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I just learnt triple integrals recently but I'm a bit confused about the way how we solve a triple integral problem. I'm not sure if I found the limits of this triple integral question correctly.

"Find the volume of the solid region B in the first octant between $x+y+z^2=1$, $x=0$, $y=0$ and $z=0$. "

Is it correct for me to put $x=0$, $y=0$ and $z=0$ into $x+y+z^2=1$ to get the x, y, z-limits, that gives us $\int_0^1$$\int_0^1$$\int_0^1$$dxdydz$? If it's wrong, how can I find the correct limits to the integrals?

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  • $\begingroup$ first, get a feel of how the surface $z=\sqrt{1-x-y}$ looks click this $\endgroup$ – AgentS Oct 31 '19 at 16:16
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If you use $\int_0^1 \int_0^1 \int_0^1 dx dy dz,$ then you are integrating over the unit cube $[0,1] \times [0,1] \times [0,1].$ With the region you described, $x,y,$ and $z$ depend on each other. Even though each of $x, y,$ and $z$ can sometimes be any given value in $[0,1]$ with the region you described, they cannot, for example, all be $1$ at the same time, because $1 + 1 + 1^2 > 1.$

A nice trick to find the bounds of integration in a problem like this is to work outside-in with one variable at a time. If we are going to integrate $\int_?^?\int_?^?\int_?^?dxdydz,$ then after we integrate $dz$ last, our answer needs to be a number, not a variable. This tells us that the bounds for integration $dz$ cannot have $x$ or $y$ in them, so they must just be numbers. Then we use $\int_0^1\int_?^?\int_?^?dxdydz,$ because $z$ can be anything in $[0,1]$ if it is independent of $x$ and $y$.

But now that we have chosen our bounds for $z$, $y$ has to live with this choice; $y$ can still be as small as $0$, but since we've already chosen a $z$ value, we have to have $y \leq 1 - z^2.$ This gives us $\int_0^1\int_0^{1-z^2}\int_?^?dxdydz.$

Now that we have chosen $z$ and $y$, $x$ has to live with these choices; $x$ can still be as small as $0$, but we have to have $x \leq 1 - y - z^2,$ which means our integral is $\int_0^1\int_0^{1-z^2}\int_0^{1-y-z^2}dxdydz.$

So, in short, work outside-in. If you're integrating $dxdydz,$ then $z$ is free from $x$ and $y$. But you have to choose the bounds for $y$ based on $z$, and you have to choose the bounds for $x$ based on $y$ and $z.$

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  • $\begingroup$ Thank you so much! Our teacher told us to sketch the region with the vertical projection of the solid in the $xy$-plane, and find the bounds by drawing lines parallel to $z$-axis, $y$-axis respectively. I have sketch the surface $z$ online and it looks like a sheet of paper fold in half. But I find it really difficult to do it that way. $\endgroup$ – Dominique Nov 1 '19 at 7:04

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