Polynomial Integrals over Balls

Why Coordinate Powers over Balls Are Manageable

A polynomial can contain many different coordinate powers, but a ball centered at the origin has enough symmetry to make their integrals systematic. Spherical coordinates separate the distance from the origin from the two angular directions. For a monomial, this separation turns a triple integral into a radial factor and two angular factors.

The first useful observation is that some monomials integrate to zero: if a coordinate has an odd power, reflecting the ball across the corresponding coordinate plane reverses the sign of the integrand while leaving the region unchanged. For the remaining monomials, a general formula gives the exact integral.

Definition: Let \(B_R=\{(x,y,z):x^2+y^2+z^2\le R^2\}\), where \(R>0\). A coordinate monomial is a function \(x^a y^b z^c\), where \(a,b,c\) are nonnegative integers.

Writing a Monomial in Spherical Coordinates

Use the spherical-coordinate convention \(x=r\sin\theta\cos\phi\), \(y=r\sin\theta\sin\phi\), and \(z=r\cos\theta\), with \(0\le r\le R\), \(0\le\theta\le\pi\), and \(0\le\phi\le2\pi\). The volume element is \(dV=r^2\sin\theta\,dr\,d\theta\,d\phi\). Thus

x^a y^b z^c\,dV = r^{a+b+c+2} \sin^{a+b+1}\theta \cos^c\theta \cos^a\phi \sin^b\phi\, dr\,d\theta\,d\phi.

The extra \(r^2\sin\theta\) comes from the volume element; it is essential. Because the ball has constant radial bound \(R\) and the angular ranges are fixed, the integral separates into a radial integral, a polar-angle integral, and an azimuthal integral.

Symmetry and the Zero Case

If \(a\) is odd, reflecting \(x\) to \(-x\) maps \(B_R\) onto itself but changes the sign of \(x^a y^b z^c\). The contributions at reflected points cancel, so the integral is zero. The same argument applies if \(b\) or \(c\) is odd. In spherical coordinates, these cancellations appear in the angular integrals as well.

Theorem: If at least one of \(a,b,c\) is odd, then \(\iiint_{B_R}x^a y^b z^c\,dV=0\). If all three exponents are even, write \(a=2i\), \(b=2j\), \(c=2k\), where \(i,j,k\) are nonnegative integers, and put \(s=i+j+k\). Then

\iiint_{B_R}x^{2i}y^{2j}z^{2k}\,dV = \frac{16\pi (s+2)!(2i)!(2j)!(2k)!} {(2s+4)!\,i!\,j!\,k!}\,R^{2s+3}.

This formula includes the volume of the ball when \(i=j=k=0\): it gives \(4\pi R^3/3\). The exponent of \(R\) is \(2s+3\), matching the total degree \(2s\) of the monomial plus the three powers of length contributed by volume.

Why the Formula Works

For the even-power monomial, spherical coordinates give three independent factors:

\int_0^R r^{2s+2}\,dr \quad\cdot\quad \int_0^\pi \sin^{2(i+j)+1}\theta\cos^{2k}\theta\,d\theta \quad\cdot\quad \int_0^{2\pi}\cos^{2i}\phi\sin^{2j}\phi\,d\phi.

The radial factor is \(R^{2s+3}/(2s+3)\). To evaluate the angular factors, define the double factorial by \(n!!=n(n-2)(n-4)\cdots\), stopping at \(1\) or \(2\), and use \((-1)!!=0!!=1\). The standard angular integrals for nonnegative integers \(u,v\) are

\int_0^\pi\sin^{2u+1}\theta\cos^{2v}\theta\,d\theta = \frac{2(2u)!!(2v-1)!!}{(2u+2v+1)!!},

and

\int_0^{2\pi}\cos^{2i}\phi\sin^{2j}\phi\,d\phi = \frac{2\pi(2i-1)!!(2j-1)!!}{(2i+2j)!!}.

For the first identity, symmetry about \(\pi/2\) reduces the polar integral to twice the integral over \([0,\pi/2]\). Substituting \(t=\cos^2\theta\) gives \(\int_0^1 t^{v-1/2}(1-t)^u\,dt\). Integration by parts yields the usual recurrence for this integral, and its starting value at \(u=v=0\) is \(1\) on \([0,\pi/2]\). Repeated application gives the stated double-factorial expression. For the azimuthal identity, split the full interval into four quadrants. The quadrant integral starts at \(\pi/2\) when \(i=j=0\); integration by parts reduces either even exponent by two and gives the displayed factor.

Set \(u=i+j\) and \(v=k\) in the polar identity. Multiplying the radial and angular factors cancels \((2i+2j)!!\), leaving

\frac{4\pi R^{2s+3}}{(2s+3)(2s+1)!!} (2i-1)!!(2j-1)!!(2k-1)!!.

Finally, use \((2n)!!=2^n n!\), \((2n-1)!!=(2n)!/(2^n n!)\), and \((2s+3)(2s+1)!!=(2s+3)!!\). Simplifying gives the factorial formula in the theorem. This derivation shows why the method works: the spherical volume element and the even powers produce separate one-variable factors, each of which can be evaluated exactly.

Worked Examples

Worked Example: An Odd Power Makes the Integral Zero

Evaluate \(\iiint_{B_2}xy^2z^4\,dV\). The exponent of \(x\) is odd, so reflection across the \(yz\)-plane sends \((x,y,z)\) to \((-x,y,z)\). This preserves \(B_2\) and changes the integrand from \(xy^2z^4\) to \(-xy^2z^4\). Each pair of reflected contributions cancels:

\iiint_{B_2}xy^2z^4\,dV=0.

