# Find the sum of the family of sets

I have such family of sets: $X_{a,b} =\{ (x,y)\in \mathbb R^{+2}: ax^2<y\le \sqrt[3]{ab^2} \}$, $a,b \in \Bbb R$, $a,b>0$ and I have to find $\bigcup_{a}X_{a,b}$. Even though i had already done this in the past i forgot how. All I remember is that the result is a curve dependent from a. Could anyone give me at least a hint?

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A curve depending on $b$ not $a$, no? – 1015 Feb 9 '13 at 15:12
Oh yeah, you're right - on b – Max Feb 9 '13 at 15:13
I think it is a two-dimensional object, rather than a curve. – 1015 Feb 9 '13 at 15:17
I am not really familiar with mathemathical terms in english. What i meant is that the 2D object is outlined by a curve. – Max Feb 9 '13 at 15:20

Consider the intersection of $\{y=ax^2\}$ and $\{y=\sqrt[3]{ab^2}$ in the first the upper-right quadrant. This is $$\left(\frac{b^{1/3}}{a^{1/3}},\sqrt[3]{ab^2}\right).$$ Then convince yourself that your set is the open region delimited by the lines $\{x=0\}$ and $\{y=0\}$, and the curve $$\{\left(\frac{b^{1/3}}{a^{1/3}},\sqrt[3]{ab^2}\right)\;;\; a>0\}=\{(x,b/x)\;;\; x>0\}. .$$
Note: to prove the claim, Look for instance at the segment joining $(0,0)$ and the point above, whose interior points are all in your set, and then take the union of these over $a>0$.