Talk:Multiple integral

Repeated Integral should be separate page

Repeated Integral is made to point to this page. But a multiple integral differs from a repeated integral. Please see Repeated Integral.

SewerCat (talk) 17:44, 28 June 2016 (UTC)[reply]

I've changed it so that repeated integral is a redirect to iterated integral. Though, this article looks much better than that one. --JBL (talk) 21:36, 28 June 2016 (UTC)[reply]

Stilted

While this article seems very technically-sound, I also found it almost incomprehensible as a non-mathematician. In particular, the language seems stilted and perhaps was not written by a native English speaker. In any case, since this article is of great potential interest to many people who may not be mathematically-inclined (that is, may find the information useful/necessary but be unable to decipher the swarm of symbols here), it would benefit from an overhaul. Usually I don't like to just stamp a CleanUp tag on and keep going, but I am in no way qualified enough to properly edit this category. Mineralogy 23:44, 15 January 2006 (UTC)[reply]

I think the author is a native speaker of Spanish and not of English. Perhaps the article double integral is in some ways clearer. Michael Hardy 01:12, 7 March 2006 (UTC)[reply]

Mathematical Error

In the [Formulas of Reduction -> Normal Domains on R^2] section, upon checking the math I found the answer (7/10) to be wrong. I solved the double integral both ways noted and got (13/20) for both values. The value has been corrected.

Issue in Change of Variable Section

"There exist three main "kinds" of changes of variable (one in R2, two in R3), however is possible to hand with this method so as to to operate the substitution that more thinks good."

The above sentence doesn't make any sense. Anybody care to venture what they are talking about? user:Feinstein

Some questionable material

Agree with the above. Also, the example:

Example (3-a):

The region is

D=\{x^{2}+y^{2}\leq 9,\ x^{2}+y^{2}\geq 4,\ 0\leq z\leq 5\}

(that is the "tube" whose base is the circular crown of the 2-d example and whose height is 5); if you apply the transformation you'll get this region:

T=\{2\leq \rho \leq 3,\ 0\leq \phi \leq \pi ,\ 0\leq z\leq 5\}

(that is the parallelepiped whose base is the rectangle in 2-d example and whose height is 5).

Is questionable, as the "circular crown" of the 2-d example is not relevant, since in example 2-d, y > 0. This is falsely assumed in this example by setting the region to $T=\{2\leq \rho \leq 3,\ 0\leq \phi \leq \pi ,\ 0\leq z\leq 5\}$ instead of $T=\{2\leq \rho \leq 3,\ 0\leq \phi \leq 2\pi ,\ 0\leq z\leq 5\}$

Am I not correct? Still learning...

Rough translation

The article seems to be a translation from the Italian version of that article. The article needs to be rewritten on some parts. For example, I guess when the author spoke of dominion, he/she actually meant domain (mathematics) (a false friend translation). --Abdull 13:51, 25 May 2006 (UTC)[reply]

I've made some changes to try to improve the clarity, but it needs further work, as I can't pick out the meaning throughout! –Ian 12:38, 15 June 2006 (UTC)[reply]

Question regarding Example 2-d

Example (2-d):

The domain is

D=\{x^{2}+y^{2}\leq 9,\ x^{2}+y^{2}\geq 4,\ y\geq 0\}

, that is the circular crown in the semiplane of positive y (please see the picture in the example); you note that φ describes a plane angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following rectangle:

T=\{2\leq \rho \leq 3,\ 0\leq \phi \leq \pi \}

.

The Jacobian determinant of that transformation is the following:

{\frac {\partial (x,y)}{\partial (\rho ,\phi )}}={\begin{vmatrix}\cos \phi &-\rho \sin \phi \\\sin \phi &\rho \cos \phi \end{vmatrix}}=\rho

which has been got by inserting the partial derivatives of x = ρ cos(φ), y = ρ sin(φ) in the first column respect to ρ and in the second respect to φ, so the dx dy differentials in this transformation becomes ρ dρ dφ.

Once transformed the function and evaluated the domain, it's possible to define the formula for the change of variables in polar coordinates:

\iint _{D}f(x,y)\ dx\,dy=\iint _{T}f(\rho \cos \phi ,\rho \sin \phi )\rho \,d\rho \,d\phi .

Please note that φ is valid in the [0, 2π] interval while ρ, because it is a measure of a length, can only have positive values.

Can someone explain why the Jacobian determinant had to be used here, which then produced an additional φ to the equation? --Abdull 13:51, 25 May 2006 (UTC)[reply]

It's really quite simple. The original integration domain is a circle, which makes things very difficult to work with (you end up with lots of sqrt(x^2 + 1) which do not reduce). By using a Jacobian determinant we can transform the calculation into a double integral over a rectangular domain. In many cases this produces a much more solvable integral. 24.2.51.248 (talk) 06:50, 7 November 2010 (UTC)[reply]

Spherical Coordinates

In the section on formulae of reduction, the article makes reference to the spherical coordinate system.