In spherical coordinates, the same cancellation is visible in the factor \(\cos\phi\) from \(x\): its integral over a full azimuthal turn is zero. The answer is therefore \(0\), with no remaining terms to evaluate.

Worked Example: Integrating \(x^2\) over a Ball

Evaluate \(\iiint_{B_2}x^2\,dV\). Here \(i=1\), \(j=0\), \(k=0\), so \(s=1\). Substitution into the formula gives

\begin{aligned} \iiint_{B_2}x^2\,dV &= \frac{16\pi(3)!(2)!(0)!(0)!}{(6)!(1)!(0)!(0)!}\,2^5\\ &= \frac{16\pi\cdot6\cdot2}{720}\cdot32\\ &= \frac{4\pi}{15}\cdot32 = \frac{128\pi}{15}. \end{aligned}

As a check, spherical coordinates give \(x^2=r^2\sin^2\theta\cos^2\phi\). Including \(dV\), the radial factor is \(\int_0^2r^4\,dr=32/5\), the polar factor is \(\int_0^\pi\sin^3\theta\,d\theta=4/3\), and the azimuthal factor is \(\int_0^{2\pi}\cos^2\phi\,d\phi=\pi\). Their product is \((32/5)(4/3)\pi=128\pi/15\), as required.

Worked Example: A Mixed Even-Power Moment

Evaluate \(\iiint_{B_1}x^2y^2z^2\,dV\). The indices are \(i=j=k=1\), so \(s=3\). The formula gives

\begin{aligned} \iiint_{B_1}x^2y^2z^2\,dV &= \frac{16\pi(5)!(2)!(2)!(2)!}{(10)!(1)!(1)!(1)!}\\ &= \frac{16\pi\cdot120\cdot8}{3628800}\\ &= \frac{15360\pi}{3628800} = \frac{4\pi}{945}. \end{aligned}

The powers of \(R\) contribute \(R^9=1\) because the ball is a unit ball. This checks the scale as well as the numerical factor: for a ball of radius \(R\), the same integral would be \(4\pi R^9/945\).

Worked Example: Integrating a Polynomial

Evaluate \(\iiint_{B_3}(x^2+y^2+z^2)\,dV\). By linearity, this is the sum of the three coordinate-square integrals. The ball and the formula treat \(x\), \(y\), and \(z\) equally. Each term has \(i+j+k=1\), so

\begin{aligned} \iiint_{B_3}(x^2+y^2+z^2)\,dV &=3\left(\frac{4\pi}{15}3^5\right)\\ &=3\left(\frac{4\pi}{15}\cdot243\right)\\ &=\frac{972\pi}{5}. \end{aligned}

There is also a direct spherical check: \(x^2+y^2+z^2=r^2\), so the integral is \(4\pi\int_0^3r^4\,dr=4\pi(3^5/5)=972\pi/5\). Both computations agree.

Using the Formula for Any Polynomial

A polynomial is a finite sum of coordinate monomials. Integrate it over \(B_R\) one monomial at a time, multiplying each monomial integral by its coefficient and adding the results. Discard every term with an odd exponent in any coordinate, since its integral is zero. Apply the even-power formula to each term left over.

Expand into monomials.
Write the polynomial as a finite sum of terms \(C x^a y^b z^c\), where \(C\) is a coefficient.

Apply symmetry.
Set a term’s integral to zero if any of its three coordinate exponents is odd.

Evaluate and add.
For each remaining term, use the even-power formula and multiply by its coefficient.

A Common Mistake: Losing the Volume Element

When converting \(x^2\) to spherical coordinates, it becomes \(r^2\sin^2\theta\cos^2\phi\). A common error is to integrate this expression using \(dr\,d\theta\,d\phi\), omitting \(r^2\sin\theta\). That incorrect setup would produce

\left(\int_0^2r^2\,dr\right) \left(\int_0^\pi\sin^2\theta\,d\theta\right) \left(\int_0^{2\pi}\cos^2\phi\,d\phi\right) = \frac{8}{3}\cdot\frac{\pi}{2}\cdot\pi = \frac{4\pi^2}{3}.

This is not the integral over the ball: it omits the spherical volume element and does not give the correct volume scaling. The correction is to include \(r^2\sin\theta\), giving radial power \(r^4\) and polar power \(\sin^3\theta\). The correct value, as checked above, is \(128\pi/15\). A reliable setup always writes \(dV\) explicitly before separating the factors.

Why These Moments Matter

Integrals of coordinate powers summarize how a region is distributed relative to the coordinate axes. Even powers measure contributions that do not cancel under reflection, while odd powers often vanish on a centered ball by symmetry. The same reasoning can simplify polynomial integrals over any region with the relevant reflection symmetries. The ball formula also gives exact answers without repeatedly setting up and evaluating three separate angular integrals.

Check Your Understanding

Use reflection symmetry and the even-power formula to answer these questions.

Why is \(\iiint_{B_R}x^3y^2z^2\,dV=0\) for every \(R>0\)?
For \(\iiint_{B_R}x^2y^2\,dV\), what are the values of \(i,j,k\) and \(s\)?
What power of \(R\) appears in the integral of \(x^4y^2z^2\) over \(B_R\)?
When evaluating a polynomial integral over \(B_R\), what should be done with each term that has an odd coordinate exponent?
Which factor in the spherical volume element is missing from the incorrect \(x^2\) calculation above?