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I don't know how to seal the deal on this one. If you just read the comments on the answer by @RecklessReckoner you will see where I am still stuck.




I would appreciate if someone could overlook my method here please :). I was asked specifically to utilize $\oint_C \mathrm{F\cdot T \;ds}$

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 $C$.

Calculating:

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

Parameters:

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

Which using $T(t) = \cfrac{r'(t)}{||r'(t)||}$, we get: $$\begin{align} \\T_1(t) = \left(1,0\right) \\T_2(t) = \left(0,1\right)(*), \\T_3(t) = \left(-1,0\right)(*), \\T_4(t) = \left(0, -1\right) \end{align}$$

$\oint_C \mathrm{F\cdot T \;ds} = \int_0^1 xy \mathrm{ds} + \int_0^1 x^2\; \mathrm{ds} + \int_0^1 -xy \;\mathrm{ds} + \int_0^1 -x^2\; \mathrm{ds}$

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

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  • $\begingroup$ The integral does not equal zero. Don't forget that on the opposite "horizontal" sides of your rectangle the value of $ \ y \ $ has changed; likewise, the value of $ \ x \ $ is different on opposite "vertical" sides. (If you've had Green's Theorem, you can check to see that the value for the integral is non-zero.) $\endgroup$ – colormegone May 18 '14 at 4:15
  • $\begingroup$ @RecklessReckoner I am sorry, I don't understand. So on the right and top of my rectangle y and x are respectively evaluated differently? Meaning my T is wrong? $\endgroup$ – user142198 May 18 '14 at 4:20
  • $\begingroup$ How did you compute those four line integrals in the end? Each should evaluate to a number, not something involving $x$ or $y$. $\endgroup$ – Santiago Canez May 18 '14 at 4:32
  • $\begingroup$ @SantiagoCanez My final integral is in terms of $s$ was how I achieved those final integrals. $F \cdot T$ gave me $xy,x^2,-xy,-x^2$ for each $T_n$ respectively, evaluated with respect to $s$ over $0\leq s leq 1$ gave me that final set of values($xy + x^2 - xy - x^2 = 0$) $\endgroup$ – user142198 May 18 '14 at 4:44
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Your unit tangent vectors are fine; it's the dot products that are not evaluated properly. Your line integral should look like

$$\oint_C \ \mathbf{F} \cdot \mathbf{T} \ \ ds \ \ = \ \int_0^1 \ (3t) \ \cdot \ 0 \ \ (3 \ dt) \ + \ \int_0^1 \ 3^2 \ \ dt \ + \ \int_0^1 (3t) \ \cdot \ 1 \ \ (-3 \ dt) \ + \ \int_0^1 0^2 \ \ (-dt) \ \ . $$

On each leg, you would have:

$$ \vec{F} \ \cdot \ \vec{T_1} \ \vert_{y=0} \ = \ xy \ \vert_{y=0} \ = \ 0 \ \ ; $$

$$ \vec{F} \ \cdot \ \vec{T_2} \ \vert_{x=3} \ = \ x^2 \ \vert_{x=3} \ = \ 9 \ \ ; $$

$$ \vec{F} \ \cdot \ \vec{T_3} \ \vert_{y=1} \ = \ -xy \ \vert_{y=1} \ = \ -x \ \ ; $$

$$ \vec{F} \ \cdot \ \vec{T_4} \ \vert_{x=0} \ = \ -x^2 \ \vert_{x=0} \ = \ 0 \ \ . $$

In the last two line integral terms above, I've attached the negative orientation to the differentials.

[After you've had Green's Theorem, you can form a double integral over the rectangle, thus

$$ \iint_A \ \left( \ \frac{d}{dx} [x^2] \ - \ \frac{d}{dy}[xy] \ \right) \ \ dx \ dy \ \ = \ \ \iint_A \ ( \ 2x \ - \ x \ ) \ \ dx \ dy $$

$$ = \ \ \int_0^1 \int_0^3 \ x \ \ dx \ dy \ \ . $$

Using either method, the integral value is $ \ \frac{9}{2} \ $ . ]