In R3 some domains have a spherical symmetry, so it's possible to determinate the coordinates of every point of the integration's region by two angles and one distance. It's possible to use therefore the passage in spherical coordinates; the function is transformed by this relation: f(x,y,z) \longrightarrow f(\rho \sin \theta \cos \phi, \rho \sin \theta \sin \phi, \rho \cos \theta)\,\!

This system, while correct, has the standard American placements of theta and phi reversed. This has the ability to cause confusion, especially when viewed alongside the Spherical Coordinate System article, which uses the American system of notation.

While I understand that the American placement is not universally accepted, it would make sense to standardize notation across the various English articles. So, should we change the article to use the American notation, or add a note stating the difference, and pointing to the Spherical Coordinate System article for further information?

Plumberwill 09:45, 10 April 2007 (UTC)[reply]

I agree on standardizing notation across articles in principle. MathWorld lists several conventions, which should be noted at spherical coordinate system. I'd prefer using a mathematical convention over a physics convention in articles like multiple integral, but articles about physics that talk about spherical coordinates can rightly reverse theta and phi (going by MathWorld, that's what they do more often). As long as context is given, i.e. define r, thetha and phi before referring to them, it should be clear what's going on in each case. Although I personally prefer r, writing ρ distinguishes it from the r in cylindrical coordinates. –Pomte 10:41, 10 April 2007 (UTC)[reply]

Sorry, I think Jacobian is sin phi, not sin theta. Integral on sin from 2pi to zero is zero.. 89.138.242.197 10:52, 24 July 2007 (UTC)[reply]

Translation

I tried to improve the translation. I hope that despite being italian, it seems now clearer.

poor article

The article does not have give a clear idea of what a multiple integral is. yes, multiple integrals may be used to find volumes, but this is not their only application. also the article does not provide a method for evaluating multiple integrals. —Preceding unsigned comment added by 212.159.75.167 (talk • contribs)

as my contributions are allways deleted i shall not bother to fix this. —Preceding unsigned comment added by 212.159.75.167 (talk • contribs)

Indefinite antiderivatives?

Hello. This page states that:

"Since it is impossible to calculate the antiderivative of a function of more than one variable, indefinite multiple integrals do not exist. Therefore all multiple integrals are definite integrals."

However, Wolfram Mathworld says that

"In an indefinite multiple integral, the order in which the integrals are carried out can be varied at will; for definite multiple integrals, care must be taken to correctly transform the limits if the order is changed."

Which seems to imply the contrary. I leave the hyperlink. [[1]]

Cheers. —Preceding unsigned comment added by 200.6.195.225 (talk) 03:45, 6 February 2008 (UTC)[reply]

All I can think that Wolfram Mathworld might be talking about is treating an "indefinite multiple integral" as multiple "partial integrals" (in the sense of partial derivatives). So, for example,

$\int \int (\sin x)(\cos y)\,dx\,dy=\int (-\cos x)(\cos y)+C'(y)=(-\cos x)(\sin y)+C(y)+D(x)$

where C and D are arbitrary (differentiable) functions of a single variable. The integration process is indeed commutative with this method. See also this part of the article to observe that the process is analogous to the computation with limits.

However, I've never seen anywhere other than Mathworld mention an "indefinite multiple integral" (my first calculus reference is the Adams book, which just starts multiple integrals from the perspective of calculating volumes/hypervolumes), so for now I'm just going to reduce the strength of the statement about there being "no such thing" (yesterday I only reworded the existing statement). If anybody has another published reference (i.e. not Mathworld) for the above method, particularly one that describes the actual computation process, perhaps we should change the article accordingly. Tcnuk (talk) 08:45, 19 August 2009 (UTC)[reply]

Article Break-Up

This article is too long, but it still does not devote enough space to applications (area, plane laminas, avg. value, SA).

I propose we create new articles devoted specifically to applications of double and triple integrals and just leave brief explanations on this page. Any thoughts?

