Vector Line Integrals

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Objectives:

1.: Know what a vector line integral represents geometrically.
2.: Know the process to compute a vector line integral.
3.: Understand what happens to a vector line integral if we reverse the orientation of the curve.

Recap Video

First, take a look at this recap video going over the basics of vector line integrals.

Test your understanding of the video with the following summary of the video.

Geometrically, the vector line integral $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ represents:
The sum of the vector projections of $\textbf {F}$ along $C$ . The sum of the scalar components of $\textbf {F}$ along $C$ . The net sum of projections of the unit tangent vectors of $C$ along the vector field $\textbf {F}$ .
For a vector line integral over a curve $C$ , the orientation doesdoes not matter.
If $C$ is an oriented curve and $C'$ is the curve $C$ with the opposite orientation, and $\textbf {F}$ is a vector field, then the line integral $\displaystyle \int _{C'} \textbf {F} \cdot d\textbf {r}$ is the the line integral $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .
The procedure to compute a vector line integral is the following:
To compute a vector line integral $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ , do the following steps:
- Parametrize $C$ as $\textbf {r}(t)$ for $a \leq t \leq b$ .
- Check the orientation of $C$ against the direction specified by $\textbf {r}(t)$ .
- Compute $\int _C \textbf {F} \cdot d\textbf {r} = \int _a^b \textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t)dt.$

Example Video

Take a look at the following worked out example.

If $C$ is the intersection of the cylinder $x^2+y^2=1$ with the plane $z=x$ , oriented counterclockwise when viewed from above, and $\textbf {F} = \left \langle x,y,z \right \rangle$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Problems

If $\textbf {F} = \left \langle x^2,xy \right \rangle$ and $C$ is the portion of $x^2+y^2=1$ in the second quadrant, oriented counterclockwise, evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

We will follow the steps outlined above:

A parametrization for $C$ is given by $\textbf {r}(t) = \left \langle \cos (t),\answer {\sin (t)} \right \rangle$ .
The bounds for $t$ are $\answer {\pi /2} \leq t \leq \answer {\pi }$ (give answers between $0$ and $2\pi$ ). This parametrization and these bounds dodo not give the orientation specified in the problem.
We compute
$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \int_a^b \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_{\answer{\pi/2}}^{\answer{\pi}} \left\langle \answer{\cos^2(t)},\answer{\cos(t)\sin(t)} \right\rangle \cdot \left\langle \answer{-\sin(t)},\answer{\cos(t)} \right\rangle dt \\ &=& \int_{\pi/2}^{\pi} \answer{0} dt \\ &=& \answer{0}. \end{eqnarray*}$

In the previous problem, the $\textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t)$ worked out to be $0$ because $\textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t) = 0$ . Geometrically, why did the answer come out to be zero?

The vector field $\textbf {F}$ is always perpendicular to the curve $C$ . The vector field $\textbf {F}$ isn’t always perpendicular to the curve $C$ , but there is a cancellation of positive and negative components of $\textbf {F}$ along $C$ . Some other reason.

If $C$ is the line segment from $(0,0,0)$ to $(1,4,4)$ , evaluate $\int _C (x-y)dx+(y-z)dy+zdz.$

We can write this as a line integral $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ , where $\textbf {F} = \left \langle \answer {x-y},\answer {y-z},\answer {z} \right \rangle$ . Given this, we can follow the same steps:

One parametrization for $C$ is $\textbf {r}(t) = \left \langle t,\answer {4t},\answer {4t} \right \rangle$
The bounds for $C$ are $\answer {0} \leq t \leq \answer {1}$ . Again notice this in the direction given in the problem.
We get $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {-3t},\answer {0},\answer {4t} \right \rangle$
$\textbf {r}'(t) = \left \langle \answer {1},\answer {4},\answer {4} \right \rangle$
We get
$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \int_a^b \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_{\answer{0}}^{\answer{1}} \answer{13t}dt\\ &=& \answer{13/2}. \end{eqnarray*}$

If $C$ is the part of $y=e^x$ from $(2,e^2)$ to $(0,1)$ , and $\textbf {F} = \left \langle x^2,-y \right \rangle$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Steps:

