$\int^2_0 \int^2_\sqrt{2y} \frac{x^2}{\sqrt{x^4 – 4 y^2}} dx \ dy$

Solution:

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

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

$\therefore \quad d x \ d y \ d z=a b c$ du.dv.dw

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

$\mathbf{v}=\int_{0}^{1} \int_{0}^{1-u} \int_{0}^{1-u-v} a \ b \ c \ a^{2} \ u^{2}$ du dv dw

$=a^{3} b c \int_{0}^{1} \int_{0}^{1-u}(1-u-v) u^{2} \ d v \ d u$

$=a^{3} b c \int_{0}^{1} u^{2}\left[v-u v-\frac{v^{2}}{2}\right]_0^{1-u} d u$

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

$\begin{aligned} &=a^{3} b c\left[\frac{u^{3}}{3}-\frac{u^{4}}{2}+\frac{u^{5}}{5}-\frac{1}{2}\left(\frac{u^{3}}{3}-\frac{1}{2} u^{4}+\frac{u^{5}}{5}\right)\right]_{0}^{1} \\ \therefore V &=\frac{1}{60}\left(a^{3} b c\right) \end{aligned}$