Question Number 215604 by universe last updated on 11/Jan/25
Commented by MathematicalUser2357 last updated on 12/Jan/25
$${can}\:{you}\:{please}\:{translate} \\ $$
Answered by mr W last updated on 12/Jan/25
![=(1/π)∫∫∫(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 =(1/π)∫∫∫(6x^2 +6y^2 +6z^2 −8xy−6yz+2zx)dxdydz =(6/π)∫∫∫(x^2 +y^2 +z^2 )dxdydz =(6/π)∫_0 ^1 r^2 4πr^2 dr =24×[(r^5 /5)]_0 ^1 =24×(1^5 /5) =4.80 ✓](https://www.tinkutara.com/question/Q215634.png)
$$=\frac{\mathrm{1}}{\pi}\int\int\int\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{4}{yz}+\mathrm{2}{zx}+\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{4}{xy}+\mathrm{2}{yz}−\mathrm{4}{zx}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}{z}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{4}{yz}+\mathrm{4}{zx}\right){dxdydz} \\ $$$$=\frac{\mathrm{1}}{\pi}\int\int\int\left(\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{y}^{\mathrm{2}} +\mathrm{6}{z}^{\mathrm{2}} −\mathrm{8}{xy}−\mathrm{6}{yz}+\mathrm{2}{zx}\right){dxdydz} \\ $$$$=\frac{\mathrm{6}}{\pi}\int\int\int\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right){dxdydz} \\ $$$$=\frac{\mathrm{6}}{\pi}\int_{\mathrm{0}} ^{\mathrm{1}} {r}^{\mathrm{2}} \mathrm{4}\pi{r}^{\mathrm{2}} {dr} \\ $$$$=\mathrm{24}×\left[\frac{{r}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{24}×\frac{\mathrm{1}^{\mathrm{5}} }{\mathrm{5}} \\ $$$$=\mathrm{4}.\mathrm{80}\:\checkmark \\ $$
Commented by mr W last updated on 12/Jan/25
$${due}\:{to}\:{symmetry} \\ $$$$\underset{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant{R}^{\mathrm{2}} } {\int\int\int}{xydxdydz}=\mathrm{0} \\ $$$$\underset{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant{R}^{\mathrm{2}} } {\int\int\int}{yzdxdydz}=\mathrm{0} \\ $$$$\underset{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant{R}^{\mathrm{2}} } {\int\int\int}{zxdxdydz}=\mathrm{0} \\ $$