This may seem oddly contradictory to another current question which asks for harder integrals. Let me start off by saying that inn asking that question I was not asserting that my skill level was high enough to solve the questions given, just that I was interested in aggregating a list.

That said, I'm having trouble with the following integral which I've been told is solvable under a change of integration order but I fail to see how: $$ \int_0^2\int_0^4\int_z^2 yze^{x^3}\text{d}x\text{d}y\text{d}z $$ So, the region of integration in question is a triangle-based prism, with base equal to the triangle in the $xz$ plane bounded by $x=2-z$, $z=0$ and $x=0$, and protruding in the positive $y$ axis to $y=4$.

The problem is that $e^{x^3}$. We need an $x^2$ in the integrand in order to deal with that.

First, we know that the $y$ value is independent of the $x$ and $z$ values, meaning that our integral is equal to $$ \underbrace{\left(\int_0^4 y\text{d}y\right)}_{=8}\left(\int_0^2\int_z^2 ze^{x^3}\text{d}x\text{d}z\right) $$

So we should only have to switch $x$ and $z$: $$ \int_0^2\int_z^2ze^{x^3}\text{d}x\text{d}z = \int_0^2\int_0^{2-x}ze^{x^3}\text{d}z\text{d}x = \frac12\int_0^2(2-x)^2e^{x^3}\text{d}x $$

This last integral, however, has a terribly ugly closed form, even though the original integral supposedly has the closed form $4/3(e^8-1)$.

What exactly am I doing wrong? I can't find the fault in my logic.


1 Answer 1


When I swap $x$ and $z$ I get $$\int_0^2\int_0^x ze^{x^3}\,dz\,dx$$ which looks much more promising.

Observe that the double integral is over the region $$\{(x,z):0\le z\le x\le 2\}.$$

  • $\begingroup$ Right, I guess I was tired and had mirrored the triangle along the z-axis, making me think the reordering made it impossible. Thanks for clarifying, it makes complete sense now. $\endgroup$ Aug 19, 2017 at 12:03

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