I have the following triple integral:
$$ I = \int_0^1 dz \int_z^1 dx \int_0^{x-z} f(x, y, z) \ dy $$
I want to reiterate the integral in such a way that the integrations are performed in the following order: first (z), then (y), then (x), and sketch the following graph in space. I am a little bit new to changing the order of integration in triple integrals, as how the boundaries are changing and how to graph it correctly.
I would appreciate any kind of help. Thank you in advance.
As we can see we want to change the order of integration of the integral:
$$ I\ = \int \int \int \ f(x,y,z) \ dy\ dx\ dz $$
We want to change it into the next order:
$$ I\ = \int \int \int \ f(x,y,z) \ dz\ dy\ dx $$
I will use algebra to write iteration of the integral with the changed integration order. More exactly, I will use inequalities.
Now we see from the given iteration:
$$ I = \int_0^1 dz \int_z^1 dx \int_0^{x-z} f(x, y, z) \ dy $$
We can write three sets of inequalities satisfied by the outer variable $z$, the middle variable $x$, and the inner variable $y$. We write these in order as follows:
$$ 0 \leq \ z \ \leq \ 1 $$
$$ z \leq \ x \ \leq \ 1 $$
$$ \ \ \ \ \ \ \ 0 \leq \ y \ \leq \ x - z $$ Note that the limits for each variable can be constant or can depend only on variables whose inequalities are on lines above the line for that variable. (In this case, the limits for $z$ must both be constant, those for $x$ can depend on $z$, and those for $y$ can depend on both $x$ and $z$.) This is a requirement for iterated integrals; outer integrals cannot depend on the variables of integration of the inner integrals.
Now we want to change inequalities in the next order:
$$ ? \leq \ x \ \leq \ ? $$
$$ ? \leq \ y \ \leq \ ? $$
$$ ? \leq \ z \ \leq \ ? $$ From the above inequalities and statements, it is easy to determine $"?"$ marks. As we inspect now, we know that limits for x must both be constant, $x \leq \ 1$ and $0 \leq \ z \ \leq \ x$ so for $x$ we have $0 \leq \ x \ \leq \ 1$. Limits for $y$ can depend on $x$. $0 \leq \ y$ and $ y \leq \ x-z \ \leq \ x$ therefore we have $0 \leq \ y \ \leq \ x$ and finally for $z$ limits can depend on both $x$ and $y$. $z \leq \ x-y$ and $0 \leq \ z$ so we have $0 \leq \ z \ \leq \ x-y$
Let's see how our new ordered inequalities look:
$$ 0 \leq \ x \ \leq \ 1 $$
$$ 0 \leq \ y \ \leq \ x $$
$$ \ \ \ \ \ \ \ 0 \leq \ z \ \leq \ x - y $$ Now we know the limits and changed boundaries in the integral as we change the order of integration and our iteration will look like this:
$$ \int_0^1 \ dx \ \int_0^x \ dy \ \int_0^{x-y} \ f(x,y,z) \ dz $$
Now all we left with is to sketch the graph in 3D. For this we see at the boundaries since they are functions of the corresponding variable.
We have figure bounded by next surfaces: $z = 0$ ; $z = x-y$ ; $y = 0$ ; $y = x$ ; $x = 1$ ; $x = 0$
Our figure is tetrahedron which looks like this: