# Mass Triple Integrals

• Mar 28th 2010, 09:03 AM
ur5pointos2slo
Mass Triple Integrals
Find the mass of the solid region bounded by the planes
x+z=1
x-z = -1
y=0
and the surface
y=sqrt(z)

The density of the solid is (x,y,z)=2y+5

My analytic geometry skills are a little lost. Could someone please remind me how to graph these planes so I can find the limits for my integrals?
• Mar 29th 2010, 05:22 AM
shawsend
Hi. Use Mathematica to draw it. Oh I realize that's a lazy approach and cuts into your analytical education but if you're having problems visualizing it, then first get it (Mathematica), then figure out how the three equations give rise to the surfaces: Red is x+z=1, Blue is x-z=-1, and Green is y=Sqrt[z]. Now, in general, a triple integral is:

$\int_a^b\int_{y=f_1(x)}^{y=f_2(z)}\int_{z=g(x,y)}^ {z=h(x,y)} u(x,y,z) dzdydx$

so that's up from the surface g to the surface h, bordered in the x-y plane between the functions f1 and f2, and then finally in the range of x from a to b.

So we want that tee-pee volume in there right and it's symmetrical to the z-x plane since u=2y+5 so take half and multiply by two. So we go up from z=y^2 to the red sheet z=1-x and the foot-print in the x-y plane is when the two equations y=Sqrt[z] and x+1=z are satisfied (the red and green surfaces intersect or x=y^2-1. So integrating in the order dzdxdy, I get:

$2\int_0^1\int_0^{y^2-1}\int_{y^2}^{1-x} (2y+5)dzdxdy$

but I'm not sure about it so you go over it and see if I missed it.