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Question Number 215604 by universe last updated on 11/Jan/25

Answered by mr W last updated on 12/Jan/25

$$=\frac{1}{\pi }\int \int \int (x^2+4y^2+z^2-2xy-4yz+2zx+4x^2+y^2+z^2-4xy+2yz-4zx+x^2+y^2+4z^2-2xy-4yz+4zx)dxdydz$$$$=\frac{1}{\pi }\int \int \int (6x^2+6y^2+6z^2-8xy-6yz+2zx)dxdydz$$$$=\frac{6}{\pi }\int \int \int (x^2+y^2+z^2)dxdydz$$$$=\frac{6}{\pi }{\int }_0^1{r}^{2}4\pi r^{2}dr$$$$=24\times {\left[\frac{r^5}{5}\right]}_0^1$$$$=24\times \frac{1^5}{5}$$$$=4.80✓$$

Commented by mr W last updated on 12/Jan/25

$$duetosymmetry$$$$\underset{x^2+y^2+z^2⩽R^2}{\int \int \int }xydxdydz=0$$$$\underset{x^2+y^2+z^2⩽R^2}{\int \int \int }yzdxdydz=0$$$$\underset{x^2+y^2+z^2⩽R^2}{\int \int \int }zxdxdydz=0$$

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