Flux

**noles2188** · April 23rd 2009, 07:43 PM

Find the flux of F across the surface sigma by expressing sigma parametrically.

F(x,y,z)=xi+yj+zk; sigma is the portion of the cylinder x^2+z^2=1 between the planes y=1 and y=-2, oriented by outward unit normals.

**chainrule** · April 23rd 2009, 11:58 PM

Find the flux of F across the surface sigma by expressing sigma parametrically.

F(x,y,z)=xi+yj+zk; sigma is the portion of the cylinder x^2+z^2=1 between the planes y=1 and y=-2, oriented by outward unit normals.

Parametrize the given surface as $x = 1 cos \theta , y = t$ and $z = 1 sin \theta$ . So we have the limits as $0 \leq \theta \leq 2\pi$ and $1 \leq t \leq -2$ .

The surface is now $\Phi(\theta,t) = (cos\theta, t ,sin\theta)$ .
To calculate the normal to the surface, we have, $\left(\frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t)\right)$ .
$\frac{d\Phi}{d\theta}(\theta,t) = (-sin\theta, 0 , cos\theta)$ .

$\frac{d\Phi}{dt}(\theta,t) = (0, 1, 0)$
$\frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t) = \begin{array}{|l c r|}i&j&k\\-sin\theta&0&cos\theta\\0&1&0\end{array} = (-cos\theta, 0 , -sin\theta)$

The flux is now calculated as $\int_1^{-2}\int_0^{2\pi} F(\Phi(\theta,t)) . \left(\frac{d\Phi}{d\theta}(\theta,t) \times \frac{d\Phi}{dt}(\theta,t)\right) d\theta \ dt$

Could you substitute and complete this?

**Calculus26** · April 24th 2009, 01:16 AM

The term

has to be in the same dirction os the orientation vector n . since n is outward we would use [cos(theta),0, sin(theta)]

Also the outer limit of integration shoud be -2 to 1

Of course these cancel and you would have the correct answer --if thats what you had in mind I apologize for my remarks.

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