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Evaluate $\int\int\int x^2 \ y \ z \ dx \ dy \ dz$ over the volume bounded by the planes x = 0, y = 0, z = 0 and $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
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Solution:

Let $v=\iiint x^{2} \ d x \ d y \ d z$

Region of integration is volume bounded by the planes $x=0, y=0, z=0$

And $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Put $x=a u, y=b v, z=c w$

$\therefore \quad \ d x \ d y \ d z=a \ b \ c \ d u . d v$

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The intersection of tetrahedron with all axes is: $(1,0,0),(0,1,0),(0,0,1)$

$0 \leq w \leq(1-u-v)$

$0 \leq v \leq(1-u)$

$0 \leq u \leq 1$

The volume required is given by

$\begin{aligned} \mathbf{V} &=\int_{0}^{1} \int_{0}^{1-u} \int_{0}^{1-u-v} a b c \ a^{2} \ u^{2} \ b v . c w . \text { du dv dw } \\ &=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} \int_{0}^{1-u} u^{2} v(1-u-v)^{2} \ d v \ d u \end{aligned}$

$=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} \int_{0}^{1-u} u^{2} v\left[(1-u)^{2}-2(1-u) v+v^{2}\right] d u \ d v$

$=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} u^{2}\left[(1-u)^{2} \frac{v^{2}}{2}-2(1-u) \frac{v^{3}}{3}+\frac{v^{4}}{4}\right]_{0}^{1-u} du$

$\begin{aligned} &=\frac{a^{3} b^{2} c^{2}}{2} \int_{0}^{1} \frac{u^{2}(1-u)^{4} d u}{12} \\ &=\frac{a^{3} b^{2} c^{2}}{24} \beta(3,5) \\ &=\frac{a^{3} b^{2} c^{2}}{24}\left(\frac{2 ! 4 !}{7 !}\right) \\ \therefore I &=\frac{a^{3} b^{2} c^{2}}{2520} \end{aligned}$

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