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

Commented by JDamian last updated on 19/Feb/20

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

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

$$z\to 6-z$$$$y\to 8-y$$$$x\to 10-x$$$$I=\int \int \int \frac{log\left(xyz\right)dxdydz}{log(10x-x^2)+log(8y-y^2)+log(6z-z^2)}$$$$\iff$$$$I=\int \int \int \frac{log(6-z)+log(8-y)+log(10-x)}{log(10x-x^2)+log(8y-y^2)+log(6z-z^2)}dxdydz$$$$addthisTwointegralforme\Rightarrow$$$$2I=\int \int \int \frac{log(6z-z^2)+log(8y-y^2)+log(10x-x^2)}{log(6z-z^2)+log(8y-y^2)+log(10x-x^2)}dxdydz$$$$2I={\int }_3^7{\int }_2^6{\int }_1^{5}dxdydz=(5-1)(6-2)(7-3)=64$$$$I=32$$$$$$$$$$

Commented by mhmd last updated on 19/Feb/20

$$thankyousir$$

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

$$withepleasur$$

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