Domain of a multivariable eq'n

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Homework Help Overview

The discussion revolves around finding and sketching the domain of the multivariable function f(x,y) = (x² + y³) / (x² + y² - 1). Participants are exploring the conditions under which the function is defined, particularly focusing on the denominator.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the condition for the denominator to be non-zero and relate it to the unit circle in the xy-plane. There is a question about whether the interior of the circle is included in the domain.

Discussion Status

The discussion is active, with participants confirming each other's understanding of the domain. There is a productive exchange regarding the implications of the unit circle on the domain, though no explicit consensus is reached on the final interpretation.

Contextual Notes

Participants are considering the implications of the function being undefined at the unit circle and whether points inside the circle are included in the domain. There is an acknowledgment of the need to clarify the conditions under which the function is defined.

sjmacewan
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Ok, i think i understand this one, but it's giving me a bit of trouble in terms of comprehension, so I thought I'd get some help on it.

I need to find and sketch the domain for:

[tex]f(x,y)= \frac{x^2 + y^3}{x^2 + y^2 -1}[/tex]


The way i see it, that would only be undefined when the denominator is equal to zero. So wouldn't the domain simply be:

D={(x,y)|x^2+y^2-1 /= 0}


Right?
 
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That's right.
 
Seems okay; and x²+y²-1 = 0 <=> x²+y² = 1 is exactly the unit circle in the xy-plane, so a cylinder of that circle with variable heigth z.
 
edit:
right! i knew that, when i made the unit circle connection i forgot to take the 1 over to the other side, so it wasn't making much sense.

So the sketch would basically just include the circle, stating that THAT is the only undefined place? or would the inside of the circle be excluded from the domain as well?
 
Last edited:
f is perfectly well defined inside the unit circle, since then x^2+y^2-1 is not equal to zero.
 
Last edited:
that's what i thought, since it's [tex]x^2+y^2=1[/tex] and NOT equal to or greater than.

Anywho, thanks for the quick confirmation people!
 

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