definite integralsintegrationmultivariable-calculusreal-analysis
I've been dealing with a calculus problem that's driving me insane. I haven't been able to find the integration limits when changing the order of integration from the equation on the left to the equation on the right in
$$\int_0^1\int_0^z\int_0^zf(x,y,z)dxdydz = \int_?^?\int_?^?\int_?^?f(x,y,z)dxdzdy .$$
If anybody has any idea about solving this or a reference on how to solve this equation, I would really appreciate it.
Try to work it out from the hints before reading the spoilers.
The first hint is to make a sketch of the cross-section of a plane $z=c$ with the volume of interest and $c$ some constant for which $0 \leq c \leq 1$. What is the shape of this cross-section?
Answer : a trapezoid, bounded by $x \geq 0, x+y\geq z, y \geq 0, 1 \geq x+y$
Can you now find the integration limits?
Did you think of splitting the area in two parts to make it easier?
The simplest way would be to split the trapezoid in a parallelogram and a triangle separated by the line $y=c$. Can you solve the problem for both these areas?
$$ \int_0^1 d z \left\{ \int_0^z d y \int_{z-y}^{1-x} d x + \int_z^1 d y \int_0^{1-y} d x \right\} = \frac{1}{3} $$ Obviously not the easiest way to evaluate the integral, but not too difficult either.
Please see the region shaded in the diagram.
So with change of order, the integral will be
$\displaystyle \int_1^2 \int_y^{y^2} e^{\frac{x}{y}} \ dx \ dy$
This will require you to integrate $\displaystyle y \ e^y$ in the second integral, which can be done by Integration by parts.
Edit: If you combine your second and third integral, and then with the first, you get the same integral I wrote above. You just considered change of order for each sub-region separately and so ended up with three integrals.
Best Answer
If an integral of this form had been presented to me, I would translate the limits into three (pairs of) inequalities: the outermost is $0\le z\le1$, next is $0\le y\le z$, dependent as you see on the value of $z$, and the third also is $0\le x\le z$.
What happens for a particular $z$? You have the last two inequalities describing a square of side $z$, and it should be that achieving the integration over this square should be independent of the order: $x$ first (innermost) then $y$, or the reverse order. This may depend on the regularity of the function $f$, and I’m not up on details there. If $f$ is just a polynomial in three variables, there should be no difficulty. Hope this helps, as incomplete an answer as it is.