Change the order of integration to evaluate the triple integral of dz dy dx with z from 0 to 3 - 3x - y, y from 0 to 3(1 - x), and x from 0 to 1, integrating first with respect to y, then x, then z.

Step-by-step solution

1. 1

   Describe the solid by inequalities instead of by limits. Reading the three limits from the inside out:

   $0\le z\le 3-3x-y,\qquad 0\le y\le 3-3x,\qquad 0\le x\le 1.$

   All three collapse into the single symmetric description

   $x\ge 0,\quad y\ge 0,\quad z\ge 0,\quad 3x+y+z\le 3,$

   which is a tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,3,0)$ and $(0,0,3)$.

2. 2

   Choose the outermost variable for the new order. The requested order is $dy\,dx\,dz$, so $z$ is outermost. Projecting the tetrahedron onto the $z$-axis gives $0\le z\le 3$.

3. 3

   Fix z and find the range of x. With $z$ held, the slice is $\{3x+y\le 3-z,\ x,y\ge 0\}$. Setting $y=0$ gives the widest $x$:

   $0\le x\le \frac{3-z}{3}.$

4. 4

   Fix z and x, then read off y. From $3x+y\le 3-z$,

   $0\le y\le 3-z-3x,$

   so the reordered integral is

   $\int_{z=0}^{3}\int_{x=0}^{(3-z)/3}\int_{y=0}^{3-z-3x}dy\,dx\,dz.$

5. 5

   Evaluate from the inside out. The inner integral gives $3-z-3x$. Then

   $\int_{0}^{(3-z)/3}\left(3-z-3x\right)dx=\left[(3-z)x-\tfrac{3x^{2}}{2}\right]_{0}^{(3-z)/3}=\frac{(3-z)^{2}}{3}-\frac{(3-z)^{2}}{6}=\frac{(3-z)^{2}}{6}.$

   Finally

   $\int_{0}^{3}\frac{(3-z)^{2}}{6}\,dz=\frac{1}{6}\cdot\frac{27}{3}=\frac{27}{18}=\frac{3}{2}.$

6. 6

   Check geometrically. The integrand is $1$, so the answer is just the volume of a tetrahedron with mutually perpendicular legs $1$, $3$ and $3$:

   $V=\frac{1}{6}(1)(3)(3)=\frac{3}{2}.$

   The two routes agree, confirming the reordering was done correctly.