SamplePublisher`GrassmannCalculus`
FormIntegral  |
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Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
We show the evaluation of the integrals here but more complete information is given on the page.
One-dimensional integrals look very much like ordinary integrals.
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 | SetCoordinateVectorSpace |
The following specifies the integral domain as an inequality in an .
ImplicitRegion
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| FormIntegral |
5
,{x}],xdx] | EvaluateFormIntegrals |
Out[3]=
∫
ℴ
Out[3]=
2
Using parameters for the limits can result in Piecewise results.
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| FormIntegral |
| EvaluateFormIntegrals |
Out[4]=
∫
ℴ
Out[4]=
|
The following uses an for the domain.
Interval
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| FormIntegral |
| EvaluateFormIntegrals |
Out[5]=
∫
ℴ
Out[5]=
3
2
The following is a 2-dimensional integral using a 2-form and inequality for the domain.
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| SetCoordinateVectorSpace |
| FormIntegral |
| EvaluateFormIntegrals |
Out[6]=
∫
ℴ
Out[6]=
4
3
The following is the same integral specified with a Simplex region.
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| FormIntegral |
| EvaluateFormIntegrals |
Out[7]=
∫
ℴ
Out[7]=
4
3
The following expresses a 2-dimensional integral in polar coordinates using as the volume factor. To avoid a Piecewise result we can include an Assumptions option in the evaluation.
r
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 | SetActiveSpacePreferences |
 | PublicGrassmannAtlas |
| SwitchBasis |
| FormIntegral |
| EvaluateFormIntegrals |
Out[8]=
∫
ℴ
Out[8]=
π
2
R
But with circular or spherical region specification Mathematica automatically inserts the volume factor so we use:
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| FormIntegral |
| EvaluateFormIntegrals |
Out[9]=
∫
ℴ
Out[9]=
π
2
R
A 3-dimensional integral works analogously.
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| SetCoordinateVectorSpace |
| FormIntegral |
| EvaluateFormIntegrals |
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∫
ℴ
Out[10]=
1
8
A is sensitive to the orientation of the differential form. This case reverses the sign of the result.
FormIntegral
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| FormIntegral |
| EvaluateFormIntegrals |
Out[11]=
∫
ℴ
Out[11]=
-
1
8
In the following we have to include the volume factor in the integrand.
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| SetActiveSpacePreferences |
| PublicGrassmannAtlas |
| SwitchBasis |
| FormIntegral |
| EvaluateFormIntegrals |
Out[12]=
2
r
Out[12]=
r>0&&0<θ<π&&-π<φ≤π
Out[12]=
∫
ℴ
2
r
Out[12]=
4π
3
R
3
But here we leave it out.
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| FormIntegral |
| EvaluateFormIntegrals |
Out[13]=
∫
ℴ
Out[13]=
4π
3
R
3
The following is an indefinite integral.
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| SetCoordinateVectorSpace |
| FormIntegral |
dx
2
a
2
x
2
c
| EvaluateFormIntegrals |