1. triple integral

i'm new to triple integrals and having trouble setting this problem up.
its the equation f(x,y,z)= xy
bound by the planes x=0, y=0, z=0, x+2y+4z=8
how do i find the boundaries for integration?

2. Find out where the plane intersects the y and x-axis. You can do that from the equation $x+2y+4z=8$ first let x=0 and z=0 so then $2y=8$, then let z=0 and y=0 so then x=8 then you can find $x_a$ and $f(x)$ below and set up:

$\int_0^{x_a}\int_0^{f(x)} \int_0^{g(x,y)} dzdydx$

or:

$\int_0^{x_a}\int_0^{f(x)} 1/4(8-x-2y) dydx$

3. thanks so x is from 0 to 8, y from 0 to 4, and z from 0 to 2?

4. Originally Posted by holly123
thanks so x is from 0 to 8, y from 0 to 4, and z from 0 to 2?
No. y goes from the x-axis to where the plane x+2y+4z=8 cuts through the x-y plane right? Isn't that just y=1/2(8-x)? And z goes from the x-y plane up to the surface x+2y+4z=8 solved for z. Wouldn't that then be:

$\int_0^8 \int_0^{1/2(8-x)} \int_0^{1/4(8-x+2y)} dz dy dx$

or it's just a double integral:

$\int_0^8 \int_0^{1/2(8-x)}1/4(8-x+2y) dy dx$

5. ughhh i set it up this way, as well as many variations of this way, and am getting a completely wrong answer. i have worked on this problem for hours and done it several times. the answer in the back of the book is 256/15
i have gotten 2816/3 and 15104/15 many times. how in the world do you do this problem?!?