# Q: [Beginner] Using Triple Integral to find Volume of solid

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 $$\int_0^1\int_0^1\int_0^1dxdydz$$? If it's wrong, how can I find the correct limits to the integrals?

• first, get a feel of how the surface $z=\sqrt{1-x-y}$ looks click this – AgentS Oct 31 '19 at 16:16

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 $$z.$$
• 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. – Dominique Nov 1 '19 at 7:04