My Math Forum Stoke's theorem stuck here

 Calculus Calculus Math Forum

September 10th, 2017, 10:37 AM   #11
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Quote:
 Originally Posted by romsek oh.. they call it $p$ it's just the parametric description of the curve
No, I meant the radius. Why is from 0 to 1?

September 10th, 2017, 12:18 PM   #12
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Quote:
 Originally Posted by sarajoveska No, I meant the radius. Why is from 0 to 1?
just match up the extreme's of the component's of the position vector.

Did you look at the integration curve/surface at all?

September 10th, 2017, 10:16 PM   #13
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Quote:
 Originally Posted by romsek just match up the extreme's of the component's of the position vector. Did you look at the integration curve/surface at all?
I did but still confused. What are the extremes?

September 10th, 2017, 11:32 PM   #14
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Quote:
 Originally Posted by sarajoveska I did but still confused. What are the extremes?
The integration curve is an ellipse, the surface an elliptical disk.

The extremes are the max and min values the coordinates along the ellipse.

In defining the surface we have $-1 \leq u \leq 1,~-1\leq v \leq 1$ by design.

This corresponds to $\rho \in [0,1]$

Take a look at the curve/surface in some graphing software and it should become clear.

 Tags stoke, stuck, theorem

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