Integrals over a Curve, Multidimensional Integrals

We will look at two examples of path integrals of vector functions of position and examine their path dependence.  The first integral has a non-zero curl (and so we know that it is not the gradient of some scalar potential)

Here is a vector function (xyz, xyz, xyz) for which the curl does not vanish anywhere

In[1]:=

Out[2]=

$\left\{xyz,xyz,xyz\right\}$

Out[3]=

$\left\{x\left(-y+z\right),y\left(x-z\right),\left(-x+y\right)z\right\}$

These are the conditions that the curl is zero:

In[45]:=

$\mathrm{ConditionsOfZeroCurl}=\mathrm{Table}\left[0==\mathrm{CurlVectorFunction}\left[\left[i\right]\right],\left\{i,3\right\}\right]$

Out[45]=

$\left\{0⩵x\left(-y+z\right),0⩵y\left(x-z\right),0⩵\left(-x+y\right)z\right\}$

There is only one point where this occurs:

In[46]:=

$\mathrm{FindInstance}\left[\mathrm{ConditionsOfZeroCurl},\left\{x,y,z\right\}\right]$

Out[46]=

$\left\{\left\{x\to 0,y\to 0,z\to 0\right\}\right\}$

Let's evaluate the integral of the vector potential ( ∮ $\overrightarrow{v}$  •d$\overrightarrow{s}$  ) for any  curve that wraps around a cylinder of radius R with an axis that coincides with the z-axis

Any curve that wraps around the cylinder can be parameritized as (x(t), y(t), z(t)) = (R cos(t), R sin(t), A $P_{2\pi }$(t)) where $P_{2\pi }$(t) = $P_{2\pi }$(t + 2π) and in particular $P_{2\pi }$(0) = $P_{2\pi }$(2π).
Therefore d$\overrightarrow{s}$=    (-R sin(t), R cos(t), ${P\text{'}}_{2\pi }$(t)) dt = (-y(t), x(t), A ${P\text{'}}_{2\pi }$(t))     dt
The integrand for an integral of "VectorFunction" around such a curve is (written in terms of an arbitrary P(t):

In[4]:=

$\mathrm{vf}=\mathrm{VectorFunction}.\left\{-y,x,\mathrm{Amp}D\left[P\left[t\right],t\right]\right\}/.\left\{x\to \mathrm{Radius}\mathrm{Cos}\left[t\right],y\to \mathrm{Radius}\mathrm{Sin}\left[t\right],z\to \mathrm{Amp}P\left[t\right]\right\}//\mathrm{Simplify}$

Out[4]=

$\mathrm{Amp}{\mathrm{Radius}}^2\mathrm{Cos}\left[t\right]P\left[t\right]\mathrm{Sin}\left[t\right]\left(\mathrm{Radius}\left(\mathrm{Cos}\left[t\right]-\mathrm{Sin}\left[t\right]\right)+\mathrm{Amp}{P}^{\prime }\left[t\right]\right)$

The integral depends on the choice of P(t)

In[5]:=

$\mathrm{PathDepInt}=\mathrm{Integrate}\left[\mathrm{vf},\left\{t,0,2\mathrm{Pi}\right\}\right]$

Out[5]=

${\int }_0^{2\pi }\mathrm{Amp}{\mathrm{Radius}}^2\mathrm{Cos}\left[t\right]P\left[t\right]\mathrm{Sin}\left[t\right]\left(\mathrm{Radius}\left(\mathrm{Cos}\left[t\right]-\mathrm{Sin}\left[t\right]\right)+\mathrm{Amp}{P}^{\prime }\left[t\right]\right)dt$

Let's introduce some specific periodic functions for P. Note how the value of the integral changes as the path changes:

In[6]:=

$\mathrm{PathDepInt}/.P\to \mathrm{Sin}$

Out[6]=

$\frac{1}{4}\mathrm{Amp}\pi {\mathrm{Radius}}^2\left(\mathrm{Amp}+\mathrm{Radius}\right)$

In[7]:=

$\mathrm{PathDepInt}/.P\to \mathrm{Cos}$

Out[7]=

$-\frac{1}{4}\mathrm{Amp}\pi {\mathrm{Radius}}^2\left(\mathrm{Amp}+\mathrm{Radius}\right)$

