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Evaluate $\int\int\int_v x^2 dx \ dy \ dz$ 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 $I=\int_{0}^{2} \int_{\sqrt{2 y}}^{2} \frac{x^{2}}{\sqrt{x^{4}-4 y^{2}}} d x \ d y$

$\begin{aligned} \text { Region of integration : } & \sqrt{2 y} \leq x \leq 2 \\ 0 & \leq y \leq 2 \end{aligned}$

$\begin{aligned} \text { Curves: } & \text { (i) } x=2, y=2, y=0 \text { are lines. } \\ &(i i) x=\sqrt{2 y} \Rightarrow x^{2}=2 y \\ & \text { Parabola with vertex }(0,0) \text { opening in upward direction. } \end{aligned}$

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After changing the order of integration:

$0 \leq y \leq \frac{x^{2}}{2}$

$0 \leq x \leq 2$

$\begin{aligned} \therefore I &=\int_{0}^{2} \int_{0}^{\frac{x^{2}}{2}} \frac{x^{2}}{\sqrt{x^{4}-4 y^{2}}} d y \ d x \\ &=\frac{1}{2} \int_{0}^{2} \int_{0}^{\frac{x^{2}}{2}} \frac{x^{2}}{\sqrt{\frac{x^{4}}{4}-y^{2}}} d y\ d x \end{aligned}$

$\begin{aligned} &=\frac{1}{2} \int_{0}^{2} x^{2}\left[\sin ^{-1}\left(\frac{y}{x^{2} / 2}\right)\right]_{0}^{\frac{x^{2}}{2}} \mathrm{d} y \\ \therefore I &=\frac{1}{2} \int_{0}^{2} x^{2} \frac{\pi}{2} d x \\=& \frac{\pi}{4}\left[\frac{x^{3}}{3}\right]_{0}^{2} \\ \therefore I &=\frac{2 \pi}{3} \end{aligned}$

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