3
$$ \iiint \limits_0^{A} 1\,d\rho\,d\theta\,d\phi $$

enter image description here

The first integral is supposed to go from 0 to A, then second is from 0 to pi, and the third should be from 0 to 1. I couldn't find anything about it online.

1
  • 2
    The usual approach would be three different \int, each with their own limits. \iiint would have the limits too bunched up to tell apart, in my opinion. Are you sure you want that? (PS, you should use \[...\] instead of $$...$$; see tex.stackexchange.com/q/503/107497)
    – Teepeemm
    Nov 17, 2021 at 19:46

4 Answers 4

3

Since you appear to prefer placing the limits of integration above and below the integral symbols, I would like to recommend that you "snug up" the integral symbols by inserting \!\! (double negative thinspace) between them.

enter image description here

\documentclass{article}
\usepackage[intlimits]{amsmath}
\begin{document}

\[
\int_{0}^{A} \!\! \int_{0}^{\pi} \!\! \int_{0}^{1} \!   d\rho \, d\theta \, d\phi
\qquad
\int_{0}^{A} \!\! \int_{0}^{\pi} \!\! \int_{0}^{1} 1 \, d\rho \, d\theta \, d\phi
\]

\end{document}
3
  • Plus, if it blends with your discipline, a differential operator set upright (e.g. here on tex.se).
    – Buttonwood
    Nov 17, 2021 at 20:35
  • @Buttonwood - Thanks. Since the query was (mainly? exclusively?) about how to place the three pairs of limits of integration, I decided not to modify the OP's code for the differential operators.
    – Mico
    Nov 17, 2021 at 21:41
  • I don't see it as an error. Like setting variables italic or upright shows depends on convention (and varies locally), the comment was about an optional «if».
    – Buttonwood
    Nov 18, 2021 at 14:06
2

I'm not a fan of \limits with \int, unless one wants to just set a domain.

Here's a simplified version, as regards to user level syntax, that automatically computes the number of integral signs to use, based on the list of bounds.

I provide \INT for limits on the side and \INT* for limits above and below.

\documentclass{article}
\usepackage{amsmath}

\ExplSyntaxOn
\NewDocumentCommand{\INT}{sm}
 {
  \kourosh_int:een
   { \IfBooleanT{#1}{\limits} } % \limits for *-version
   { \IfBooleanTF{#1}{6}{9} }   % less kerning for *-version
   { #2 }
 }

\seq_new:N \l_kourosh_int_in_seq
\seq_new:N \l_kourosh_int_out_seq

\cs_new_protected:Nn \kourosh_int:nnn
 {
  \seq_set_from_clist:Nn \l_kourosh_int_in_seq { #3 }
  \seq_set_map:NNn \l_kourosh_int_out_seq \l_kourosh_int_in_seq { \int#1 ##1 }
  \seq_use:Nn \l_kourosh_int_out_seq { \mspace{-#2mu} }
 }
\cs_generate_variant:Nn \kourosh_int:nnn { ee }
\ExplSyntaxOff

\begin{document}

\[
\INT{_0^A,_0^\pi,_0^1} 1\,d\rho\,d\theta\,d\phi
\]
\[
\INT*{_0^A,_0^\pi,_0^1} 1\,d\rho\,d\theta\,d\phi
\]
\[
\INT{_0^\pi,_0^1} 1\,d\theta\,d\phi
\]
\[
\INT{_0^1} 1\,d\phi
\]

\end{document}

enter image description here

1

For a triple integral with bounds I would be more inclined to write

\[
  \int_{0}^{A}  
  \int_{0}^{\pi}
  \int_{0}^{1}
     1\,d\rho\,d\theta\,d\phi
\]

limits style is going to push your limits too close together. It makes more sense if you have something like

\[
   \iiint\limits_{(x,y,z)\in U^3} f(x,y,z) dx\, dy\, dz
\]
0

Comment came in before I could post..

Yea, that's the answer:

See also: To have two limits in double integral?

MWE

\documentclass[12pt]{article}

\usepackage{amsmath}
\newcommand{\Int}{\int\limits}
\begin{document}

$$\Int_{0}^{A} \Int_{0}^{\pi} \Int_{0}^{1} d\rho\; d\theta\; d\phi$$

\end{document}

2
  • I would use thinspace (\,), not thickspace (\;), between the dummy variables of integration.
    – Mico
    Nov 17, 2021 at 20:23
  • Alright thanks it worked now! Nov 17, 2021 at 22:52

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