In[8]:=

$\mathrm{PathDepInt}/.\left\{P\left[t\right]\to t\left(t-2\mathrm{Pi}\right),P\text{'}\left[t\right]\to D\left[t\left(t-2\mathrm{Pi}\right),t\right]\right\}$

Out[8]=

$\mathrm{Amp}{\mathrm{Radius}}^2\left(\frac{3\mathrm{Amp}\pi }{2}-\frac{8\pi \mathrm{Radius}}{9}\right)$

However, here is  curious result which shows that some special paths can ``accidentally'' have zero integrals : let P(t) = cos(n t),

In[9]:=

$\mathrm{pdigen}=\mathrm{PathDepInt}/.\left\{P\left[t\right]\to \mathrm{Cos}\left[nt\right],P\text{'}\left[t\right]\to D\left[\mathrm{Cos}\left[nt\right],t\right]\right\}$

Out[9]=

$-\frac{\mathrm{Amp}{\mathrm{Radius}}^2\left(8\left(-3+n^2\right)\mathrm{Radius}{\mathrm{Sin}\left[n\pi \right]}^2+n\left(-8\mathrm{Radius}+\mathrm{Amp}\left(-9+n^2\right)\mathrm{Cos}\left[2n\pi \right]\right)\mathrm{Sin}\left[2n\pi \right]\right)}{4\left(9-10n^2+n^4\right)}$

In[10]:=

$\mathrm{Simplify}\left[\mathrm{pdigen},n\in \mathrm{Integers}\right]$

Out[10]=

$0$

In[14]:=

$\mathrm{thecurves}=\mathrm{ParametricPlot3D}\left[\left\{\left\{\mathrm{Cos}\left[t\right],\mathrm{Sin}\left[t\right],\mathrm{Cos}\left[3t\right]\right\},\left\{\mathrm{Cos}\left[t\right],\mathrm{Sin}\left[t\right],\mathrm{Cos}\left[t\right]\right\}\right\},\left\{t,0,2\mathrm{Pi}\right\}\right]$

Out[14]=

$⁃\mathrm{Graphics3D}⁃$

In[24]:=

$\mathrm{Show}\left[\left\{\mathrm{Graphics3D}\left[\mathrm{Thickness}\left[0.01\right]\right],\mathrm{Graphics3D}\left[\mathrm{Hue}\left[0.25,0.5,0.5\right]\right],\mathrm{thecurves}\right\}\right]$

Out[24]=

$⁃\mathrm{Graphics3D}⁃$

Apparently, the symmetry of the vector function causes cancelation (note results for P = Sin and P=Cos, differ by a minus sign)
But, why doesn't n=1 give us the correct result above? Note that the denominator goes to zero as n→1

Try the same thing with a conservative (curl free, or exact) Vector Function:

Start with a scalar potential

In[25]:=

$\mathrm{temp}=\mathrm{Grad}\left[\mathrm{Exp}\left[xyz\right],\mathrm{Cartesian}\left[x,y,z\right]\right]$

Out[25]=

$\left\{e^{xyz}yz,e^{xyz}xz,e^{xyz}xy\right\}$

Create another vector function that should have a zero curl

In[26]:=

Out[26]=

$\left\{e^{xyz}yz,e^{xyz}xz,e^{xyz}xy\right\}$

Out[27]=

$\left\{0,0,0\right\}$

In[28]:=

$\mathrm{anothervf}=\mathrm{AnotherVFunction}.\left\{-y,x,D\left[P\left[t\right],t\right]\right\}/.\left\{x\to \mathrm{Radius}\mathrm{Cos}\left[t\right],y\to \mathrm{Radius}\mathrm{Sin}\left[t\right],z\to P\left[t\right]\right\}//\mathrm{Simplify}$

Out[28]=

$\frac{1}{2}{e}^{{\mathrm{Radius}}^2\mathrm{Cos}\left[t\right]P\left[t\right]\mathrm{Sin}\left[t\right]}{\mathrm{Radius}}^2\left(2\mathrm{Cos}\left[2t\right]P\left[t\right]+\mathrm{Sin}\left[2t\right]P^{\prime }\left[t\right]\right)$

The integral depends doesn't on the choice of P(t)

In[29]:=

$\mathrm{PathDepInt}=\mathrm{Integrate}\left[\mathrm{anothervf},t\right]$

Out[29]=

$e^{{\mathrm{Radius}}^2\mathrm{Cos}\left[t\right]P\left[t\right]\mathrm{Sin}\left[t\right]}$

