# Integral of a polynomial over a three-dimensional ball

Let $$f$$ be a polynomial of total degree at most three in $$(x,y,z)\in\mathbb{R}^3$$. Prove that $$\int\limits_{x^2+y^2+z^2\leq1}f(x,y,z)\,dx\,dy\,dz = \frac{4\pi f((0,0,0))}{3} + \frac{2\pi(\Delta f)((0,0,0))}{15}$$ Here $$\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$ is the Laplacian operator on $$\mathbb{R}^3$$.

I even don't know how to approach this problem. Since the LHS of the equality is given by a volume integral, I thought about applying the divergence theorem, but wasn't able to do it. Any help?

• Just for curiosity: Where this question is from? – DiegoMath Jun 27 at 18:11
• @DiegoMath, this is from the PhD exam in Analysis at the University of Pittsburgh: mathematics.pitt.edu/sites/default/files/… – Hasek Jun 27 at 19:14

This probably isn't the slickest way of presenting the answer, but here goes. We're going to repeatedly use the change of variables theorem in what follows. Let $$B$$ denote the unit ball in $$\Bbb{R}^3$$. Then, by the change of variables formula, if $$\alpha,\beta,\gamma$$ are non-negative integers, then $$$$\int_B x^{\alpha}y^{\beta}z^{\gamma} \, dV = - \int_B x^{\alpha}y^{\beta}z^{\gamma} \, dV = 0,$$$$ provided that atleast one of $$\alpha,\beta,\gamma$$ is odd. For example if $$\alpha$$ is odd, use the change of variables $$(x,y,z) \mapsto (-x,y,z)$$. If $$\beta$$ is odd, use $$(x,y,z) \mapsto (x,-y,z)$$, and use $$(x,y,z) \mapsto (x,y,-z)$$ if $$\gamma$$ is odd. For example, $$\int_B x \, dV = \int_B yz^2 \, dV = 0$$, etc. This works because the region of integration is still the unit ball $$B$$, the absolute value of the determinant is $$1$$, but the actual integrand picks up a minus sign.

This observation immediately implies that we can ignore "cubic" terms, the "cross-quadratic terms" (like $$xy$$) and linear terms in $$f(x,y,z)$$, because these terms all integrate to $$0$$. So, if we write $$$$f(x,y,z) = \text{cubic terms} + a_1x^2 + a_2y^2 + a_3 z^2 + \text{cross quadratic terms} + \text{linear terms} + c,$$$$

then to quickly integrate such an expression, notice that (by symmetry in change of variables) that $$$$\int_B x^2 \, dV = \int_B y^2 \, dV = \int_B z^2 \, dV = \dfrac{1}{3} \int_B (x^2 + y^2 + z^2) \, dV$$$$ The last expression can be easily integrated using spherical coordinates, and the answer is $$\dfrac{4\pi}{15}.$$

Lastly integrating the constant term means we simply multiply by the volume of the unit ball. Hence, putting this all together, we find that \begin{align} \int_B f \, dV = (a_1 + a_2 + a_3) \cdot \dfrac{4 \pi}{15} + c \cdot \dfrac{4\pi}{3} \tag{*} \end{align}

What this shows is that if $$f$$ is a polynomial of degree at most $$3$$, then integrating over the unit ball only depends on the constant term, and the coefficients of the "pure quadratic terms" (like $$x^2,y^2,z^2$$). It is easy to verify that \begin{align} a_1+a_2+a_3 = \dfrac{\Delta f(0,0,0)}{2} \quad \text{and} \quad c= f(0,0,0). \tag{**} \end{align}

So, substituting $$(**)$$ into $$(*)$$, we find that $$$$\int_B f\, dV = \dfrac{4 \pi}{3} \cdot f(0,0,0) + \dfrac{2 \pi}{15} \cdot \Delta f (0,0,0).$$$$

I found this question pretty interesting, so I tried to generalize this result to $$n$$-dimensions, and here's what I came up with so far. I'll prove that

