Triple Integral with bounds in first octant

I am really confused on how to get my integrating function because I don't know, even after graphing, how the tetrahedron intersects the x-y-z axis.

I am supposed to find the triple integral for the volume of the tetrahedron cut from the first octant by the plane $6x + 3y + 2z = 6$.

I have found the bounds of integration by isolating $x,y,z$ in the tetrahedron equation. So I know how the bounds will be in any order of integration that I want. I am now stuck on what the integrating function will be.

So for example, if I was integrating w.r.t dxdydz then, according to the bounds I found through a numerical method, should be:

$$\int_{z=0}^3\int_{y=0}^{\frac{6-2z}{3}} \int_{x=0}^{\frac {6-3y-2z}{6}} dx dy dz$$

Now these are the bounds I have found. How do I get the integrating equation? Would the integrating equation be x for the first definite integral?

so something like this?

$$x \rvert _{0}^{\frac{6-3y-2z}{6}}$$

Is that right?

So then to find the volume this would be the whole equation with the bounds.

{0,3} {0, (6-2z)/3 } {0, (6-3y-2z)/6} dx dy dz As you can see in the graph above, it is bounded by the given plane so are my bounds correct? is the function a constant "1" ?

• I improved the formatting of your post; check it and see if it's correct. Also try improving your last equation. By the way, the function in this case(Cartesian coordinates) happens to be $1$; in other coordinate systems it can be different.
– Ali
Nov 29 '13 at 12:53
• @Ali how do you put your SO posts into that mathematical form? Is there some code to type to do that? Nov 29 '13 at 12:59
• Some of these sites support latex codes via mathjax I believe.
– Ali
Nov 29 '13 at 13:51
• There is a large FAQ in the meta section on the Latex/Mathjax notation used here. meta.math.stackexchange.com/questions/5020/… Nov 29 '13 at 16:58

You have a plane $P:~6x+3y+2z=6$ which cut the axes in the first octant as you see through the below plot. $P$, clearly, intersects $z=0$ in a line: $$z=0\to 6x+3y+2\times 0=6\to 6x+3y=4$$ I don't want to tell you that all triple integrals will start from this point that we did above, but we do for many certain triples. From the line above we can get the proper limits for $x$ and $y$: $$y=0\to x=2/3\Longrightarrow x|_0^{2/3}\\ y=\frac{4-6x}{3}=4/3-2x\Longrightarrow y|_0^{\frac{4}3-2x}$$ Therefore we have: $$V=\large{\int_x\int_y\int_0^{\frac{6-6x-3y}{2}}}dzdydx$$ Find the intersections with the plane $6x + 3y + 2z = 6$ and the coordinate axes in the first place, giving a tetrahedron with vertices $(0,0,0)$ , $(1,0,0)$ , $(0,2,0)$ , $(0,0,3)$ . Then introduce normed coordinates $(\xi,\eta,\zeta)$ defined by $(x,y,z) = (\xi,2\eta,3\zeta)$ and find the integral: $$\iiint dx\,dy\,dz = \iiint 1\,d\xi\;2\, d\eta\;3\, d\zeta = 6 \times \iiint d\xi\,d\eta\,d\zeta$$ $$\iiint d\xi\,d\eta\,d\zeta = \int_{\xi=0}^{\xi=1}\left[\int_{\eta=0}^{\eta=1-\xi}\left\{\int_{\zeta=0}^{\zeta=1-\xi-\eta}d\zeta\right\}d\eta\right]d\xi = \int_{\xi=0}^{\xi=1}\left[\int_{\eta=0}^{\eta=1-\xi}\left\{1-\xi-\eta\right\}\,d\eta\right]d\xi = \int_{\xi=0}^{\xi=1}\left[(1-\xi)-\xi(1-\xi)-\frac{1}{2}(1-\xi)^2\right]d\xi = \int_{\xi=0}^{\xi=1}\left[\frac{1}{2}-\xi+\frac{1}{2}\xi^2\right]d\xi = \frac{1}{2}-\frac{1}{2}+\frac{1}{2}\frac{1}{3} = \frac{1}{6}$$ So the integral is $6 \times 1/6 = 1$ . This can be seen immediately, though, by noting that $\; \iiint dx\,dy\,dz\;$ is the volume of a pyramid (: take a look at the picture by B.S.).