# triple integral: how do I graph the solid?

Let's say I'm given a triple integral such as $\int_0^1\int_0^{4-2y}\int_0^{4-2y-z} dx dz dy$ and I need to graph the solid based on the bounds... is there some "algorithm" I can follow?

I understand y "moves" between $0$ and $2$, z "moves" between $0$ and $4-2y$ and x "moves" between $0$ and $4-2y-z$ but how do I plot that?

## 1 Answer

The lower bounds $x=0,y=0,z=0$ are the $yz,xz,$ and $xy$ planes respectively. The upper bounds $y=1$, $z=4-2y$, and $x=4-2y-z$ are also planes. Here I plotted the first three, then added the next three. We see that the intersection bounds some region. The final plot shows that region more clearly.

The Mathematica code for generating these plots:

ContourPlot3D[
{x == 0, y == 0, z == 0, x == 4 - 2 y - z, y == 1,z == 4 - 2 y},
{x, 0, 5}, {y, 0, 5}, {z, 0, 5}, AxesLabel -> Automatic]

RegionPlot3D[
x < 4 - 2 y - z && y < 1 && z < 4 - 2 y,
{x, 0, 5}, {y, 0, 1.5}, {z,0, 5}, AxesLabel -> Automatic, Mesh -> 5]

• Thank you! However, I'm still confused when it comes to the 3rd dimension. Since the equation is $4-2y-z$ wouldn't the plane be $x=4-2y-z$? And in that case, when I plot it it would be $z=4-2y-x$? – Floella Dec 13 '16 at 11:44