# Parametrization of a surface

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1. Mar 24, 2016

### CGMath

An area A in the (x,y) plane is limited by the y-axis and a parabola with the equation x=6-y^2. Further, is a surface F given by the part of the graph for the function h(x,y)=6-x-y^2 which satisfies the conditions x>=0 and z>=0.

Determine a parametrization for A and for F.

So far i've got the parametrization for A, which i got to r(u,v)=(6-v^2,v), v ∈ [0,6].

My attempt of a solution for F is r(u,v)=(u,v, 6-u-v^2), but i am not sure about the limits of each parameter and if it's the correct parametrization. Could someone help me out?

Thanks!

2. Mar 24, 2016

### LCKurtz

First, let's talk about F. I agree with what you have done, but I don't see any reason to rename the parameters so I would have written$$\vec r(x,y) = \langle x,y,6-x-y^2\rangle$$Your parameterization for A isn't correct because it has only one variable, thus describing a curve instead of an area. The easy way to parameterize A would be just to take the z coordinate equal zero in the parameterization of the surface. Using a different letter that would give$$\vec R(x,y) = \langle x,y,0\rangle$$I know, that doesn't seem correct because x and y could vary all over the place, which brings us to your original question: what are the limits? Well, what limits would you use for a double intgral over that xy region if you were calculating its area? You will find your answer there.

Last edited: Mar 24, 2016
3. Mar 25, 2016

### CGMath

Thanks a lot for your reply and help!

I think i got it now :) I got a new problem now, i'll have to find the volume between those two parameterizations, but i'll let my brain struggle with that one for a bit longer.

Take care!

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