Typographics

Typographically this page needs some work. There is a very great mixture of how in-line math styles are done. I have started moving them to HTML as per WP:MOS (it wasn't all in either HTML or PNG, so I've gone with moving all of them to HTML) but have only got through some of the article before tiring! I'll keep cleaning the article as I can, but others are welcome to help. Xantharius (talk) 22:18, 18 August 2008 (UTC)[reply]

-Nmoo (talk) 06:24, 2 April 2008 (UTC)[reply]

Multiple versus iterated integrals

The current introduction confuses the two concepts of multiple and iterated integrals, and should be edited to remove this confusion. Also,I think that the section "multiple integrals and iterated integrals," or at least some of the content found in that section, should be moved to just after the definition of a multiple integral to show that in most cases multiple integrals can be evaluated as iterated integrals. The distinction between multiple and iterated integrals is important theoretically, and the theorem that links the two types -Fubini's theorem- should be more stressed. Also, Stokes theorem (in it's most general sense) is essentially a generalization of the fundamental theorem of calculus to multiple integrals and should be mentioned somewhere. Specifically, Stokes theorem can be used to evaluate some multiple integrals (and in fact could give some meaning the the concept of "indefinite integral" for multiple integrals).

Holmansf (talk) 01:22, 3 December 2008 (UTC)[reply]

Remove merge?

The merge suggestion has been there for over a year. Was there a reason why volume integral should be merged?--128.119.194.156 (talk) 07:49, 6 November 2009 (UTC)[reply]

Question Regarding Example (4-a)

Note that in the following examples the roles of φ and θ have been reversed

Why the roles of φ and θ have been reversed? I was following the article and confused at this point. I think they should not be reversed or an explanation should be added on why such reversing is needed. Thanks. —Preceding unsigned comment added by 69.181.139.155 (talk • contribs)

Article Improvement

I have just begun trying to improve this article, beginning with the formal definition, which needed a rewrite. Before you alter anything in the Mathematical Definition section, or revert my edits, please be aware that I am planning on adding a simpler example and explanation of the concept in two dimensions, to make the article more approachable for non-mathematicians. I am also looking at re-writing other parts of the article, but figured the formal definition was a good place to start. 24.2.51.248 (talk) 08:25, 7 November 2010 (UTC)[reply]

Example 4c

f(x,y,z) = x^2+y^2 over the ball of radius 3a. Isn't the result wrong? A quick run through mathematica gives 648pi/5 * a^5. Using polar coordinates the problem reduces to [triple int.]r^4 (sin s)^3 drdsdt = 2pi*243a^5/5 * [int] (sin s)^3 ds over [0,Pi]. This integral is easy to calculate (introducing cylindrical coordinates isn't necessary; chain rule is key) and the answer is 4/3. Thus we have 2pi * 243a^5/5 * 4/3 = 648pi/5 * a^5.

Spherical coordinates, points on the z axis

I believe the following line is incorrect :

Note that points on z axis do not have a precise characterization in spherical coordinates, so φ can vary between 0 to π .

It's not φ that can vary from 0 to π but θ that can vary from 0 to 2π. Notice that z depends on φ but not on θ, so θ can be anything.

There is a comment in the discussion that says that φ and θ were exchanged in the past to follow another convention, so the sentence in question must have been left unchanged. —Preceding unsigned comment added by 24.37.247.233 (talk) 14:25, 12 December 2010 (UTC)[reply]

I just corrected it. —Preceding unsigned comment added by 24.37.247.233 (talk) 14:34, 12 December 2010 (UTC)[reply]

Ambiguous definition of the partition

When the article states:

"Then the finite family of subrectangles C given by

C=I_{1}\times I_{2}\times \cdots \times I_{n}

is a partition of T"

Does it mean $C=\{\prod _{j=1}^{n}I_{j\alpha }\}_{\alpha =1}^{n}$ , with each block containing the αnd block of each interval, or $C=\{\prod _{\alpha =1}^{n}I_{j\alpha }\}_{j=1}^{n}$ , with each block containing all the jth interval's blocks? ᛭ LokiClock (talk) 22:51, 4 March 2013 (UTC)[reply]

What is "normal domain"?

"Normal domain" is used but not defined. Boris Tsirelson (talk) 06:05, 19 March 2017 (UTC)[reply]

Muddled edits

Recent edits have removed the discussion of normal domains, and replaced the example of integration over normal domains. I do not believe this makes logical sense. It is a theorem that the integral of a continuous function over a normal domain is given as an iterated integral. The edits in question appear to take this theorem for granted, and appear to be under the mistaken impression that the multiple integral is defined as an iterated integral. (Not to mention, important facts like Fubini's theorem appear to be dismissed as mere "technicalities".) The normal domains section should stay, as the example given in that section is an example of using the normal domain method for evaluating a multiple integral. Sławomir Biały (talk) 15:54, 19 March 2017 (UTC)[reply]

I agree. Conflating independent and dependent variables into this mix is extremely confusing to readers when both x and y are independent variables on which the function f depends. @Onmaditque: Be careful about saying "non-rectangular domain". General domains of integration need not be normal with respect to either independent variable.--Jasper Deng (talk) 17:52, 19 March 2017 (UTC)[reply]

@Jasper Deng: @Sławomir Biały: Thank you both very much for your help. With the help of your suggestions, I can now fully understand the concept of normal domains and the difference between multiple integrals and iterated integrals, both of which I was confused about.Onmaditque (talk) 14:24, 2 April 2017 (UTC)[reply]

Glad to help. Really @Tsirel: also should be thanked ;-) Sławomir Biały (talk) 14:48, 2 April 2017 (UTC)[reply]

Same mess again?

