Use the divergence theorem to compute the integral: which is correct

Use the divergence theorem to compute the integral?
Calculate the flux integral through the surface of the solid.
(x^2 +2yz, 2xy + z^2, 2xz +y^2)
divergence = 6x?
the region is a box from 0 to 2 in the x and y directions and from 0 to 3 in the z direction with a unit box cut out of this
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By the Divergence Theorem, ∫∫s F · dS = ∫∫∫ div F dV.

Since div F = 2x + 2x + 2x = 6x, the triple integral equals
∫(x = 1 to 2) ∫(y = 1 to 2) ∫(z = 1 to 3) 6x dz dy dx
= ∫(x = 1 to 2) (2 - 1) * (3 - 1) * 6x dx
= 6x^2 {for x = 1 to 2}
= 18.
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If I took the integral from 0to 2 for x and y then 0 to 3 for z and got 72, then took the integral from 0 to 1 fo x, y, and z for the unit box and subtracted from the 72, giving 69, is this incorrect? If it is, why does this way not work?

Hello again.

First of all, a 'culpa mea' is in order for the bounds of integration; I'll fix that below.
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Secondly, what do you mean by the unit box?
I'll assume that you mean that x,y,z are in [0, 1].

By the Divergence Theorem, ∫∫s F · dS = ∫∫∫R div F dV.
Here, div F = 2x + 2x + 2x = 6x.

We compute the triple integral in two stages.

(i) Over the given box (with no unit box removed):

∫(x = 0 to 2) ∫(y = 0 to 2) ∫(z = 0 to 3) 6x dz dy dx
= ∫(x = 0 to 2) (2 - 0) * (3 - 0) * 6x dx
= 18x^2 {for x = 0 to 2}
= 72.

(ii) Over the unit box

∫(x = 0 to 1) ∫(y = 0 to 1) ∫(z = 0 to 1) 6x dz dy dx
= ∫(x = 0 to 1) (1 - 0) * (1 - 0) * 6x dx
= 3x^2 {for x = 0 to 1}
= 3.

Since R is the region of the given rectangular box with the unit box removed,
we subtract the result from (i) and (ii):
∫∫∫R div F dV = 72 - 3 = 69.

(That is, your answer is correct!!)

I hope this helps!