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Let $u(x,y,z)$ be a continuous function, it has continuous second order partial derivatives at $M(x_0,y_0,z_0)$. $\Sigma$ is a sphere centered at $M$ with radius $R$, and $$T(R)=\frac{1}{4\pi R^2}\iint_{\Sigma}u(x,y,z)dS.$$

I have proved that $\lim_{R\to 0}\frac{1}{4\pi R^2}\iint_{\Sigma}u(x,y,z)dS=u(u_0,y_0,z_0)$. Then I need to find the main part of $\frac{1}{4\pi R^2}\iint_{\Sigma}u(x,y,z)dS-u(x_0,y_0,z_0)$ when $R\to 0$. I'm trying to do a Taylor Expand on $u$, $$\frac{1}{4\pi R^2}\iint_{\Sigma}u(x,y,z)dS=\frac{1}{4\pi R^2}\iint_{\Sigma}\left(u(x_0,y_0,z_0)+[u_{x}(x-x_0)+u_{y}(y-y_0)+u_{z}(z-z_0)]+\frac{1}{2}[u_{xx}(x-x_0)^2+u_{yy}(y-y_0)^2+u_{zz}(z-z_0)^2+2(u_{xy}(x-x_0)(y-y_0))+2(u_{xz}(x-x_0)(z-z_0))+2(u_{yz}(y-y_0)(z-z_0))]+R_k\right)dS.$$

But I don't know how to make a further calculation. Can anyone help me? Thanks in advance!

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  • $\begingroup$ What is the "main part" ? $\endgroup$ Nov 15, 2023 at 8:21
  • $\begingroup$ The coefficient of the first term of its taylor expansion.@NinadMunshi $\endgroup$
    – Ychen
    Nov 15, 2023 at 8:28

1 Answer 1

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Use the linearity of integral, compute the first integrals and bound above the last one. By symmetry $$\iint_{\Sigma} (x-x_0) dS = \iint_{\Sigma} (y-y_0) dS = \iint_{\Sigma} (z-z_0) dS = 0$$ and \begin{eqnarray*} \iint_{\Sigma} (x-x_0)^2 dS = \iint_{\Sigma} (y-y_0)^2 dS &=& \iint_{\Sigma} (z-z_0)^2 dS \\ &=& \frac{1}{3}\iint_{\Sigma} [(x-x_0)^2+(y-y_0)^2+(z-z_0)^2] dS \\ &=& \frac{1}{3}\iint_{\Sigma}R^2 dS = \frac{1}{3} 4\pi R^4. \end{eqnarray*} Last, the remainder in the Taylor expansion can be written $[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2] \phi(x,y,z)$ where $\phi(x,y,z) \to 0$ as $(x,y,z) \to (x_0,y_0,z_0)$. Given $\epsilon>0$, one can find $\delta>0$ such that for all $(x,y,z)$ in the closed ball $\bar{B}((x_0,y_0,z_0),\delta)$, one has$|\phi(x,y,z)| \le \epsilon$. Thus, when $R \le \delta$,
\begin{eqnarray*} \Big|\iint_{\Sigma}[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2] \phi(x,y,z) dS \Big| \le \iint_{\Sigma}R^2 |\phi(x,y,z)| dS \le 4\pi R^4\epsilon\end{eqnarray*}

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