Integrals on arbitrary (bounded) domains

[Dr. Seafood]

Homework Statement

Let $A = \{(x, y, z) \in \mathbb{R}^n : 0 \lt x \leq 1, 0 \lt y \leq 1 - x^2, 0 \lt z \leq x^2 + y\}$. Define $f : A \rightarrow \mathbb{R}$ by $f(x, y, z) = y$ for each $(x, y, z) \in A$. Accept that Fubini's theorem is applicable here. Find $\int_A f$.

Homework Equations

Fubini's theorem must be used. Here, I will give a verbose statement of this theorem. Let $D \subseteq \mathbb{R}^n$ be rectangular (i.e. a Cartesian product of intervals), and suppose $f : D \rightarrow \mathbb{R}$ is integrable. Suppose n = p + q and let $P \subseteq \mathbb{R}^p$, $Q \subseteq \mathbb{R}^q$ such that $D = P \times Q$. For each $x \in \mathbb{R}^p$, define $f_x : Q \rightarrow \mathbb{R}$ by $f_x(y) = f(x, y)$ and suppose this function is integrable on Q. Now define $F: P \rightarrow \mathbb{R}$ by $F(x) = \int_Q f_x$. Then F is integrable and $\int_P F = \int_P \int_Q f_x = \int_P \int_Q f(x, y) dy dx = \int_D f$.

Another very relevant is as follows. Let $A \subseteq \mathbb{R}^n$ be a bounded set, not necessarily rectangular. Let $D \subseteq \mathbb{R}^n$ be rectangular such that $A \subseteq D$. We are interested in the integral of an integrable function $f : A \rightarrow \mathbb{R}$. Let $f_0 : D \rightarrow \mathbb{R}$ be defined as $f_0(x) = f(x)$ when $x \in A$, and $f_0(x) = 0$ when $x \in D \setminus A$. Then $\int_D f_0 = \int_A f$.

The Attempt at a Solution

The question asks for an integral of a function defined on a non-rectangular region, so we will begin by enclosing the domain in a rectangle. Let $D = (0, 1] \times (0, 1] \times (0, 2]$ so that D is rectangular and $A \subseteq D$. Extend f to D by defining $f_0 : D \rightarrow \mathbb{R}$ by $f_0(x, y, z) = f(x, y, z) = y$ when $(x, y, z) \in A$ and $f_0(x, y, z) = 0$ when $(x, y, z) \in D \setminus A$. By the above result, we get that f₀ is integrable on D and $\int_D f_0 = \int_A f$, so now the problem is about an integral on a rectangular region. Thus we can (attempt to) apply Fubini's theorem.

Let $P, Q, R \subseteq \mathbb{R}$ be the intervals composing D so that $D = P \times Q \times R$. Define $f_x : Q \times R \rightarrow \mathbb{R}$ by $f_x(y, z) = f_0(x, y, z)$, then define $F : P \rightarrow \mathbb{R}$ by $F(x) = \int_{Q \times R} f_x$. By Fubini's theorem, we get that

$$\int_P F = \int_{0}^{1} F = \int_{0}^{1} \int_{Q \times R} f_0$$

Now I'm stuck. I think I've gotten the bounds on the first integral wrong. I think I have the right general idea with the set-up so far but there must be more to it. It has something to do with using 1 - x² and x² + y as the bounds on the integral, but I'm really not sure. There are a lot of technicalities and formalities to get out of the way before we can get to the nuts and bolts of the integral.

 

[mighty2000]

To continue, we need to find the bounds for the inner integral, which will give us the bounds for the outer integral. Let's start with the inner integral:

\int_{Q \times R} f_0 = \int_{0}^{1-x^2} \int_{0}^{x^2 + y} y dz dy

Here, we have used the bounds for y and z based on the definition of A. Now, we can simplify this by integrating with respect to z first:

\int_{Q \times R} f_0 = \int_{0}^{1-x^2} y(x^2 + y) dy

Integrating this with respect to y, we get:

\int_{Q \times R} f_0 = \int_{0}^{1-x^2} (x^2y + y^2) dy = \frac{1}{3}x^2(1-x^2)^3

Now, we can plug this back into our original integral:

\int_P F = \int_{0}^{1} \frac{1}{3}x^2(1-x^2)^3 dx

Simplifying this, we get:

\int_P F = \frac{1}{15}

Therefore, the integral of f over A is equal to 1/15.