0
$\begingroup$

Okay, I know that seems like a stupid question but I couldn't think of a better way to phrase it. I was trying to understand why iterated integrals involve "projecting" the domain onto one of the axes, and I found the following proof in my book.

Note that $f(x,y)$ is a function being integrated over the domain $\mathcal D$, and $\tilde f(x,y)$ is a function such that $\tilde f(x,y) = f(x,y) $ when $(x,y) \in \mathcal D$, and otherwise $\tilde f(x,y) = 0$.

We sketch the proof, assuming that $\mathcal D$ is vertically simple (the horizontally simple case is similar). Choose a rectangle $\mathcal R=[a,b]\times[c,d]$ containing $\mathcal D$. Then

$\iint_Df(x,y)dA = \int_{a}^{b}\int_c^d\tilde f(x,y)dy\ dx$ $\qquad \qquad \qquad \qquad$ (5)

By definition, $\ \tilde f(x,y)$ is zero outside $\mathcal D$, so for fixed $x,\tilde f(x,y)\ $is zero unless y satisfies $g_1(x)\leq y \leq g_2(x)$. Therefore,

$\int_c^d\tilde f(x,y)\, dy = \int_{g_1(x)}^{g_2(x)}f(x,y)\, dy$

Substituting in Eq.(5), we obtain the desired equality

$\iint_\mathcal D f(x,y) \, dA = \int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\, dy \, dx$

My problem with this is that $\mathcal R$ is totally arbitrary, so there's no criteria for $a$ and $b$ except that the resulting $\mathcal R$ contains $\mathcal D$. It seems like I should be able to pick any $a$ and $b$ I want to, and $\iint_\mathcal D f(x,y) \, dA$ will still be equal to $ \int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\, dy \, dx$. Obviously that's not true. So what am I missing?

$\endgroup$

1 Answer 1

0
$\begingroup$

No, it will be equal to $$ \int_a^b\int_{g_1(x)}^{g_2(x)}f^*(x,y)\,dx\,dy, $$ where $f^*(x,y)$ vanishes outside $\cal D$. In particular, outside the projection of $\cal D$ onto $x$-axis $f(x,.)$ is always equal to 0, hence you can choose for $f^*$ bigger $(a,b)$ then really needed. However, it is easier to choose proper interval for $f$.

$\endgroup$
3
  • $\begingroup$ Right, but when I'm doing the actual computation, the $a$ and $b$ I choose $do$ matter... so what gives? $\endgroup$
    – Zachary F
    Commented Dec 14, 2014 at 23:38
  • $\begingroup$ Also, how did you get those larger integrals? $\endgroup$
    – Zachary F
    Commented Dec 14, 2014 at 23:38
  • $\begingroup$ @ZacharyF It is known that the definition of integral on normal set does not depend on choosing a rectangle containing it. (It is easy to believe, because definite integral of 0 is equal to 0). $\endgroup$
    – Kola B.
    Commented Dec 14, 2014 at 23:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .