Circulation

**Apprentice123** · June 8th 2009, 08:48 AM

Determine the circulation of $F(x,y,z) = (z-x)i +(x-y)j + (e^{xyz})k$ around the curve $\alpha$ traveled in a counter-clockwise direction when viewed from the axis OZ, knowing that this curve is the edge of the surface: S1 $x^2+y^2+z^2 = 4; z > 0$

**Calculus26** · June 8th 2009, 10:03 AM

Disregard this attachment

See correct version on later post

**Random Variable** · June 8th 2009, 10:05 AM

EDIT: fixed errors

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

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

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

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

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

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

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

**Apprentice123** · June 8th 2009, 10:13 AM

Originally Posted by Random Variable

EDIT: fixed errors

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

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

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

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

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

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

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