Using Stokes theorem to integrate $\vec{F}=5y \vec{\imath} −5x \vec{\jmath} +4(y−x) \vec{k}$ over a circle

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!

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

• 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. – Dmoreno Aug 13 '14 at 18:06