written 4.9 years ago by |
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 $x=a u, y=b v, z=c w$
$\therefore \quad \ d x \ d y \ d z=a \ b \ c \ d u . d v$
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
$\begin{aligned} \mathbf{V} &=\int_{0}^{1} \int_{0}^{1-u} \int_{0}^{1-u-v} a b c \ a^{2} \ u^{2} \ b v . c w . \text { du dv dw } \\ &=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} \int_{0}^{1-u} u^{2} v(1-u-v)^{2} \ d v \ d u \end{aligned}$
$=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} \int_{0}^{1-u} u^{2} v\left[(1-u)^{2}-2(1-u) v+v^{2}\right] d u \ d v$
$=\frac{1}{2} a^{3} b^{2} c^{2} \int_{0}^{1} u^{2}\left[(1-u)^{2} \frac{v^{2}}{2}-2(1-u) \frac{v^{3}}{3}+\frac{v^{4}}{4}\right]_{0}^{1-u} du$
$\begin{aligned} &=\frac{a^{3} b^{2} c^{2}}{2} \int_{0}^{1} \frac{u^{2}(1-u)^{4} d u}{12} \\ &=\frac{a^{3} b^{2} c^{2}}{24} \beta(3,5) \\ &=\frac{a^{3} b^{2} c^{2}}{24}\left(\frac{2 ! 4 !}{7 !}\right) \\ \therefore I &=\frac{a^{3} b^{2} c^{2}}{2520} \end{aligned}$