0
$\begingroup$

I am a student and I have a conflict with a given answer in the textbook. The question is the following:

Evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ for the given vector field $\mathbf{F}$ and the specified curve $C$.

$\mathbf{F} = \mathbf{a} \times \mathbf{r}$, where $\mathbf{a}$ is a constant vector, $\mathbf{r} = \langle x, y, z \rangle$, and $C$ is a straight line segment from $\mathbf{r}_1$ to $\mathbf{r}_2$.

Here is my solution:

$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (\mathbf{a} \times \mathbf{r}) \cdot d\mathbf{r} = 0$$ because the triple product of coplanar vectors vanishes.

However, the solution given is $\mathbf{r}_2 \cdot (\mathbf{a} \times \mathbf{r}_1)$.

$\endgroup$
  • 1
    $\begingroup$ I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration? $\endgroup$ – Lord Shark the Unknown Dec 10 '18 at 6:54
  • $\begingroup$ @LordSharktheUnknown So I suppose I've learned that $\mathbf{r}$ and $d\mathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem. $\endgroup$ – Davis Rash Dec 10 '18 at 7:44
  • $\begingroup$ @LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $\mathbf{r}$ and $d\mathbf{r}$ are not parallel. Shame. $\endgroup$ – Davis Rash Dec 10 '18 at 7:47
0
$\begingroup$

With the help of the comment by Lord Shark the Unknown, I let \begin{align} \mathbf{r} & = \mathbf{r}_1 + t(\mathbf{r}_2 - \mathbf{r}_1) \\ \Longrightarrow \quad d\mathbf{r} & = (\mathbf{r}_2 - \mathbf{r}_1)\,dt. \end{align}

We now have \begin{align} \mathbf{a} \times \mathbf{r} & = \mathbf{a} \times \mathbf{r}_1 + t(\mathbf{r}_2 - \mathbf{r}_1) \\ & = \mathbf{a} \times \mathbf{r}_1 + t\mathbf{a} \times (\mathbf{r}_2 - \mathbf{r}_1). \end{align}

And finally \begin{align} \int_C \mathbf{F} \cdot d\mathbf{r} & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1 + t\mathbf{a} \times (\mathbf{r}_2 - \mathbf{r}_1)) \cdot (\mathbf{r}_2 - \mathbf{r}_1)\,dt \\ & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1) \cdot (\mathbf{r}_2 - \mathbf{r}_1)\,dt \\ & = \int_0^1 (\mathbf{a} \times \mathbf{r}_1) \cdot \mathbf{r}_2\,dt = (\mathbf{a} \times \mathbf{r}_1) \cdot \mathbf{r}_2. \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.