Find $\oint_C \vec{F} \cdot d \vec{r}$ where $C$ is a circle of radius $2$ in the plane $x+y+z=3$, centered at $(2,4,−3)$ and oriented clockwise when viewed from the origin, if $\vec{F}=5y \vec{\imath} −5x \vec{\jmath} +4(y−x) \vec{k}$

Relevant equations:

Stokes theorem: $$\int_S \operatorname{curl}{F} \cdot \mathbf{n} \, dS = \oint_{\partial S} F \cdot d\mathbf{r}$$

My attempt:

  • For the curl I get $(4,4,-10)$.
  • For $d\vec{S}$ I get $(1,1,1)$ from $z = 3-x-y$
  • Dotted together its $-2$.
  • So: $-2 \iint_S dA$.
  • Area of circle is $4\pi$.

    My answer would be $-8\pi$ but the online homework system says it's not correct. Please help!

  • $\begingroup$ You seem to have forgotten to normalize the normal vector. $\endgroup$ – Harald Hanche-Olsen Aug 13 '14 at 16:56
  • $\begingroup$ Use $\mathbf{n} = \frac{1}{\sqrt{3}}(1,1,1)$. $\endgroup$ – Dmoreno Aug 13 '14 at 16:57
  • $\begingroup$ Can someone explain why its 1/sqrt(3)<vector> I thought the sqrt(3) cancels out. $\endgroup$ – Adam Aug 13 '14 at 17:05
  • $\begingroup$ Because $\mathbf{n}$ is defined as the outer normal unit vector of the surface $S$. In this case $\mathbf{n} = (1,1,1)/|(1,1,1)| = \frac{1}{\sqrt{3}}(1,1,1)$. $\endgroup$ – Dmoreno Aug 13 '14 at 17:07
  • $\begingroup$ Here you have some examples: mathinsight.org/stokes_theorem_examples $\endgroup$ – Dmoreno Aug 13 '14 at 17:10

$x+y+z=3 \implies z = -x-y+3$

$\begin{align} \\ d\vec{S} =\hat{n}dS &= \langle -f_x,-f_y,1 \rangle dxdy\\~\\&= \langle 1,1,1 \rangle dA \end{align}$

you get : $-2\iint_S dA$

So far your work is correct ! there is a mistake in your next step :

$-2\iint_{\color{green}{\mathbb S}} dA \color{red}{\ne} 4\pi $

why ? because you're assuming that the given circle itself is a shadow in the $xy$ plane, which is wrong.

since $dA=dxdy$, for the range of $x$ and $y$, you need to take shoadow of $\color{green}{\mathbb S} $ in $xy$ plane :

enter image description here

If $\alpha$ is the angle between $xy$ plane and $\color{green}{\mathbb S}$, then clearly the area scales by a factor of $\cos \alpha $ :

$\text{Area of ellipse in xy plane} = (\text{Area of circle in } \color{green}{\mathbb S})\times \cos \alpha = 4\pi \cos \alpha = 4\pi \dfrac{ \langle 1,1,1 \rangle . \langle 0,0,1 \rangle}{|| \langle 1,1,1 \rangle||} = \dfrac{4\pi}{\sqrt{3}}$


well, I think that it is rather a question of getting C in its parametric form. Your plane is $\Pi :x+y+z=3$. I will try here to find the orthogonal vectors to the $\Pi$:

let's say that $\underline e_1=(a_1,a_2,0)$ and $\underline{e_2}=(-a_1,a_2,0)$. Dot product of those vectors with the normal vector of $\Pi$ gives you these 2 eqautions:

$a_1+a_2+a_3=0, -a_1+a_2=0$.

So you can say that $\underline{e_1}= \left(\begin{array}{c} 1 \\ 1 \\ -2 \end{array} \right)$ and $\underline{e_2}=\left(\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right)$

So that your curve parametric form is: $$C = \sqrt 3 +4 \cos t\cdot \left(\begin{array}{c} 1 \\ 1 \\ -2 \end{array}\right) -4\sin t\cdot \left(\begin{array}{c} -1 \\ 1 \\ 0 \end{array} \right)$$ where $t\in[0,2\pi]$ And from here it becomes much easy. I hope I helped!

  • $\begingroup$ Since the vector field $\vec{F}$ has a constant curl and, furthermore, the outer normal vector is also constant, I think that this approach is not advantageous and we should therefore take advantage of Stoke's theorem. $\endgroup$ – Dmoreno Aug 13 '14 at 18:06

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.