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Let $R$ be the region in the first octant bounded by a surface $F(x,y,z) = 0$ and the coordinate planes. The projection of $R$

  1. on the $xy$-plane is bounded by the coordinates axes and the curve $y = 9 - x^2$ ,
  2. on the $xz$-plane is bounded by the coordinates axes and the curve $x =\sqrt{9-z}$.
  3. on the $yz$-plane is bounded by the coordinates axes and the curve $z=9-y$.

Now, the volume of the region $R$ is asked. When I sketch the region, I get $1/8$ of a sphere cutted by the plane $z=9-y$. Then, I write a triple integral as follows. $$\int_{0}^{9}\int_{0}^{9-z}\int_{0}^{\sqrt{9-z}}dx \, dy \, dz$$

However, the solution says the inner integral has bounds $$\int_{0}^{\sqrt{9-y-z}} dx .$$ What is the part that I am missing?

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  • $\begingroup$ If I understand the region correctly, neither integrals seem correct. You will have to split the region into two sub-regions at $y = z$. $\endgroup$
    – Math Lover
    Apr 23, 2022 at 15:54
  • $\begingroup$ I agree with MathLover. This is a challenging question because we're not given actual equations of surfaces that determine the region. The upper bound on $x$ should be the smaller of $\sqrt{9-z}$ and $\sqrt{9-y}$. Where they got $\sqrt{9-y-z}$ is ... a mystery. $\endgroup$ Apr 24, 2022 at 0:32

1 Answer 1

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Note that $y=9-x^2$, $z=9-x^2$ and $y+z=9$. So, the region $R$ is symmetric with respect to the plane $z=y$, which allows the volume integration to be set up as follows $$V=2\int_0^{\frac92}dz \int_z^{9-z}dy \int_0^{\sqrt{9-y}}dx $$

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  • $\begingroup$ Thanks, but I still do not get it why the correct answer is as above, unfortunately. $\endgroup$
    – Ninja
    Apr 23, 2022 at 16:49

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