In[30]:=

$\left(\mathrm{PathDepInt}/.t\to 2\mathrm{Pi}\right)-\left(\mathrm{PathDepInt}/.t\to 0\right)$

Out[30]=

$0$

For a last example, suppose the curl vanishes on the cylindrical surface defined above:

Suppose we can find a function that has a non-vanishing curl on this surface

In[79]:=

$\mathrm{VanishOnCylinder}=x^2+y^2-\mathrm{Radius}^2$

Out[79]=

$-{\mathrm{Radius}}^2+x^2+y^2$

In[80]:=

$\mathrm{CurlOfOneStooge}=\left\{0,0,\mathrm{VanishOnCylinder}\right\}$

Out[80]=

$\left\{0,0,-{\mathrm{Radius}}^2+x^2+y^2\right\}$

It is easy to see that this is the curl of Stooge, where

In[81]:=

$\mathrm{Stooge}=\left\{-1/2\mathrm{Integrate}\left[\mathrm{VanishOnCylinder},y\right],1/2\mathrm{Integrate}\left[\mathrm{VanishOnCylinder},x\right],0\right\}$

Out[81]=

$\left\{\frac{1}{2}\left({\mathrm{Radius}}^{2}y-x^{2}y-\frac{y^3}{3}\right),\frac{1}{2}\left(-{\mathrm{Radius}}^{2}x+\frac{x^3}{3}+x{y}^2\right),0\right\}$

In fact, we could add to Stooge, any vector function that has vanishing curl--there are an infinite number of these

In[82]:=

$\mathrm{Simplify}\left[\mathrm{Curl}\left[\mathrm{Stooge},\mathrm{Cartesian}\left[x,y,z\right]\right]\right]$

Out[82]=

$\left\{0,0,-{\mathrm{Radius}}^2+x^2+y^2\right\}$

Its integral doesn't care which path around the cylinder it takes, the integrand doesn't depend on P(t)

In[83]:=

$\mathrm{WhyIOughta}=\mathrm{Stooge}.\left\{-y,x,D\left[P\left[t\right],t\right]\right\}/.\left\{x\to \mathrm{Radius}\mathrm{Cos}\left[t\right],y\to \mathrm{Radius}\mathrm{Sin}\left[t\right],z\to P\left[t\right]\right\}//\mathrm{Expand}$

Out[83]=

$-\frac{1}{2}{\mathrm{Radius}}^4{\mathrm{Cos}\left[t\right]}^2+\frac{1}{6}{\mathrm{Radius}}^4{\mathrm{Cos}\left[t\right]}^4-\frac{1}{2}{\mathrm{Radius}}^4{\mathrm{Sin}\left[t\right]}^2+{\mathrm{Radius}}^4{\mathrm{Cos}\left[t\right]}^2{\mathrm{Sin}\left[t\right]}^2+\frac{1}{6}{\mathrm{Radius}}^4{\mathrm{Sin}\left[t\right]}^4$

In[84]:=

$\mathrm{Integrate}\left[\mathrm{WhyIOughta},\left\{t,0,2\mathrm{Pi}\right\}\right]$

Out[84]=

$-\frac{\pi {\mathrm{Radius}}^4}{2}$

Multidimensional Integral over Irregular Domains

We will attempt to model the energy of ion just above one half of a triangular capacitor.  Suppose there is a uniformly charged surface  (σ≡charge/area=1) occupying an equilaterial triangle in the z=0 plane:
    
    what is the energy (voltage) of a unit positive charge located at (x,y,z)

The electrical potential goes like $\frac{1}{r}$, therefore the potential of a unit charge located at (x,y,z) from a small surface patch at (ξ,η,0) is $\frac{\sigma \mathrm{d\xi }\mathrm{d\eta }}{r}$= $\frac{\mathrm{d\xi }\mathrm{d\eta }}{\sqrt{{\left(x-\xi \right)}^2+{\left(y-\eta \right)}^2+z^2}}$

