How to Integrate a Polynomial over a Sphere

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How to Integrate a Polynomial over a Sphere
Author(s): Gerald B. Folland
Source: The American Mathematical Monthly, Vol. 108, No. 5 (May, 2001), pp. 446-448
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2695802
Accessed: 03-06-2015 16:13 UTC
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REFERENCE
1. I. Panakis,Plane Trigonometry,(in Greek) Vol. II, privatepublication,Athens, 1973.
81 PatmouStreet,Athens, 11144, Greece
[email protected]
Florida Atlantic University,Boca Raton, FL 33431
[email protected]
How to Integrate A Polynomial Over
A Sphere
Gerald B. Folland
Severalrecent articlesin the MONTHLY ([1], [2], [4]) have involvedfindingthe
area of n-dimensionalballs or spheres or integratingpolynomialsover such sets.
None of these articles,however,makesuse of the most elegantand painlessmethod
for performingsuch calculations,which is to reversethe usual trick for computing
2
flOe
dx.
Thereis nothingnew in this idea. It was shownto me in 1971 by V. Bargmannand
E. Nelson,andI includedit as an exercisein my book [3], whose firsteditionappeared
in 1984.But the evidencesuggeststhatit is not as universallyknownas it shouldbe.
First,some notation:For x = (xi, ... xn) E Rn, we set Ix = (X2+ * + x2)1/2,
and we denote the sphereand ball of radiusr aboutthe origin in Rnby Sn(r) and
Bn(r), respectively:
,
Sn(r)= {x E Rn: lxl = r},
Bn(r)= {x E Rn: lxI < r}.
We also set
Sn = Sn(1),
Bn = Bn(l).
Any nonzerox E Rn can be writtenuniquelyin "polarcoordinates":
x
= rx',
x
where r = lxI and x' =-E
Sn.
lxl
Let a denotethe (n - l)-dimensional surfacemeasureon Sn;thenthe formulafor
integrationin polarcoordinatesis
f
f(x)dx =
f(rx') rn-1dr da(x').
f
(1)
(A briefderivationis sketchedin Remark2.) Ourobjectis to computefS P du where
P is a polynomial,andfor this it sufficesto considerthe case whereP is a monomial,
P(X)
446
?
= X
= xx22. 2
..
n
(al,
aXn E {O,1 2,...})
(2)
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The answeris expressedin termsof the gammafunction,
(t)
=
J
e-s
stl
ds = 2 J
dr.
r2t1e-r2
The relevantvaluesof t are integersor half-integers;we recall that F(k) = (k - 1)!
and F(k +
2) =
(k - 2)
when k is an integer.
)
***
Theorem. Let P be given by (2), and let f3j =
2
(aj
+ 1). Then
if some aj is odd,
I 0
2FP(j)F(/2)...P(/On)
+/32+ *-+ P)
=
]Pd
Jr(V1
if all aj areeven.
Proof. If some aj is odd,.the integralvanishesby symmetry.Assume thatall oj are
even, andconsiderthe integral
I=
P(x)e-x12 dx.
j
Evaluationof I in Cartesiancoordinatesgives
0, o
n
n
I=H]I
}
xi e-
dxj =
,00
2,
On the other hand, since P(rx')
=
e-xj dxj = F(/1)
xJ
ral+
+nnP(x'),
(
..)
.(rfn).
(3)
evaluationin polar coordinates
gives
p
00
P(rx')e f
rrn- dr da(x')
raul+
+aUn+n-1 er2
+
+ PO)
r(/1
2
dr
P (x') du
c(x')
p (x') d c(x').
(4)
s
Comparisonof (3) and(4) yields the desiredresult.
U
Theintegralof a polynomialP overa ball is nowreadilyobtainedby takingf (x) =
P(x)x (x) in (1), where x is the characteristicfunctionof the ball. Namely,if P is
givenby (2),
R
P(x) dx
]
Bn(R)
=
rul+
+Un+n-l
dr
JOl
In particular,for the case P
+fn+n
rRol
Pda =
+
+
n+ n
/
P d.
1, we have:
Corollary. The (n - 1)-dimensional area of Sn is 2wTn/2/r (n/2), and the n-dimensional
volume of Bn is 2r n2/n2 (n/2).
May 2001]
NOTIES
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447
Remark 1. The samecalculationsyield the integralof IxiI1"... Ix n over a sphere
nonnegativerealnumbers.Thisgeneralizationwas
or a ball whenthe aj's arearbitrary
also consideredin [2].
Remark2. Here is a sketchof the constructionof the surfacemeasurea and the
proofof (1). The startingpointis the fact thatdilationby a factorof r > 0 multiplies
if c is the volumeof Bn(1), then
n-dimensionalvolumesby a factorof r . In particular,
the volumeof Bn(r) is cr . It follows thatthe volumeof a sphericalshell of radiusr
andinfinitesimalthicknessdr is d(crn) = ncr n-1 dr. On the otherhand,this volume
is the productof the thicknessdr andthe surfaceareaof the sphereSn(r), so thatarea
must be ncr-1.
The samereasoningappliesto a sectorratherthana whole ball. Thatis, considera
subsetU of Sn= Sn(1) andthe correspondingsectorQ?u(r)of Bn(r),
Qu(r) = {sx': 0 < s < r, x' E U} .
Then if curn is the volume of Qu (r), the surface area of U must be (d/dr) (curn)lr1 =
ncU. We are therefore led to define surface measure a on Snby a (U) = n *vol(Qu (1)).
of the radianmeasureof an angle thatis used
(Forn = 2, this is the characterization
in the standardgeometricproofthatlim,0 (sin x)/x = 1.) Finally,if we take U to be
an infinitesimalbit of Snwith measureda, we see thatthe volumeof the infinitesimal
box
{sx' r <s < r + dr, x'
E
U}
is rn-1 dr da, from which (1) follows.
To see these calculationsfully clothedin rigorousmeasuretheory,the readermay
consult [3, Theorem2.49], where (1) is proved for Lebesgue integrablefunctions
on Rn. A moreelementarytreatmentof the case of continuousfunctionscan be found
in [2].
REFERENCES
1. L. Badger, Generatingthe measure of n-balls, this MONTHLY 107 (2000) 256-258.
2. J. A. Baker, Integrationover spheres and the divergence theorem for balls, this MONTHLY 104 (1997)
36-47.
3. G. B. Folland, Real Analysis (2nd ed.), John Wiley, New York, 1999.
4. 0. Hijab, The volume of the unit ball in Cn, this MONTHLY 107 (2000) 259.
Universityof Washington,Seattle, WA98195-4350
[email protected]
448
(
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[Monthly 108
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