Calculus 3, Triple Integrals

[cosmonautics]

I'm really bad at visualizing things, so naturally, I'm terrible at finding the bounds for triple integrals.
Find the mass of the solid bounded by the xy-plane, yz-plane, xz-plane, and the plane (x/3) + (y/2) + (z/6) = 1
if the density of the solid is given by rho(x,y,z) = x+y
I think the innermost integral is from 0 to 6-2x-3y.
And since
the xy-plane is z=0
the yz-plane is x=0 and
the xz-plane is y=0
I thought I needed to find out where the slanted plane intersected each y=0 and x=0
So I set y=0=z=6-2x-3y [0=6-2x-3y] and solved for x

x=3-(3/2)y
So I think the middle integral is from 0 to 3-(3/2)y
but then I'm really stuck on how to find the outer bounds. Please also tell me if what I already did was complete junk.

Thanks

EDIT: After looking at it, do I just set x=3-(3/2)y equal to x=0 to find y (which gives y=2) so then my outer bound would be from 0 to 2?

 

[HallsofIvy]

Yes, that figure is a 'tetrahedron' with vertices at (0, 0, 0), (3, 0, 0), (0, 2, 0), and (0, 0, 6).

Projecting down into the xy-plane, we have the triangle with vertices at (0, 0, 0), (3, 0, 0), and (0, 2, 0). So, we can take x running from 0 to 3. The upper side of that triangle is the line x/3+ y/2= 1 so that, for each x, y= 2- (2/3)x. The next integral has y from 0 to 2- (2/3)x (NOT "y= 3- (3/2)x!). And, of course, the upper plane, x/3+ y/2+ z/6= 1 can be solved for z as z= 6- x/2- y/3 so that the "inner" integral is from 0 to 6- x/2- y/3.

The mass is given by \(\displaystyle \int_{x=0}^3\int_{x=0}^{3- (3/2)x}\int_{z= 0}^{6- x/2- y/3} (x+ y)dzdydx\)