Evaluate the triple integral of 2x - 4yz over the region E where 2 <= x <= 3, 0 <= y <= 2x-4, and 0 <= z <= 5x+5y-3.

Step-by-step solution

1. 1

   Read the order of integration off the description. Each bound depends only on the variables to its left, so the region is already set up for $dz\,dy\,dx$:

   $\int_{2}^{3}\int_{0}^{2x-4}\int_{0}^{5x+5y-3}(2x-4yz)\,dz\,dy\,dx$

2. 2

   Integrate with respect to $z$ first. Treat $x$ and $y$ as constants. An antiderivative is $2xz - 2yz^2$, so evaluating from $0$ to $5x+5y-3$ gives

   $2x(5x+5y-3)-2y(5x+5y-3)^2$

3. 3

   Expand the integrand. With $(5x+5y-3)^2 = 25x^2+25y^2+9+50xy-30x-30y$, the bracket becomes

   $10x^2+70xy-6x-50x^2y-100xy^2+60y^2-50y^3-18y$

4. 4

   Integrate with respect to $y$ from $0$ to $a = 2x-4$. An antiderivative is

   $(10x^2-6x)y+35xy^2-25x^2y^2-\tfrac{100x}{3}y^3+20y^3-\tfrac{25}{2}y^4-9y^2$

   Substituting $y=a$ leaves a polynomial in $x$ alone (the $y=0$ end contributes nothing).

5. 5

   Note a useful check before finishing. Because $a = 2x-4$ vanishes at $x=2$, the entire integrand vanishes at the lower limit — if your expression does not, the algebra above went wrong somewhere.

6. 6

   Integrate the resulting polynomial in $x$ from $2$ to $3$. Every term is a power of $x$, so this last step is routine. Carrying the exact rational arithmetic through gives

   $\iiint_E (2x-4yz)\,dV = -268$

7. 7

   Key takeaway. The only real difficulty here is bookkeeping: integrate in the order the inequalities dictate, substitute one limit at a time, and expand fully before moving to the next variable. Keep everything as exact fractions — rounding partway through a triple integral is where most sign and magnitude errors creep in.