We can parametrize $C$ as $\textbf {r}(t) = \left \langle t,\answer {e^t} \right \rangle$ .
The point $(0,1)$ corresponds to $t = \answer {0}$ and the point $(2,e^2)$ corresponds to $t = \answer {2}$ .
The problem is that $\textbf {r}(t)$ goes the wrong direction. However, we can just add a minus sign to account for this. Namely,
$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& -\int_0^2 \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& -\int_0^2 \answer{t^2-e^{2t}}dt \\ &=& \answer{-\frac{19}{6}+\frac{1}{2}e^4}. \end{eqnarray*}$

If $C$ is the triangle with vertices $(0,0)$ , $(2,0)$ , and $(0,1)$ , oriented clockwise, then $\int _C xy dx + (x+y)dy = \answer {-1/3}.$

We can write this as a vector line integral with $\textbf {F} = \left \langle xy,x+y \right \rangle$ . Since $C$ is a triangle, we will have to calculate three different integrals and add them up.

The line segment $C'$ from $(0,0)$ to $(0,1)$ (since we’re going clockwise): We can parametrize this line segment as $\textbf {r}(t) = \left \langle \answer {0},t\right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {0},\answer {t} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {0},\answer {1} \right \rangle .$ Therefore,
$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_0^1 \answer{t}\,dt \\ &=& \answer{1/2}. \end{eqnarray*}$
Now we have to repeat for the line segment $C'"$ from $(0,1)$ to $(2,0)$ , which we can parametrize as $\textbf {r}(t) = \left \langle 2t,\answer {1-t} \right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {2t(1-t)},\answer {t+1} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {2},\answer {-1} \right \rangle .$ Therefore,
$\begin{eqnarray*} \int_{C''} \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_0^1 \answer{4t(1-t)-t-1}\,dt \\ &=& \answer{-5/6}. \end{eqnarray*}$
Finally, we calculate the line integral over the line segment $C'''$ between $(2,0)$ and $(0,0)$ , which we can parametrize as $\textbf {r}(t) = \left \langle 2-t,\answer {0} \right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {0},\answer {2-t} \right \rangle ,$ and $\textbf {r}'(t) = \left \langle \answer {-1},\answer {0} \right \rangle .$ Therefore,
$\begin{eqnarray*} \int_{C'''} \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_0^1 \answer{0}\,dt \\ &=& \answer{0}. \end{eqnarray*}$

Now add up all the answers over the individual segments to get the line integral over $C$ .

If $\textbf {F} = \left \langle x+y,x-y \right \rangle$ , and $C$ is the ellipse $4x^2+y^2=16$ , oriented counterclockwise, then $\displaystyle \int _C \textbf {F} \cdot dr = \answer {0}$ .

Parametrize $C$ as $\textbf {r}(t) = \left \langle 2\cos (t),4\sin (t) \right \rangle$ for $0 \leq t \leq 2\pi$ .

As shown in the hint, we can parametrize the curve as $\textbf {r}(t) = \left \langle 2\cos (t),4\sin (t) \right \rangle$ for $0 \leq t \leq 2\pi$ . This is oriented correctlyincorrectly . Therefore, the integral is

$\begin{eqnarray*} \int_{C} \textbf{F} \cdot d\textbf{r} &=& \int_0^{2\pi} \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)dt \\ &=& \int_0^{2\pi} \left\langle 2\cos(t)+4\sin(t),\answer{2\cos(t)-4\sin(t)} \right\rangle \cdot \left\langle \answer{-2\sin(t)},\answer{4\cos(t)} \right\rangle \,dt \\ &=& \int_0^{2\pi} \answer{-4\cos(t)\sin(t)-8\sin^2(t)+8\cos^2(t)-16\cos(t)\sin(t)}\,dt \\ &=& \answer{0}. \end{eqnarray*}$

True/False

Suppose $\textbf {F}$ is an everywhere defined vector field and $C$ is an oriented curve with $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} = 0$ . Then $\textbf {F}$ is always perpendicular to the curve $C$ .

True False

True/False, and explain your answer: Suppose $\textbf {F}$ is an everywhere defined vector field and $C$ is an oriented curve such that $\textbf {F}$ always makes an acute angle with the unit tangent vector of $C$ . Then $\displaystyle \int _C \textbf {F} \cdot d\textbf {r} >0$ .

True False