This is the statement that was given me. I believe $x$ denotes an ordered pair $(x_1,x_2)$ but then the mass comes out to zero because below the $x$-axis it's negative density, and above it's positive. Are there any other interpretations? My answer seems odd.
Mass of unit circle centered at origin, with density $\rho(x) = x_2$; is this saying negative density exists?
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multivariable-calculus
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0I'm going to hazard a GUESS as to what you're trying to ask: You meant that the $x$ in $\rho(x)$ was a pair $(x_1,x_2),$ so that $\rho(x_1,x_2)=x_2,$ and later when you refer to $x$ and $y$ you meant that $x$ is the same thing as $x_1$ and $y$ is the same thing as $x_2$. So you altered the meaning of "$x$" along the way, so it first meant the pair $(x_1,x_2)$ and later meant the first component of the pair $(x,y).$ Is that what you meant? If so I suggest you choose your notation and edit the question so as to follow it consistently throughout the whole of the short paragraph. – 2017-01-28
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0I'm very sorry, I should've clarified - I totally recognize the notation is unclear, but what's in the title is exactly the question that was given me. So I've been trying to figure it out myself; does it make sense that $x2 = y$? Is $\rho(x)$ indicating that $x$ is just a general coordinate? But then I get that mass = 0 which is odd – 2017-01-28
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1This is mathematics, not physics. If you don't like the idea of negative mass density, just replace mass by charge. – 2017-01-28
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If you're talking about a probability density or a mass density, then it should never be negative. A negative density could occur in other contexts, and one of those, as pointed out in a comment by achille hui, is electric charge density (although then I suppose there is the question of whether the particular density you mention is physically realistic).
The information given does imply negative mass below the $x_1$-axis, and a total mass of $0$.
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0Sounds good, thank you! – 2017-01-28