If $$f: \Bbb{R}^n \to \Bbb{R}$$ is a polynomial of degree at most $$3$$, $$B = \{\xi \in \Bbb{R}^n: \lVert \xi\rVert^2 \leq 1\}$$ is the closed unit ball, and $$dV$$ denotes the $$n$$-dimensional volume element, then \begin{align} \int_B f \, dV &= f(0) \cdot \text{vol}(B) + \dfrac{\text{trace}(D^2f_0)}{2}\cdot \lambda \\ &= f(0) \cdot \text{vol}(B) + \dfrac{\Delta f(0)}{2}\cdot \lambda, \end{align} where $$D^2f_0$$ is the second differential of $$f$$ at $$0$$ (a symmetric bilinear form), and $$\Delta = \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$$ is the Laplacian, and $$\lambda$$ is a constant, computed by $$$$\lambda := \dfrac{1}{n} \int_B \lVert \xi \rVert^2 \, dV$$$$

The proof of this is very similar to the one I gave above in the special case. First, note that since $$f$$ is a polynomial by assumption, it equals its own Taylor polynomial: $$$$f(\xi) = f(0) + Df_0(\xi) + \dfrac{1}{2}D^2f_0(\xi)^2 + \dfrac{1}{6}D^3f_0(\xi)^3$$$$ where $$D^kf_0$$ is a symmetric $$k$$-linear map from $$\Bbb{R}^n \times \cdots \times \Bbb{R}^n$$ into $$\Bbb{R}$$ and $$(\xi)^k$$ denotes the element $$(\xi,\dots, \xi) \in \Bbb{R}^n \times \cdots \times \Bbb{R}^n$$ (k-products).

Hence to compute $$\int_B f \, dV$$, we have to sum up $$4$$-terms. By an almost identical argument I gave above, the cubic and linear terms all vanish after integration: $$$$\int_B Df_0(\xi) \, dV = \int_V D^3f_0(\xi)^3 \, dV = 0.$$$$ So, we have that \begin{align} \int_B f \, dV &= \int_B \left [f(0) + \dfrac{1}{2} D^2f_0(\xi)^2 \right] \, dV \\ &= f(0)\cdot \text{vol}(B) + \dfrac{1}{2} \int_B D^2f_0(\xi)^2 \, dV \end{align}

In the second term, $$D^2f_0(\xi)^2$$ is a sum of terms of the form $$\xi_i \xi_j \cdot (\partial_i \partial_j f)(0)$$. But now notice that (again change of variables) if $$i \neq j$$ then $$$$\int_B \xi_i \xi_j \, dV = - \int_B \xi_i \xi_j \, dV = 0$$$$

Hence, the only contribution to the integral comes from terms where $$i=j$$ "the diagonal terms". More precisely, \begin{align} \int_B D^2f_0(\xi,\xi) \, dV &= \sum_{i=1}^n (\partial_i^2 f)(0) \cdot \left( \int_B (\xi_i)^2 \, dV \right) \tag{\ddot{\smile}} \end{align} But now notice that (again symmetry in change of variables) that \begin{align} \int_B (\xi_1)^2 \, dV = \dots = \int_B (\xi_n)^2 \, dV = \dfrac{1}{n} \int_B \lVert \xi \rVert^2 \, dV =: \lambda \tag{\ddot{\smile} \ddot{\smile}} \end{align}

Substituting $$\ddot{\smile} \ddot{\smile}$$ into $$\ddot{\smile}$$ yields the result \begin{align} \int_B D^2f_0(\xi)^2 \, dV &= \lambda \cdot \sum_{i=1}^n (\partial_i^2f)(0) \\ &= \lambda \cdot \Delta f(0) = \lambda \cdot \text{trace}(D^2f_0) \end{align}

This proves that

$$$$\int_B f \, dV = f(0) \cdot \text{vol}(B) + \dfrac{\text{trace}(D^2f_0)}{2}\cdot \lambda$$$$

In the case $$n=3$$, everything was nice because we could easily compute $$\text{vol}(B)$$ and $$\lambda$$ explicitly using spherical coordinates. In higher dimensions, this will necessarily be more complicated, and I think it will involve the use of gamma functions and stuff. Also, if we allow for higher order polynomials, I'm pretty sure we can get a formula involving $$f(0),D^2f_0, D^4f_0,D^6f_0 \dots$$, although things will probably get more messy.

• While we're here: there is an analogous formula for all polynomials $f$, involving only values at $0$, value of $\Delta f$ at $0$, value of $\Delta^2 f$ at $0$, and so on, from various ways of thinking... Representation theory? Spherical harmonics? – paul garrett Jun 27 at 20:04
• @paulgarrett That's interesting to know. I think integration over a ball is something which has been studied in some detail (although not covered in university lectures too much). I have just added my own slight generalisation to $n$-dimensions, for polynomials up to the third degree – peek-a-boo Jun 27 at 20:45