Therefore it remains to integrate this function over the domain η∈(0,$\frac{\sqrt{3}}{2}$) and ξ∈ ($\frac{\eta }{\sqrt{3}}$- $\frac{1}{2}$) , ($\frac{1}{2}$-$\frac{\eta }{\sqrt{3}}$))  
${\int }_0^{\frac{\sqrt{3}}{2}}$${\int }_{\frac{\eta }{\sqrt{3}}-\frac{1}{2}}^{\frac{1}{2}-\frac{\eta }{\sqrt{3}}}$$\frac{\mathrm{d\xi }\mathrm{d\eta }}{\sqrt{{\left(x-\xi \right)}^2+{\left(y-\eta \right)}^2+z^2}}$dξdη

Mathematica integrates over the last iterator first:

We will try to find the potential due to a triangular patch on a particle located at (x,y,z=1)

In[31]:=

$\mathrm{TrianglePotentialDirect}=\mathrm{Integrate}\left[\frac{1}{\sqrt{{\left(x-\xi \right)}^2+{\left(y-\eta \right)}^2+z^2}},\left\{\eta ,0,\frac{\sqrt{3}}{2}\right\},\left\{\xi ,\frac{\eta }{\sqrt{3}}-\frac{1}{2},\frac{1}{2}-\frac{\eta }{\sqrt{3}}\right\},\mathrm{Assumptions}\to \left\{x\in \mathrm{Reals},y\in \mathrm{Reals},z>0\right\}\right]\text{}$

Out[31]=

$\mathrm{\$Aborted}$

Trying to do this directly either takes too long or there is no closed form! We have to work around it by using Indefinite Integrals

In[96]:=

$\mathrm{TrianglePotentialNumeric}\left[\text{x\_},\text{y\_},\text{z\_}\right]:=\mathrm{NIntegrate}\left[\frac{1}{\sqrt{{\left(x-\xi \right)}^2+{\left(y-\eta \right)}^2+z^2}},\left\{\eta ,0,\frac{\sqrt{3}}{2}\right\},\left\{\xi ,\frac{\eta }{\sqrt{3}}-\frac{1}{2},\frac{1}{2}-\frac{\eta }{\sqrt{3}}\right\}\right]$

In[97]:=

$\mathrm{TrianglePotentialNumeric}\left[1,3,\text{.01}\right]$

Out[97]=

$0.15025179102646993$

In[90]:=

$\mathrm{Plot}\left[\mathrm{TrianglePotentialNumeric}\left[x,x,1/40\right],\left\{x,-1,1\right\}\right]$

$\mathrm{NIntegrate}\text{::}\mathrm{slwcon}:\text{Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.}\mathrm{More\dots }$

$\mathrm{NIntegrate}\text{::}\mathrm{slwcon}:\text{Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.}\mathrm{More\dots }$

Out[90]=

$⁃\mathrm{Graphics}⁃$

In[92]:=

$\mathrm{Table}\left[\mathrm{ContourPlot}\left[\mathrm{TrianglePotentialNumeric}\left[x,y,h\right],\left\{x,-1,1\right\},\left\{y,-0.5,1.5\right\},\mathrm{Contours}\to \mathrm{Table}\left[v,\left\{v,\text{.25},2,\text{.25}\right\}\right],\mathrm{ColorFunction}\to \left(\mathrm{Hue}\left[1-0.66*\text{\#}/2\right]\&\right),\mathrm{ColorFunctionScaling}->\mathrm{False}\right],\left\{h,\text{.025},\text{.5},\text{.025}\right\}\right]$

$\mathrm{NIntegrate}\text{::}\mathrm{slwcon}:\text{Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.}\mathrm{More\dots }$

$\mathrm{NIntegrate}\text{::}\mathrm{slwcon}:\text{Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.}\mathrm{More\dots }$

$\mathrm{NIntegrate}\text{::}\mathrm{slwcon}:\text{Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate.}\mathrm{More\dots }$

$\mathrm{General}\text{::}\mathrm{stop}:\text{Further output of}\mathrm{NIntegrate}\text{::}''slwcon''\text{will be suppressed during this calculation.}\mathrm{More\dots }$

Out[92]=

$\left\{⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃,⁃\mathrm{ContourGraphics}⁃\right\}$

The plot above is for a relatively small height z = 1/20 so the contours reveal the triangular shape of the plate at z = 0.

Now look at a somewhat larger value of z = 1/2. The plot below shows contours that are very nearly circular, indicating that the plate is behaving approximately like an equivalent point charge;

Created by Mathematica  (October 17, 2005)