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- TL;DR Summary
- The integral is not iterated integral. Will this fact prevent us from swapping the order of surface and volume integrals? Why? Why not?

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- Thread starter Beelzedad
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- #1

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- TL;DR Summary
- The integral is not iterated integral. Will this fact prevent us from swapping the order of surface and volume integrals? Why? Why not?

- #2

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There is nothing which wouldn't allow us to swap integrals in a finite case over a smooth function. One has only to be aware of the fact, that integral limits might change!

- #3

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https://www.physicsforums.com/threads/multiple-integral-jacobian-confusion.978277/

The consensus there was: While you can swap the order of integration of individual integrals with appropriate change of limits, you can't simply extend that to swap a surface integral with a volume integral.

https://mathinsight.org/double_integral_change_order_integration_examples

From Fubini's theorem:

https://en.wikipedia.org/wiki/Fubini's_theorem

Failure of Fubini's theorem for non-integrable functions[edit]

Fubini's theorem tells us that (for measurable functions on a product of σ-finite measure spaces) if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect toxand then with respect toy, we get the same result as if we integrate first with respect toyand then with respect tox. The assumption that the integral of the absolute value is finite is "Lebesgue integrability", and without it the two repeated integrals can have different values.

A simple example to show that the repeated integrals can be different in general is to take the two measure spaces to be the positive integers, and to take the functionf(x,y) to be 1 ifx=y, −1 ifx=y+1, and 0 otherwise. Then the two repeated integrals have different values 0 and 1.

Another example is as follows for the function

The iterated integrals

and

have different values. The corresponding double integral does not converge absolutely (in other words the integral of the absolute value is not finite):

- #4

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I not knowledgeable enough to help you with that. @fresh_42 @Mark44 or @HallsofIvy might be able to explain it better.

- #5

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- #6

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One like this: ##\int_V f(x, y, z)dV##What do you mean by "not iterated integral"?

This can also be written as ##\iiint_V f(x, y, z)dV##. I've seen both styles in textbooks.

Here is an example of an iterated integral:

##\int_{y=0}^\pi\int_{x = 0}^\pi \sin(x + y)dx~dy##

- #7

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I don't see a difference. What is a "not iterated integral"? Is it simple versus multiple? ##\int_V f\,dV## and ##\int_{V_z}\int_{V_y}\int_{V_x} f\, dx\, dy\, dz ## is the same thing.One like this: ##\int_V f(x, y, z)dV##

This can also be written as ##\iiint_V f(x, y, z)dV##. I've seen both styles in textbooks.

Here is an example of an iterated integral:

##\int_{y=0}^\pi\int_{x = 0}^\pi \sin(x + y)dx~dy##

- #8

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One that is not an iterated integral...What is a "not iterated integral"?

Maybe, or maybe not. A double integral such as ##\int_R f\,dA## (where R is the region in the plane over which integration is to be performed) could be rewritten as two different iterated integrals: one in Cartesian form or one in polar form.Is it simple versus multiple? ##\int_V f\,dV## and ##\int_{V_z}\int_{V_y}\int_{V_x} f\, dx\, dy\, dz ## is the same thing.

- #9

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- #10

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- #11

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I disagree. No calculus textbook that I've ever seen would write an integral like this: ##\int dx\,dy\,dz##.

Again, it's the difference between this triple integral ##\int_D f(x, y, z) dV## or ##\iiint_D f(x, y, z) dV## (not iterated) and this iterated integral ##\int_{z = z_1}^{z_2}\int_{y = y_1}^{y_2}\int_{x = x_1}^{x_2} f(x, y, z) dx~dy~dz##.

- #12

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Yes, I forgot to triple the integration symbol.I disagree. No calculus textbook that I've ever seen would write an integral like this: ##\int dx\,dy\,dz##.

Yes, but how is this not purely notational? To distinguish it linguistically appears hair splitting to me.Again, it's the difference between this triple integral ##\int_D f(x, y, z) dV## or ##\iiint_D f(x, y, z) dV## (not iterated) and this iterated integral ##\int_{z = z_1}^{z_2}\int_{y = y_1}^{y_2}\int_{x = x_1}^{x_2} f(x, y, z) dx~dy~dz##.

- #13

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Because the iterated form contains information about the order in which integration is to be performed, in addition to possibly distinguishing between Cartesian coordinates or polar/cylindrical coordinates are to be used.Yes, but how is this not purely notational?

- #14

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This is how my current understanding is regarding my question:

If we change ##\mathbf{r} ∈ S## from a point ##\mathbf{r}_1 ∈ S## to another point ##\mathbf{r}_2 ∈ S##, the origin of our primed coordinate changes. Thus the bounds of integration w.r.t. primed coordinate changes. Thus the bounds of integration w.r.t. primed coordinates are functions of unprimed coordinates. Thus the integral is iterated integral.

Also, if wechange ##\mathbf{r} ∈ S## from a point ##\mathbf{r}_1 ∈ S## to another point ##\mathbf{r}_2 ∈ S##, the origin of our primed coordinate changes. Thus the bounded function ##ρ'## changes. Thus the bounded function ##ρ'## is function of unprimed coordinates. Hence when computing the iterated integral, ##ρ'## should be a function of ##(x,y,z,r',θ',φ')##

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