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$\int_E \frac{y^2}{x^2}d\lambda^2(x,y)$ for $E=\{ (x,y)\in\mathbb R^2 : x\le y, xy\ge 1, y\le2 \}$

I often struggle to find the boundaries of integration from a given area.

My approach was to use Fubini as the function is positive and compute the iterated integral

$\int_{-\infty}^{2}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy=...=\infty$

My questions is whether the boundaries $\frac{1}{y} \le x\le y$ and $-\infty \le y\le 2$ are correct and if there is a general/elegant approach to find them from a given arbitrary area.

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  • $\begingroup$ See that while $y \in (-\infty , -1) \cup (0,1)$, we have that $\frac{1}{y} > y$, so this parts should be discarded. Also note that $y = 0$ also can't be the situation. So in general $y \in [1, 2] \cup (-1,0)$. $\endgroup$ – kolobokish Feb 18 '17 at 22:14
  • $\begingroup$ I meant in previous comment the sitation where you write $\int_{\frac{1}{y}}^{y}$. And for the case where $y\in (\infty, -1) \cup (0,1)$, it should be written $\int_{y}^{\frac{1}{y}}$. $\endgroup$ – kolobokish Feb 18 '17 at 22:24
  • $\begingroup$ @kolobokish So $\int_E \frac{y^2}{x^2}d\lambda^2(x,y)=\int_{1}^{2}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy+\int_{-1}^{0}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy+\int_{-\infty}^{-1}\int_{y}^{1/y} \frac{y^2}{x^2} dx dy+\int_{0}^{1}\int_{y}^{1/y} \frac{y^2}{x^2} dx dy$ ?? $\endgroup$ – Blablablu Feb 18 '17 at 22:35
  • $\begingroup$ Yes. It seem to me, it must be so. $\endgroup$ – kolobokish Feb 18 '17 at 22:40
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Wolfram Alpha produces this plot for x <= y, xy >= 1, y <= 2:

Domain of integration, with small wedge to upper right of the origin and infinite wedge to lower left

So it makes sense that you would need to break the integration up into at least two pieces. Since $xy > 0$, $x$ and $y$ are always of the same sign.

When $x > 0, y> 0$, then $y = 2$ is the top, $y = x$ is bottom and left, and $y = 1/x$ is bottom and right. $y = 1/x$ and $y = x$ cross at $(1,1)$, so if you integrate by $y$ first, you need $y \in [1/x, 2]$ for $x \le 1$ and $y \in [x, 2]$ for $x \ge 1$. $y = 1/x$ crosses $y = 2$ at $x = 1/2$, so you get $\int_{1/2}^1\int_{1/x}^2 \, dydx$ for the first and $\int_1^2\int_{x}^2 \, dydx$ for the second.

But if you integrate $x$ first, then this piece can be done in a single integration: $\int_1^2\int_{1/y}^y \, dxdy$

The lower left piece where $x < 0, y < 0$ is simpler, as $y \le 1/x$ implies $y \le 2$ here. The integral is $\int_{-\infty}^{-1}\int_{x}^{1/x}\, dydx$

So the simplest form is $$\int_E \frac{y^2}{x^2}d\lambda^2(x,y) = \int_{-\infty}^{-1}\int_{x}^{1/x}\frac{y^2}{x^2}\,dydx + \int_1^2\int_{1/y}^y \frac{y^2}{x^2}\, dxdy$$

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