Parametrization for Surface F and Area A

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SUMMARY

The discussion focuses on the parametrization of an area A in the (x,y) plane defined by the parabola x=6-y^2 and the surface F represented by the function h(x,y)=6-x-y^2. The correct parametrization for area A is given as r(u,v)=(6-v^2,v) with v ∈ [0,6]. For surface F, the appropriate parametrization is expressed as r(x,y) = ⟨x,y,6-x-y^2⟩. The participants emphasize the importance of defining limits for the parameters based on the double integral used to calculate the area.

PREREQUISITES
  • Understanding of parametric equations in multivariable calculus
  • Familiarity with the concept of double integrals
  • Knowledge of the Cartesian coordinate system and its applications
  • Basic understanding of parabolas and their equations
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  • Research the limits of integration for double integrals in the context of area A
  • Explore techniques for calculating volumes between parametrized surfaces
  • Learn about the applications of parametrization in 3D modeling
  • Study the properties of parabolic surfaces and their implications in calculus
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CGMath
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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!
 
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CGMath said:
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!
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:
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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