Find the volume of the solid in $\mathbb{R}^3 $ under paraboloid $\{x_1^2+x_2^2-x_3=0\}$ and above the square $[0,1]^2$.

I want to calculate $\int_{\{x_1^2+x_2^2-x_3 \le 0 \ and \ 0 \le x_1,x_2 \le 1\}}1 dx_1dx_2dx_3$ , but I'm not sure how to split it to 3 integrals...any help?

  • $\begingroup$ $$\iint_{(0,1)^2}\left(x_1^2+x_2^2\right)\,dx_1\,dx_2 = \frac{2}{3}.$$ $\endgroup$ – Jack D'Aurizio Jul 1 '17 at 16:13
  • $\begingroup$ @JackD'Aurizio I know that the final answer is $2/3$ , I already had it.What I don't know is,what should be the boundaries for $x_3$? $\endgroup$ – ChikChak Jul 1 '17 at 16:17

Calculate: $$ \text{Volume}=\int_0^1 \int_0^1 \int_0^{x_1^2+x_2^2} dx_3 dx_1 dx_2. $$

  • $\begingroup$ Why the upper boundary of $x_3$ is $x_1^2+x_2^2$? $\endgroup$ – ChikChak Jul 1 '17 at 16:21
  • 1
    $\begingroup$ @ChikChak This is because the solid is bounded above by the surface $x_3=x_1^2+x_2^2$ and bounded below by the plane $x_3=0$. It might help if you convert your variables $x_1$, $x_2$, $x_3$ to $x$, $y$, $z$ and repeat the visualization. $\endgroup$ – Mee Seong Im Jul 1 '17 at 16:24

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