Solving Inequalities for Regions

Question

I have several tasks to solve where a set of inequalities is used to describe a region. I should then calculate the area or volume of that region.

Let's say we have the following inequality (for $x,y,z \geq 0$):

$x+2y+3z \leq 1$

Now I need to find out the boundaries for the triple integral. From a few examples I have here, I wasn't able to derive a way to solve such a problem.

The example solution suggests that

$0 \leq y \leq \frac{1-x}{2}$

and

$0 \leq z \leq \frac{(x - 2y)}{3}$.

for

$0 \leq x \leq 1$

leading to the integral:

$ \int_{0}^{1} \int_{0}^{\frac{1-x}{2}} \int_{0}^{\frac{x-2y}{3}} { 1 \; dz dy dx } $

How do I find out these boundaries? (Moreover, how do I find out the boundaries in a general approach)

PS: I'm not sure if the solution is correct at all, that's probably why I'm confused.

Answer 1

Take the variables one by one.

For your example, start with $z$ and move all the other variables to the other side:

$0 \leq 3z \leq 1 - (x+2y)$

Divide by $3$ and you have your bound.

For $y$, disregard $z$, but do the same thing, obtaining:

$0 \leq 2y \leq 1 - x$

Same thing for $x$ immediately gives

$0 \leq x \leq 1$.

I think the general approach is pretty evident from this.

With these limits, you get the integral

$\int_0^1 \int_0^\frac{1-x}{2} \int_0^\frac{1-x-2y}{3} dzdydx = \frac{1}{36}$

EDIT: This should be correct, since the volume described can be seen as a pyramid with base area $\frac{1}{12}$ (triangle with height $\frac{1}{2}$, base $\frac{1}{3}$) and height 1. The volume of a pyramid is $\frac{1}{3}Bh$.

Answer 2

always draw a picture! you'll see the volume lying between the $xy$-plane and the plane $x+2y+3z=1$, a tetrahedron. It's projection on the $xy$-plane is a right triangle with hypotenuse $x+2y=1$ (obtained by setting $z=0$). IMHO, the geometric approach is better than messing with a bunch of inequalities!