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I want to setup a triple integral for the volume of the surface in the ordering $dy \hspace{1mm} dx \hspace{1mm} dz$: enter image description here

So far I have that for $0\leq z \leq 1, 0 \leq y \leq x$ and for $1 \leq z \leq 2, 0 \leq y \leq \sqrt{2-z}$. I'm having trouble setting up bounds for $x$. It looks from the projection like $0 \leq x \leq 1$ for both integrals, but it doesn't give me the right value for volume (should be $\frac{11}{12}$ based on the other differential orderings.)

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    $\begingroup$ You'll need three triple integrals with the ordering $dxdydz$ $\endgroup$
    – user801306
    Commented Mar 7, 2021 at 15:57

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For $0 \leq z \leq 1$, the volume is simply bound by $y = 0$, $y = x$ and $x=1$.

For any given value of $z \in (1,2)$, there are two bounds of $y$: below $x = \sqrt{2-z}$, it is simply bound by plane $y = x$ and above $x = \sqrt{2-z}$, by the parabolic cylinder.

So the integral should be

$\displaystyle \int_0^1 \int_0^1 \int_0^x \ dy \ dx \ dz$ +

$\displaystyle \int_1^2 \int_0^{\sqrt{2-z}} \int_0^x \ dy \ dx \ dz$ +

$\displaystyle \int_1^2 \int_{\sqrt{2-z}}^1 \int_0^{\sqrt{2-z}} \ dy \ dx \ dz$

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  • $\begingroup$ Let me know if you are ale to visualize it looking at the bounds now and if it is clear to you. $\endgroup$
    – Math Lover
    Commented Mar 7, 2021 at 16:24
  • $\begingroup$ Thank you! Is $x=\sqrt{2-z}$ the projection on the xz plane? Since this is true only when y = x, I pictured this curve off of the xz plane (at some positive y value), somehow touching the y = x plane. $\endgroup$
    – Buddhapus
    Commented Mar 9, 2021 at 2:55
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    $\begingroup$ In the picture you have attached with your question, you notice there is a dotted line at $z=1$. That is the upper bound of the triangular prism - first integral. Above $z = 1$, You can see a curve in the plane $y=x$. That is intersection of $z \leq 2-y^2$ and the plane. Given the curve is in plane $y=x$, we have $z=2-x^2 = 2-y^2$. At any $z$, all values of $x$ where $x \leq \sqrt{2-z}$, you can see in the picture that, a shell along $y$ will be bound by the plane $y=x$ whereas beyond that curve ($x \geq \sqrt{2-z}$), you can see parabolic cylinder surface and that is the upper bound of $y$. $\endgroup$
    – Math Lover
    Commented Mar 9, 2021 at 5:26

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