Difference between a closed solid and a Cartesian surface

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Homework Help Overview

The discussion revolves around the distinction between a closed solid and a Cartesian surface, particularly in the context of an exercise involving a paraboloid and an inclined plane. The original poster expresses confusion regarding the nature of the surfaces described by specific equations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of equations defining surfaces, questioning how the presence of an equal sign affects dimensionality and volume. There is a focus on the specific case of the equation ##z = x^2 + y^2## and its interpretation as a surface rather than a solid.

Discussion Status

Some participants have provided clarifications regarding the definitions of surfaces and solids, noting that the equation given does not generally represent a closed volume. The conversation includes references to specific planes and inequalities that further complicate the understanding of the surfaces involved.

Contextual Notes

There are references to specific exercises and conditions under which the surfaces are defined, including the implications of using equal signs versus inequalities in the equations. The original poster's confusion highlights the need for clarity in distinguishing between different types of geometric representations.

Amaelle
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Homework Statement
look at the image
Relevant Equations
-closed solid
-cartesian surface
-Divergence theorem
Greetings All!

I have hard time to make the difference between the equation of a closed solid and a cartesian surface.

For example in the exercice n of the exam I thought that the equation was describing a closed solid " a paraboloid locked by an inclined plane (so I thought I could use the divergence theorem) while it was describing a broken paraboloid (a cartesian surface). I joined the exercice to be more precise.

I hope I could explain my problem.

Thank you
exercice.png


Best regards.
 
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You are told that ##z= x^2 + y^2## on ##\Sigma##. This is not generally true on the plane that would close the surface.
 
Any time a set in ##R^n## has an equal sign, "=", in the definition, there is no n-dimensional volume to the set, so it is a surface (boundary?). So ##z = x^2 + y^2## implies that it is a surface.
 
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FactChecker said:
Any time a set in ##R^n## has an equal sign, "=", in the definition, there is no n-dimensional volume to the set, so it is a surface (boundary?). So ##z = x^2 + y^2## implies that it is a surface.
thank you very much Z= something means surface !
 
Orodruin said:
You are told that ##z= x^2 + y^2## on ##\Sigma##. This is not generally true on the plane that would close the surface.
the plane that would close the surface would have been z=2y+3
 
Amaelle said:
the plane that would close the surface would have been z=2y+3
Note: If you had been told z=2y+3 instead of ##z\leq 2y+3##, your set would have been a one-dimensional curve (the intersection of the two surfaces). Here you are told that you have a single surface (the parabola) and that it is the part of the parabola inside a given volume (given by the inequality).
 
Orodruin said:
Note: If you had been told z=2y+3 instead of ##z\leq 2y+3##, your set would have been a one-dimensional curve (the intersection of the two surfaces). Here you are told that you have a single surface (the parabola) and that it is the part of the parabola inside a given volume (given by the inequality).
yes thanks a million!
 

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