My question: Am I meant to sub in something for $x$ and $y$ below?

I believe I have now obtained the correct answer: $\oint_C \mathrm{F\cdot T \;ds} \;= 4xy + 4x^2 - 4xy - 4x^2 = 0 $

The working below:

Consider the vector field:

$F(x,y) = xy \;\boldsymbol{i} + x^2 \;\boldsymbol{j} $

Let $C$ be the rectangle with vertices $(0,0),(3,0),(3,1),(0,1)$, let $T$ denote the unit tangent vector to $C$ directed anticlockwise around $C$.


$\oint_C \mathrm{F\cdot T \;ds}$

I want to do this directly without Green's Theorem

My attempt:


$$\begin{align} \\r_1(t) = \left(\frac{3t}4,0\right), 0\leq t\leq 4 \\r_2(t) = \left(3,\frac t4\right), 0\leq t\leq 4 \\r_3(t) = \left(3-\frac{3t}4, 1\right), 0\leq t\leq 4 \\r_4(t) = \left(0, 1 - \frac t4\right), 0\leq t\leq 4 \end{align}$$

Which using $T(t) = \frac{r(t)}{||r(t)||}$, we get: $$\begin{align} \\T_1(t) = \left(1,0\right) \\T_2(t) = \left(0,1\right)(*), \\T_3(t) = \left(-1,0\right)(*), \\T_4(t) = \left(0, -1\right) \end{align}$$

$\oint_C \mathrm{F\cdot T \;ds} = \int_0^4 xy \mathrm{ds} + \int_0^4 x^2\; \mathrm{ds} + \int_0^4 -xy \;\mathrm{ds} + \int_0^4 -x^2\; \mathrm{ds}$

$ \;= 4xy + 4x^2 - 4xy - 4x^2 = 0 $

Therefore this path was conservative.

  • $\begingroup$ don't need to get the tangent vector, just take $dr$ .. and I think you can parametrize your lines better. $\endgroup$ – Santosh Linkha May 17 '14 at 10:01

I'm not entirely sure why you always want $0\le t \le 4$ but you are free to choose this if you so desire.

The work is then (for each segment):

$$\int_0^4 \textbf{F}(\textbf{r}(t)) \textbf{. r}'(t) \mathrm{d}t$$

i.e. substitute $x$ and $y$ in terms of $t$ into $\textbf{F}$ for the given segment, dot product with the time derivative of $\textbf{r}(t)$ and integrate over the domain of $t$ you have chosen.

  • $\begingroup$ Is there a way to do this via the unit tangent vector? I think I am missing something? Is $\oint_C F \cdot T ds = \int F(r(t)) \cdot r't(t) dt$? $\endgroup$ – user142198 May 17 '14 at 10:38
  • $\begingroup$ I've actually just noticed that you defined your tangent vector incorrectly. It should be defined in terms of the \emph{derivative} of $\textbf{r}(t)$ i.e.$\textbf{r}'(t)$. In response to your above comment though, yes the integrals are equivalent (excusing a small typo in the rhs where you have a random $t$ included) $\endgroup$ – john May 17 '14 at 10:44
  • $\begingroup$ Ahaha nice catch on that typo xD. So the tangent vector has: $T(t) = \frac{r'(t)}{||r'(t)||}$ which honestly makes the question seem possible to solve now, thank you for that. I will have another go at it!! $\endgroup$ – user142198 May 17 '14 at 10:58
  • $\begingroup$ Exactly right. Note the denominator will be cancelled out by the change of integration variable from $s$ to $t$ (assuming you make this change). $\endgroup$ – john May 17 '14 at 11:22

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