1. ## Circulation

Given the vector field $\displaystyle F(x,y,z) = (x-y)i +(y-z)j + (z-x)k$ and the curve $\displaystyle \alpha$ which is the intersection of surfaces:

$\displaystyle S1 = \{ (x,y,z) \in R^3; x+y+z=1 \}$ and $\displaystyle S2 = \{ (x,y,z) \in R^3; x^2+y^2=1 \}$

Find the circulation of F arrond $\displaystyle \alpha$

2. Originally Posted by Apprentice123
Given the vector field $\displaystyle F(x,y,z) = (x-y)i +(y-z)j + (z-x)k$ and the curve $\displaystyle \alpha$ which is the intersection of surfaces:

$\displaystyle S1 = \{ (x,y,z) \in R^3; x+y+z=1 \}$ and $\displaystyle S2 = \{ (x,y,z) \in R^3; x^2+y^2=1 \}$

Find the circulation of F arrond $\displaystyle \alpha$
Paramterize alpha by

$\displaystyle x=\cos(\theta),y=\sin(\theta),z=1-\cos(\theta)-\sin(\theta)$

$\displaystyle \vec r(\theta)= \cos(\theta) \vec i +\sin(\theta) \vec j + [1-\cos(\theta)-\sin(\theta)] \vec k$

$\displaystyle r'(\theta) = [-\sin(\theta) \vec i +\cos(\theta) \vec j + [\sin(\theta)-\cos(\theta)] \vec k]d\theta$

From here sub in dot product ect and integrate from 0 to 2 Pi.

You can also use Stokes theorem

$\displaystyle \iint_S (\nabla \times \vec F)\cdot d\vec S=\oint_{\partial S} \vec F \cdot d\vec r$

$\displaystyle \nabla \times F = \vec i + \vec j +\vec k$

$\displaystyle d\vec S = (\vec i +\vec j +\vec k )dA$

$\displaystyle \iint_{\mbox{ unit circle}}3dA=3\pi$

3. Originally Posted by TheEmptySet
Paramterize alpha by

$\displaystyle x=\cos(\theta),y=\sin(\theta),z=1-\cos(\theta)-\sin(\theta)$

$\displaystyle \vec r(\theta)= \cos(\theta) \vec i +\sin(\theta) \vec j + [1-\cos(\theta)-\sin(\theta)] \vec k$

$\displaystyle r'(\theta) = [-\sin(\theta) \vec i +\cos(\theta) \vec j + [\sin(\theta)-\cos(\theta)] \vec k]d\theta$

From here sub in dot product ect and integrate from 0 to 2 Pi.

You can also use Stokes theorem

$\displaystyle \iint_S (\nabla \times \vec F)\cdot d\vec S=\oint_{\partial S} \vec F \cdot d\vec r$

$\displaystyle \nabla \times F = \vec i + \vec j +\vec k$

$\displaystyle d\vec S = (\vec i +\vec j +\vec k )\sqrt{3}dA$

$\displaystyle \iint_{\mbox{ unit circle}}3\sqrt{3}dA=3\pi \sqrt{3}$

I find:
$\displaystyle \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt = 3 \pi + \frac{3}{4}$
What is my mistake?

4. Originally Posted by Apprentice123
I find:
$\displaystyle \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt = 3 \pi + \frac{3}{4}$

What is my mistake?
the integrand is correct

$\displaystyle \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt =$

$\displaystyle \int_{0}^{2\pi}-2\cos(t)+\frac{3}{2}+\frac{3}{2}\cos(2t)+\sin(t)dt =-2\sin(t)+\frac{3t}{2}+\frac{3}{4}\sin(2t)-\cos(t) \bigg|_{0}^{2\pi}=3\pi$

5. Originally Posted by TheEmptySet
the integrand is correct

$\displaystyle \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt =$

$\displaystyle \int_{0}^{2\pi}-2\cos(t)+\frac{3}{2}+\frac{3}{2}\cos(2t)+\sin(t)dt =-2\sin(t)+\frac{3t}{2}+\frac{3}{4}\sin(2t)-\cos(t) \bigg|_{0}^{2\pi}=3\pi$
Thank you