find $ \int_0^4\int_0^4\int_0^4 \sqrt{x^2+y^2+z^2}\,dx\,dy\,dz$ I am looking for an approximation to the nearest integer of
$$ \int_0^4\int_0^4\int_0^4 \sqrt{x^2+y^2+z^2}\,dx\,dy\,dz.$$
Wolfram alpha gives up and says "computation time exceeded".
I tried, unsuccessfully, to approximate the measure of the sets
$$ E_N = \{(x,y,z):x,y,z\in[0,4],\; x^2+y^2+z^2\in[N,N+1]\},$$
but in some cases the volume of the intersection between a cube and a spherical shell is very hard to compute.
 A: And now for something entirely random:
Monte Carlo integration gives
Monte Carlo (n=1000) estimate=245.876995211
Monte Carlo (n=10000) estimate=246.605709455
Monte Carlo (n=100000) estimate=245.440557832
Monte Carlo (n=1000000) estimate=245.841098533
Monte Carlo (n=10000000) estimate=245.946260986

Generated from the following:
import math
import random

def norm(x):
    return math.sqrt(sum([xi*xi for xi in x]))

# returns a uniformly distributed sample in [0, 4]^3
def unif():
    return [random.uniform(0, 4) for i in (1, 2, 3)]

area = 4*4*4
for total_count in (1000, 10000, 100000, 1000000, 10000000):
    sum_func = 0
    for i in xrange(total_count):
        x_samp = unif()
        sum_func += norm(x_samp)

    est_volume = sum_func/float(total_count)*area

    print("Monte Carlo (n=%s) estimate=%s" % (total_count, est_volume))

A: Numerical integration gives that the closest integer is $246$, since
$$I=\int_{[0,4]^3}\sqrt{x^2+y^2+z^2}\,d\mu=245.91154\ldots$$
The best way to achieve a good accuracy "by hand" is to study the pdf of $\sqrt{X^2+Y^2+Z^2}$ when $X,Y,Z$ are three independent random variable with a uniform distribution over $[0,4]$, by eventually using some quantitative form of the Central Limit Theorem.
Since $\mathbb{E}[X^2]=\frac{16}{3}$, the Jensen's inequality gives $I\leq 4^4=256$. 
Anyway, such integral also admits the closed form:
$$ I = \frac{32}{3} \left(6 \sqrt{3}-\pi +\log\left(3650401+2107560 \sqrt{3}\right)\right).$$
