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?

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- Nov 1st 2009, 10:23 AMholly123triple 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? - Nov 1st 2009, 11:28 AMshawsend
Find out where the plane intersects the y and x-axis. You can do that from the equation $\displaystyle x+2y+4z=8$ first let x=0 and z=0 so then $\displaystyle 2y=8$, then let z=0 and y=0 so then x=8 then you can find $\displaystyle x_a$ and $\displaystyle f(x)$ below and set up:

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

or:

$\displaystyle \int_0^{x_a}\int_0^{f(x)} 1/4(8-x-2y) dydx$ - Nov 1st 2009, 11:39 AMholly123
thanks so x is from 0 to 8, y from 0 to 4, and z from 0 to 2?

- Nov 1st 2009, 12:13 PMshawsend
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:

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

or it's just a double integral:

$\displaystyle \int_0^8 \int_0^{1/2(8-x)}1/4(8-x+2y) dy dx$ - Nov 1st 2009, 05:05 PMholly123
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?!?(Headbang)