Approximating $\int_1^3 \int_1^3 \int_1^3 x^{y^z} \mathrm dx\mathrm dy\mathrm dz$? I was given a really nasty integral to approximate, which was
$$\int_1^3 \int_1^3 \int_1^3 x^{y^z}\mathrm dx\mathrm dy\mathrm dz$$
I was completely clueless; and my interviewer gave me some hints but I got nowhere. I was told that the order of magnitude is roughly $10^9$. Could someone help me out? Thank you!
 A: Note that the integrand gets huge at the $(3,3,3)$ corner.  At that point it is $3^{27}\approx 7.6\cdot 10^{12}$.  Most of the contribution will come from near that point, so you want to form some estimate of how much volume there is where the integrand is large.  
I think I would arbitrarily try to find how far you have to move in each direction to reduce the integrand by a factor $9$.  In the $y,z$ directions that means reducing $y^z$ to $25$.  Alpha tells me that $\log_3(25) \approx 2.93$ and 
$25^{1/3}\approx 2.92$.  Finally for $x$ we want $x_0^{27}=3^{25}$ and we find $(3^{25})^{(1/27)}\approx 2.77$, so our integrand is large in a volume of $0.07\cdot 0.08 \cdot 0.23 \approx 0.0013$.  This is really too large because the bottom corner has three factors of $9$ divided out, but using it the integral is somewhere near $3^{26}\cdot 0.0013\approx 3\cdot 10^9$ Alpha did it when I reduced the range in $x$ to $[2.7,3]$ and in $y,z$ to $[2.9,3]$.  It gave $8.2\cdot 10^8$ for the integral of that small volume, which includes all of mine.  
I should probably use the volume of a tetrahedron instead of a box, which divided my estimate by $6$, but that is an afterthought.
