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Weird Vector Question watch

1. I need to show that, for the cone x^2+y^2=z^2, z>0, k.ds=dxdy

I did this by first saying

ds= where r(u,v) parameterises the cone (this is the definition of ds I have been given).

I parameterised the cone by letting r(u,v)=(ucos(v),usin(v),u) so I got

dr/du=(cos(v),sin(v),1)

dr/dv=(-usin(v),ucos(v),0)

crossing these and dotting with k gives u

so far I have k.ds=ududv

I then did dx/du=cos(v) and dy/dv=ucos(v) so dudv=dxdy/cos^2(v)

so putting these together I get k.ds=dxdy/cos^2(v) which is inncorrect.

What have I done wrong?
2. You need to use the jacobian when changing from dudv to dxdy or alternatively parametrise the cone as r(x,y) = (x,y,(x^2+y^2)^0.5)
3. (Original post by Ulgoo)
You need to use the jacobian when changing from dudv to dxdy or alternatively parametrise the cone as r(x,y) = (x,y,(x^2+y^2)^0.5)
Ah of course. I'll see if that works.

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Updated: April 12, 2013
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