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Let $0<x,y,t,z<1$ with the additional condition:

$$\begin{align*} x &< t\\ \wedge & \ \\ y &<z \end{align*}$$

Call the set of all $x,y,t,z$ satisfying the above conditions $S$. I want to evaluate $\int_S dxdydtdz$. One way of doing it is first integrating out $z$ and then integrating along columns down up. This ordering gives:

$$\int_0^1\int_0^t\int_x^1\int_y^1 dzdydxdt=\int_0^1\int_0^t\int_x^1(1-y) dydxdt=\frac{1}{8}.$$

Another way is to just integrate up along columns without first integrating $z$:

$$\int_0^1\int_0^1\int_0^t\int_x^zdydxdzdt=\frac{1}{12}.$$

Why do these not agree? It seems like the second way is wrong. I think maybe in the second way one needs to integrate $x$ from $0$ to $\min(x,z)$, but then again I thought the $y$ variable ensures this.

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It is just what you said (except there is what appears to be a typo in the question where you introduced a variable $x_0$ that should be $t$) -- in the second integral [from the inside], the upper bound for $x$ must be $\text{min}(z, t)$.

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  • $\begingroup$ The way to read this out loud to yourself to avoid confusion: "My outside variable is t. That can be anything from 0 to 1. My next variable is z. For a fixed t, z can be anything from 0 to 1. My next variable is x. For t and z fixed, x can be anything from 0 to the minimum of t and z. My final variable is y. For x, z, and t fixed, y can be anything from x to z." $\endgroup$ – hunter Oct 16 '14 at 21:12
  • $\begingroup$ Plop it into Wolfram Alpha with min(z, t). You'll get 1/8. Not sure what "integrating out y" means. $\endgroup$ – hunter Oct 17 '14 at 10:37
  • $\begingroup$ (The argument about min(z, t) making the integral even smaller is wrong because the function you are integrating at this stage in the computation is (z - x)-- again, you can see the result by doing the integrals.) $\endgroup$ – hunter Oct 17 '14 at 10:43

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