Or \le : $\displaystyle \le$

The other one is $\displaystyle \ge$ \ge (or $\displaystyle \geq$ \geq)

(he has posted a topic for this)

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- Jul 1st 2008, 08:51 AMMoo
- Jul 1st 2008, 12:39 PMReckoner
You can add "negative space" with \!. Try this:

Code:`\iiiint\!\!\!\int dx\,dy\,dz\,dt\,du`

$\displaystyle \iiiint\!\!\!\int dx\,dy\,dz\,dt\,du$

Also, note the use of the \, spacer. I don't think I have ever seen you use it, but it will make your integrals a bit nicer by putting space between the integrand and the variable of integration. e.g.

Instead of $\displaystyle \int xdx$, you'd get $\displaystyle \int x\,dx$. - Jul 1st 2008, 01:31 PMMathstud28
- Jul 1st 2008, 09:38 PMChris L T521
$\displaystyle \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C

{{\mathbf{F}} \cdot d{\mathbf{r}}} = \iint\limits_R {{\text{curl}}\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}

$ - Jul 1st 2008, 09:54 PMMathstud28
- Jul 1st 2008, 10:40 PMReckoner
Actually, I believe he was trying to express a counterclockwise contour integral (look closely and notice the arrow).

I found this really old post on comp.text.tex by Donald Arseneau that provides a method to get it to look right, but I do not think these forums have the right packages. - Jul 1st 2008, 10:49 PMMathstud28
Hmm..I still do not see an arrow indicating direction.

$\displaystyle

\def\Xint#1{\mathchoice

{\XXint\displaystyle\textstyle{#1}}%

{\XXint\textstyle\scriptstyle{#1}}%

{\XXint\scriptstyle\scriptscriptstyle{#1}}%

{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}%

\!\int}

\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$}

\vcenter{\hbox{$#2#3$}}\kern-.5\wd0}}

\def\dashint{\Xint-}

\def\ddashint{\Xint=}

\def\clockint{\Xint\circlearrowright}

\def\counterint{\Xint\rotcirclearrowleft}

\def\rotcirclearrowleft{\mathpalette{\RotSymbol{-30}}\circlearrowleft}

\def\RotSymbol#1#2#3{\rotatebox[origin=c]{#1}{$#2#3$}}$ - Jul 1st 2008, 10:57 PMflyingsquirrel
- Jul 1st 2008, 11:01 PMReckoner
- Jul 1st 2008, 11:57 PMChris L T521Code:
`\mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C`

{{\mathbf{F}} \cdot d{\mathbf{r}}}

= \iint\limits_R {{\text{curl}}

\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}

You probably don't see the arrow because the arrow is at the top of the circle: $\displaystyle \circlearrowleft$

...The integral sign blocks it... - Jul 2nd 2008, 10:08 AMThePerfectHacker
Thread Closed.

Getting too long, start a Part II continuation.