Section 4.2 "Double integral over a normal domain": "Then the problem arises: There is still one unknown!" What is it about? Does it make sense? I fail to understand it. Do you? Does it say that this two-dimensional integral cannot be calculated via "normality w.r.t. x-axis" (but only via "normality w.r.t. y-axis")?? And finally: "...and obtain the same value" — the same as what? as the failure above??

Also I wonder about this "normal domain" terminology. Where is it defined? Is it in use outside this article? Boris Tsirelson (talk) 06:36, 20 March 2017 (UTC)[reply]

Doh! I thought I had replaced the section with the pre-Onmaditque version entirely, but it was still the same bad copy. Now I have replaced it with the earlier revision. I'm not sure about the "normal domain" terminology, but it is less awkward than calling them "type I, type II, type III" domains, which I have seen used in recent textbooks. I doubt it is a neologism. Sławomir Biały (talk) 10:43, 20 March 2017 (UTC)[reply]

Thanks for fixing the example.

About the terminology, the article is reluctant to say "this is the definition" in the beginning of Sect.3.3, and nevertheless says "This definition is the same for..." at the end of that section. The reader is supposed to guess that it was the definition, and also guess, is it borrowed from the sources or introduced locally for convenience. Ridiculously, some days ago I deleted the link from here to "Normal (geometry)"; and I am afraid something alike will appear again. Boris Tsirelson (talk) 16:07, 20 March 2017 (UTC)[reply]

By some googling I have found this terminology outside Wikipedia: stackexchange, and page 5 in Technická univerzita Ostrava, and Item 4.2.8 "Terminology" in book "Several Real Variables" by Shmuel Kantorovitz, and Def.17.7 in book "Integration of Functions of Two Variables" by Michael Oberguggenberger and Alexander Ostermann. Boris Tsirelson (talk) 16:19, 20 March 2017 (UTC)[reply]

Thanks. I've added a few more words of introduction. I'm glad to see that "normal domain" at least enjoys some use outside Wikipedia, since I privately approve of this terminology. The standard terms in English appear to be "type I" and "type II" domain, which go back at least to Apostol's text. I found an interesting approach to the same problem in Courant and John: instead of "normal domains", they consider convex domains. These have the added benefit of always being Jordan measurable, but that approach to integration is a singificant departure from what seems to be the modern standard. Sławomir Biały (talk) 16:32, 20 March 2017 (UTC)[reply]

Nice. "In all cases, the function to be integrated must be continuous on the domain"? Well, I understand that we just restrict ourselves to this case; but generally, Riemann integrability is enough (with some reservations). Boris Tsirelson (talk) 17:58, 20 March 2017 (UTC)[reply]

Yes, I've now clarified this too. Sławomir Biały (talk) 18:06, 20 March 2017 (UTC)[reply]

No Mention of the difficulty of indefinite integration of multivariable functions.

Perhaps the term "Multiple Integral" is defined technically in such a way that a section like this may be silly, as the term's definition may make this implication. Regardless, this article and nothing I can find on wikipedia mentions why we rarely deal with indefinite multiple integrals, because most multivariable functions do not have a sequence of partiaL antiderivatives. I know somebody even proved the set of all multivariable functions with who possess some sequence of PARTIAL antiderivatives is smaller relatively speaking than the set of all single variable functions with an antiderivative. Im merely an undergrad student but I hope I've proved that this idea has the required depth to be a section in this article.

Jarnacle (talk) 13:34, 19 December 2020 (UTC)Josh[reply]

Confusion about the partition

$I_{j}$ is defined here as a family of subintervals. Taking their product to get $C$ would give you a set of n-tuples of subintervals, so there is no partition because how do you union n-tuples? Aren't we supposed to take the product of the subintervals themselves or am I missing something? Okoyos (talk) 01:32, 11 August 2023 (UTC)[reply]

Wiki Education assignment: 4A Wikipedia Assignment

This article was the subject of a Wiki Education Foundation-supported course assignment, between 12 February 2024 and 14 June 2024. Further details are available on the course page. Student editor(s): Emilyk80 (article contribs). Peer reviewers: EArvizu.

— Assignment last updated by Ahlluhn (talk) 00:57, 31 May 2024 (UTC)[reply]