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Question Number 82232 by mhmd last updated on 19/Feb/20

Commented by JDamian last updated on 19/Feb/20

I guess eventually we will have to bear ads

$$\mathrm{I}\:\mathrm{guess}\:\mathrm{eventually}\:\mathrm{we}\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{bear}\:\mathrm{ads} \\ $$

Answered by mind is power last updated on 19/Feb/20

z→6−z  y→8−y  x→10−x  I=∫∫∫((log(xyz)dxdydz)/(log(10x−x^2 )+log(8y−y^2 )+log(6z−z^2 )))  ⇔  I=∫∫∫((log(6−z)+log(8−y)+log(10−x))/(log(10x−x^2 )+log(8y−y^2 )+log(6z−z^2 )))dxdydz  add this Two integral forme⇒  2I=∫∫∫((log(6z−z^2 )+log(8y−y^2 )+log(10x−x^2 ))/(log(6z−z^2 )+log(8y−y^2 )+log(10x−x^2 )))dxdydz  2I=∫_3 ^7 ∫_2 ^6 ∫_1 ^5 dxdydz=(5−1)(6−2)(7−3)=64  I=32

$${z}\rightarrow\mathrm{6}−{z} \\ $$$${y}\rightarrow\mathrm{8}−{y} \\ $$$${x}\rightarrow\mathrm{10}−{x} \\ $$$${I}=\int\int\int\frac{{log}\left({xyz}\right){dxdydz}}{{log}\left(\mathrm{10}{x}−{x}^{\mathrm{2}} \right)+{log}\left(\mathrm{8}{y}−{y}^{\mathrm{2}} \right)+{log}\left(\mathrm{6}{z}−{z}^{\mathrm{2}} \right)} \\ $$$$\Leftrightarrow \\ $$$${I}=\int\int\int\frac{{log}\left(\mathrm{6}−{z}\right)+{log}\left(\mathrm{8}−{y}\right)+{log}\left(\mathrm{10}−{x}\right)}{{log}\left(\mathrm{10}{x}−{x}^{\mathrm{2}} \right)+{log}\left(\mathrm{8}{y}−{y}^{\mathrm{2}} \right)+{log}\left(\mathrm{6}{z}−{z}^{\mathrm{2}} \right)}{dxdydz} \\ $$$${add}\:{this}\:{Two}\:{integral}\:{forme}\Rightarrow \\ $$$$\mathrm{2}{I}=\int\int\int\frac{{log}\left(\mathrm{6}{z}−{z}^{\mathrm{2}} \right)+{log}\left(\mathrm{8}{y}−{y}^{\mathrm{2}} \right)+{log}\left(\mathrm{10}{x}−{x}^{\mathrm{2}} \right)}{{log}\left(\mathrm{6}{z}−{z}^{\mathrm{2}} \right)+{log}\left(\mathrm{8}{y}−{y}^{\mathrm{2}} \right)+{log}\left(\mathrm{10}{x}−{x}^{\mathrm{2}} \right)}{dxdydz} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{3}} ^{\mathrm{7}} \int_{\mathrm{2}} ^{\mathrm{6}} \int_{\mathrm{1}} ^{\mathrm{5}} {dxdydz}=\left(\mathrm{5}−\mathrm{1}\right)\left(\mathrm{6}−\mathrm{2}\right)\left(\mathrm{7}−\mathrm{3}\right)=\mathrm{64} \\ $$$${I}=\mathrm{32} \\ $$$$ \\ $$$$ \\ $$

Commented by mhmd last updated on 19/Feb/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mind is power last updated on 19/Feb/20

withe pleasur

$${withe}\:{pleasur} \\ $$

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