Difference between a closed solid and a Cartesian surface

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The discussion clarifies the distinction between a closed solid and a Cartesian surface, specifically addressing the equation z = x^2 + y^2, which represents a surface rather than a solid. It emphasizes that any equation with an equal sign describes a surface without n-dimensional volume, thus indicating a boundary. The conversation also notes that if a different equation, such as z = 2y + 3, were used, it could define a one-dimensional curve instead. The participants highlight the importance of understanding the implications of inequalities and equalities in defining surfaces and solids. This understanding is crucial for correctly applying concepts like the divergence theorem in mathematical problems.
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!