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ArcLengthIntegral (1.0.0) current version: 2.0.0 »

Source Notebook

Generate an inactive integral or sum of integrals used for computing the arc length of an expression with given bounds

Contributed by: Wolfram|Alpha Math Team

ResourceFunction["ArcLengthIntegral"][expr,{x,a,b}]

returns an inactive integral for computing the arc length of expr for a≤x≤b.

ResourceFunction["ArcLengthIntegral"][expr,{θ,a,b},"polar"]

returns an inactive integral for computing the arc length of expr for a≤θ≤b in polar coordinates.

ResourceFunction["ArcLengthIntegral"][{expr₁,expr₂,…},{t,a,b}]

returns an inactive integral for computing the arc length of the parametric curve defined by {x₁[t],x₂[t],…}⩵{expr₁,expr₂,…} for a≤t≤b.

ResourceFunction["ArcLengthIntegral"][eqtn,{{x,a,b},{y,c,d}}]

returns an inactive integral for computing the arc length of the curve given by implicit equation eqtn for a≤x≤b and c≤y≤d.

Details and Options

ResourceFunction["ArcLengthIntegral"][expr,{x,lo,hi}] accepts a math expression, an equation of the form y==f[x] or a list of two or more math expressions in terms of x.

In ResourceFunction["ArcLengthIntegral"][{expr₁,expr₂,…},{x,lo,hi}], the expr_i are expected to be math expressions, not equations.

Examples

Basic Examples (2)

Set up the integral for computing the arc length of a sine wave:

In[1]:=

Out[1]=

Activate to compute the full result:

In[2]:=

Out[2]=

Set up the integral that computes the arc length of a circle:

In[3]:=

$ResourceFunction["ArcLengthIntegral"][y^2 + x^2 == 1, {x, y}]$

Out[3]=

Scope (4)

Return the integral for the arc length of a curve given in polar coordinates:

In[4]:=

$ResourceFunction["ArcLengthIntegral"][Sin[x^2], {x, 0, Pi}, "polar"]$

Out[4]=

Return the integral for the arc length of a parametric curve:

In[5]:=

Out[5]=

Return the integral for the arc length of the unit circle that is contained within the region bounded by 0≤x≤1 and 0≤y≤1/2:

In[6]:=

$ResourceFunction["ArcLengthIntegral"][ x^2 + y^2 == 1, {{x, 0, 1}, {y, 0, 1/2}}]$

Out[6]=

You can also leave x or y unbounded:

In[7]:=

$ResourceFunction["ArcLengthIntegral"][x^2 + y^2 == 1, {{x, 0, 1}, y}]$

Out[7]=

Return the integral for the arc length of an asteroid bounded by -1≤x≤1:

In[8]:=

$ResourceFunction["ArcLengthIntegral"][ x^(2/3) + y^(2/3) == 1, {{x, -1, 1}, y}]$

Out[8]=

Applications (3)

Define and display an implicit region representing two disconnected segments of the unit circle:

In[9]:=

$Region@ImplicitRegion[x^2 + y^2 == 1, {x, {y, 0, 1/2}}]$

Out[9]=

Set up the integrals for disconnected segments of the region:

In[10]:=

$ResourceFunction["ArcLengthIntegral"][ x^2 + y^2 == 1, {x, {y, 0, 1/2}}]$

Out[10]=

Then compute the result:

In[11]:=

Out[11]=

Possible Issues (1)

ArcLengthIntegral returns the integral for computing the unsigned arc length:

In[12]:=

Out[12]=

Publisher

Wolfram|Alpha Math Team

Version History

2.0.0 – 23 March 2023
1.0.0 – 01 April 2020

Related Resources

License Information

This work is licensed under a Creative Commons Attribution 4.0 International License