) = 2.$$

Solution 1

We already know that $\mathbf{F}$ is the gradient of $f(x,y)=x^2y^3+y$. We can also see that $C$ moves from $(0,0)$ to $(1,1)$ over the given time interval. Thus, $$\int_C \mathbf{F}\cdot d\mathbf{r} = f(1,1)-f(0,0) = 2.$$

Solution 2

Our vector field is $\mathbf{F}(x,y) = \langle 2xy^3, 3x^2y^2 + 1 \rangle$ and our path is $\mathbf{r}(t) = (x(t),y(t)) = (t^2,t)$. We can substitute in directly to get

$$\begin{align} \int_C \mathbf{F}\cdot d\mathbf{r} &= \int_0^1 \langle 2(t^2)(t)^3,3(t^2)^2(t)^2 + 1\rangle \cdot \langle 2t,1 \rangle \, dt \\ &= \int_0^1 \langle 2t^5,3t^6 + 1\rangle \cdot \langle 2t,1 \rangle \, dt \\ &= \int_0^1 (4t^6 + 3t^6 + 1)\,dt \\ & = \int_0^1 (7t^6 + 1) \, dt = 2. \end{align}$$