Difference between a closed solid and a Cartesian surface

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SUMMARY

The discussion clarifies the distinction between a closed solid and a Cartesian surface, specifically using the example of the equation ##z = x^2 + y^2##, which defines a surface rather than a volume. The confusion arose when interpreting a paraboloid intersected by an inclined plane, leading to the misapplication of the divergence theorem. It is established that any equation with an equal sign in R^n indicates a surface, while inequalities define volumes.

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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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