Write an integral that represents the volume of the function I have a function
$$8x+2y+4z=8$$
I need to set up a triple integral dx dy dz for the volume of the tetrahedron of the first quadrant. I tried to isolate x giving me
$$\frac{-2y-4z+8}{8}$$
So I figured I'd begin integrating with respect to x from 0 to that fraction shown as:
$$\int_0^\frac{-2y-4z+8}{8}1dx$$
However I'm not sure how to proceed to the next step as the computer states the next integral would be 
$$\int_0^{4-\frac{4z}{2}}Answer dy$$
Followed by
$$\int_0^2Answerdz$$
Help??
 A: Your computer's answer looks correct, so I'll explain each step:


*

*You are right that the inner integral should be $$\int_0^{(-2y-4z+8)/8}1\ dx$$

*Now, what are the bounds of $y$? Well $y$ is bounded below by the restriction that it all must be in the first quadrant, i.e. $y\ge 0$. $y$ is bounded above by the plane $8x+2y+4z=8$. Solving for $y$ we get $$y=\frac{8-8x-4z}{2}$$ The maximum value of $y$ occurs when $x=0$, so we get the upper bound for $y$ is $$\frac{8-4z}{2}=4-2z$$ Hence, our two inmost integrals are $$\int_0^{4-2z}\int_0^{(-2y-4z+8)/8}1\ dx\ dy$$

*Similarly to step 2, we find that $z$ is bounded below by $0$ and above by the plane. Solving for $z$ when $x=0,\ y=0$ we get an upper bound of $2$. So the final triple integral is $$\int_0^2 \int_0^{4-2z}\int_0^{(-2y-4z+8)/8}1\ dx\ dy\ dz$$
Whenever I do these kinds of problems, I find it very helpful to draw a graph of the region of integration in the $xy$-plane. This way, it is easy to come up with the outer integral. In our case, we graph $8x+2y\le 8$ in the first quadrant:

From here, it is easy to determine that the integral is: $$\int_0^1\int_0^{4-4x}\int_0^{f(x, y)} 1\ dz\ dy\ dx$$
where $f(x, y)$ is just the $z$ value such that $(x,y,z)$ is in the plane. i.e. $$f(x,y)=\frac{8-8x-2y}{4}$$ If you want to use a graph to help you determine the triple integral in a different order (e.g. $dx\ dy\ dz$) then just plot the corresponding plane (e.g. the $yz$-plane)
