# Circulation

• Jun 8th 2009, 08:48 AM
Apprentice123
Circulation
Determine the circulation of $\displaystyle F(x,y,z) = (z-x)i +(x-y)j + (e^{xyz})k$ around the curve $\displaystyle \alpha$ traveled in a counter-clockwise direction when viewed from the axis OZ, knowing that this curve is the edge of the surface: S1 $\displaystyle x^2+y^2+z^2 = 4; z > 0$
• Jun 8th 2009, 10:03 AM
Calculus26
Disregard this attachment

See correct version on later post
• Jun 8th 2009, 10:05 AM
Random Variable
EDIT: fixed errors

$\displaystyle \int \int_{S} curl F \cdot dS = \int_{C} F \cdot dr$

I think they mean that the boundary curve is $\displaystyle x^{2}+y^{2} = 4$

which can be parametrized by $\displaystyle r(\theta) = 2cos(\theta) \hat{i} + 2sin(\theta) \hat{j} + 0 \hat{k}$

where $\displaystyle 0 \le \theta \le 2 \pi$

then $\displaystyle \int_{C} F \cdot dr = \int_{0}^{2 \pi} F(r(\theta)) \cdot r'(\theta) \ d \theta$

where $\displaystyle F(r(t)) = -2cos(\theta) \hat{i} + 2\big(cos(\theta) -sin (\theta)\big) \hat{j}+ \hat{k}$

and $\displaystyle r'(t) = -2sin(\theta) \hat{i} + 2cos(\theta) \hat{j} + 0 \hat{k}$
• Jun 8th 2009, 10:13 AM
Apprentice123
Quote:

Originally Posted by Random Variable
EDIT: fixed errors

$\displaystyle \int \int_{S} curl F \cdot dS = \int_{C} F \cdot dr$

I think they mean that the boundary curve is $\displaystyle x^{2}+y^{2} = 4$

which can be parametrized by $\displaystyle r(t) = 2cos(\theta) \hat{i} + 2sin(\theta) \hat{j} + 0 \hat{k}$

where $\displaystyle 0 \le \theta \le 2 \pi$

then $\displaystyle \int_{C} F \cdot dr = \int_{0}^{2 \pi} F(r(\theta)) \cdot r'(\theta) \ d \theta$

where $\displaystyle F(r(t)) = -2cos(\theta) \hat{i} + 2\big(cos(\theta) -sin (\theta)\big) \hat{j}+ \hat{k}$

and $\displaystyle r'(t) = -2sin(\theta) \hat{i} + 2cos(\theta) \hat{j} + 0 \hat{k}$

How you found $\displaystyle F(r(t))$ ?
• Jun 8th 2009, 10:21 AM
Random Variable
Quote:

Originally Posted by Apprentice123
How you found $\displaystyle F(r(t))$ ?

$\displaystyle r(\theta)$ can be written as

$\displaystyle x = 2cos(\theta)$
$\displaystyle y = 2sin(\theta)$
$\displaystyle z = 0$

Do you know the final answer? I get $\displaystyle 4 \pi$
• Jun 8th 2009, 10:34 AM
Calculus26
Verification with Stokes Thm

previous post corrected

See attachment
• Jun 8th 2009, 10:48 AM
Apprentice123
$\displaystyle 4 \pi$ thanks