First, we are going to set up the triple integral for the given function f(x, y, z) = 36(x + y) over the given region W with the limits for y which are 0 and x, for z which are y and x, and for x which are 0 and 1:
∭ (36(x + y)) dV, with limits 0 ≤ y ≤ x ≤ 1, and y ≤ z ≤ x.
Starting with the innermost integral we have:
∫ (from y to x) 36(x + y) dz.
When we integrate this with respect to z, the integral does not contain a z term, so the antiderivative is 36z(x + y). Evaluating this at z = x and z = y, and then subtracting, we get:
[36x(x + y)] - [36y(x + y)] = 36x^2 + 36xy - 36yx - 36y^2 = 36x^2 - 36y^2
Now we have a new integral which with respect to y is,
∫ (from 0 to x) (36x^2 - 36y^2) dy
The antiderivative of this is 36x^2y - 12y^3. Evaluating this at y = x and y = 0 and then subtracting, we get:
[36x^3 - 12x^3] - [36*0 - 12*0] = 24x^3
Lastly, we perform the integral with respect to x over [0, 1]:
∫ (from 0 to 1) 24x^3 dx
The antiderivative of this is 6x^4. Evaluating this at x = 1 and x = 0, we finally get:
[6*1^4 - 6*0^4]= 6
So, the volume of the given region W under the function f(x, y, z) = 36(x + y) is equal to 6.