Use a triple integral in rectangular coordinates to find the mass of the region bounded by the planes x+y+z=1, x=0, y=0, and z=0 in octant 1, if the density at each point (x,y,z) is p(x,y,z)=kx
How do you set this up??
Always try to draw it first. If $\displaystyle x+y+z=1$ then when y=0, you have the line z=1-x in the z-x plane and if x=0, you have z=1-y in the z-x plane. Now just connect the lines in the x-y plane between y=1 and x=1. Try and draw those three lines in an x-y-z coordinate axis on paper then try to understand why the integral is (I believe):
$\displaystyle \int_0^1\int_0^{-x+1}\int_0^{1-y-x} kx dzdydx$
It's been a while since I've done these, but I think it's:
The solid shape we are integrating over is a tetrahedron; having a geometric sense can help with determining the limits of the definite integrals. For general region R, the triple integral is
$\displaystyle \iiint\limits_R f(x,y,z)\, dV\ = \int\limits_{a}^{b}\int\limits_{c}^{d}\int\limits_ {e}^{f}f(x,y,z)\,dx\,dy\,dz$
where we can integrate in any order. Here we will want to integrate with respect to x first, because p(x,y,z) only depends on x.
Anyway I get:
$\displaystyle \int\limits_{0}^{1}\int\limits_{0}^{(1-z)}\int\limits_{0}^{(1-y-z)}(kx)\,dx\,dy\,dz$
I'm pretty sure I got the upper limits right but not 100%, anyway you might find this site helpful, example 4 deals with a more generalized tetrahedral region defined similarly.
EDIT: maybe shawsend is right and it's easier to integrate dx last... it's been too long... sorry
EDIT 2: integrating with respect to x last is definitely easier.