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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?

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    $\begingroup$ Just for curiosity: Where this question is from? $\endgroup$ – DiegoMath Jun 27 at 18:11
  • $\begingroup$ @DiegoMath, this is from the PhD exam in Analysis at the University of Pittsburgh: mathematics.pitt.edu/sites/default/files/… $\endgroup$ – Hasek Jun 27 at 19:14
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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 \begin{equation} \int_B x^{\alpha}y^{\beta}z^{\gamma} \, dV = - \int_B x^{\alpha}y^{\beta}z^{\gamma} \, dV = 0, \end{equation} 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 \begin{equation} 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, \end{equation}

then to quickly integrate such an expression, notice that (by symmetry in change of variables) that \begin{equation} \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 \end{equation} 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 \begin{equation} \int_B f\, dV = \dfrac{4 \pi}{3} \cdot f(0,0,0) + \dfrac{2 \pi}{15} \cdot \Delta f (0,0,0). \end{equation}


Additional Remarks:

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 \begin{equation} \lambda := \dfrac{1}{n} \int_B \lVert \xi \rVert^2 \, dV \end{equation}

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: \begin{equation} f(\xi) = f(0) + Df_0(\xi) + \dfrac{1}{2}D^2f_0(\xi)^2 + \dfrac{1}{6}D^3f_0(\xi)^3 \end{equation} 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: \begin{equation} \int_B Df_0(\xi) \, dV = \int_V D^3f_0(\xi)^3 \, dV = 0. \end{equation} 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 \begin{equation} \int_B \xi_i \xi_j \, dV = - \int_B \xi_i \xi_j \, dV = 0 \end{equation}

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

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

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.

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  • $\begingroup$ 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? $\endgroup$ – paul garrett Jun 27 at 20:04
  • $\begingroup$ @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 $\endgroup$ – peek-a-boo Jun 27 at 20:45

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