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  • $\begingroup$ No, the given parametric equations are fine. Moreover, I don't see how your suggestion that $r_2$ should have zero $x$-component and $r_3$ zero $y$-component result in the integrals you write down. (They're correct, but they don't seem to match up with what you're saying.) $\endgroup$ – Santiago Canez May 18 '14 at 4:40
  • $\begingroup$ @SantiagoCanez I believe that is because I went counterclockwise, and he went clockwise perhaps? $\endgroup$ – user142198 May 18 '14 at 4:40
  • $\begingroup$ Oh, the $ \ r \ $ 's aren't vectors (that wasn't clear to me from the presentation -- I'll remove that remark). But you would just use your tangent vectors "dotted" with the function $ \ \vec{F} \ $ over each interval along the $ \ x-$ and $ y-$ directions, keeping in mind that $ \ y \ $ has a constant value on the "horizontal" sides of the rectangle, and $ \ x \ $ is constant on the "vertical" sides. Also, the sign of the "dot products" are negative on the top and left sides. $\endgroup$ – colormegone May 18 '14 at 4:45
  • $\begingroup$ @RecklessReckoner, usually in these types of computations $T$ denotes a unit tangent vector (which is what the OP uses), whereas you're not requiring it have length $1$. (With length $1$, you would just get $-x$ for the third leg and the factor of $3$ would come from the $ds$ term.) You should make this clear. $\endgroup$ – Santiago Canez May 18 '14 at 4:59
  • $\begingroup$ Yes, I think that's it: I've embedded the 3 into the $ \ dx\$'s. I should just write everything in terms of $ \ dt\ $. Thank you -- I'll correct that. $\endgroup$ – colormegone May 18 '14 at 5:08
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Your computations are correct up until the end. The notation

$$\int_0^1 xy\,ds$$

for instance does not mean to anti-differentiate with respect to $s$ to get $xys$ and then plug in $1$ and $0$, it denotes the line integral of the function $xy$ over the bottom of the rectangle. You need to use the general formula

$$\int_C f(x,y)\,ds = \int_a^b f(x(t),y(t))\lVert r'(t)\rVert\,dt$$

to compute these.

Edit: Or rather, perhaps you're just forgetting to plug in for $x$ and $y$ before you evaluate the integrals.

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  • $\begingroup$ Aren't $x,y$ varied over the integral? What value would I plug into them? $\endgroup$ – user142198 May 18 '14 at 4:49
  • $\begingroup$ The value your parametric equations say they should have in terms of $t$; this is what the $f(x(t),y(t))$ part denotes in the general formula above. $\endgroup$ – Santiago Canez May 18 '14 at 4:49
  • $\begingroup$ Ahhh I see $\oint_C F(x(t),y(t)) \cdot T \; \mathrm{ds}$! $\endgroup$ – user142198 May 18 '14 at 4:50
  • $\begingroup$ Yes, and you also need to substitute in for $ds = \lVert r'(t)\rVert\,dt$, which is what gives the extra factor of $3$ in the line integral along the top edge. $\endgroup$ – Santiago Canez May 18 '14 at 4:55
  • $\begingroup$ Can I just clarify, $\int_a^b f(x(t),y(t)) \cdot T * ||r'(t)|| dt$ for the first case is $\int_0^1 3t*0 + 9t^2 \cdot (1,0) * 3 dt = \int_0^1 0*3 dt$? $\endgroup$ – user142198 May 18 '14 at 5:15
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$\oint_C F \cdot T \; \mathrm{ds} = \int_a^bf(x(t),y(t)) ||r'(t)|| dt$

Where we have $r(t) = x(t)\boldsymbol{i} + y(t)\boldsymbol{j},\; a\leq t\leq b$

From this I am sure you can appreciate: $$\begin{align}\\x_1(t) = 3t, y_1(t)=0\\x_2(t)=3,y_2(t)=t\\x_3(t)=3-3t,y_3(t)=1\\x_4(t)=0,y_4(t)=1-t\end{align}$$

So we can now see that: $$\begin{align}1) F \cdot T = (3t,0)\cdot(1,0)=3t\end{align}\\2)F \cdot T = (3,t)\cdot(0,1)=t\\3)F \cdot T = (3-3t,1)\cdot(-1,0)=3t-3\\4)F \cdot T = (0,1-t)\cdot(0,-1)=t-1$$

Which gives us $\int_0^1 3t * 3 dt + \int_0^1 t * 1 dt + \int_0^1 3t-3 * -3 dt + \int_0^1 t-1 * -1 dt$

= $\frac92+\frac12-\frac92+6-\frac12+1=7$

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  • $\begingroup$ By @RecklessReckoner's calculations this is wrong. He got $\frac92 \ne 7$, sorry, I don't know who is right. $\endgroup$ – user142198 May 18 '14 at 12:42
  • $\begingroup$ The answer should be $9/2$. $\endgroup$ – Santiago Canez May 18 '14 at 13:25
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    $\begingroup$ @Katie Your expressions for $ \ \vec{F} \ \cdot \ \vec{T} \ $ are not correct. You are neglecting factors in the components of $ \ \vec{F} \ $ ; for instance, on the first side, $ \ y \ = \ 0 \ $ , so $ \ \vec{F} \ \cdot \ \vec{T} \ = 0 \ $ along that entire side. Similarly, $ \ x \ = \ 0 \ $ on the fourth side, so the dot product is also zero there. On side (2) , $ \ x \ $ is fixed at 3 , and on side (3) , $ \ y \ $ is fixed at 1 . $\endgroup$ – colormegone May 18 '14 at 18:07

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