Bounds for Triple Integral?

Question

Hello -

I'm trying to figure out the bounds for the following integral.

Triple Integral xy dV, where E is the solid tetrahedron with vertices (0,0,0), (1,0,0), (0,2,0), (0,0,3).

Please help. Thank you.

Answer 1

Note that the plane containing (1,0,0), (0,2,0) and (0,0,3) has the equation:
x + y/2 + z/3 = 1

Also, in the xy plane, the line containing (1,0,0) and (0,2,0) has the equation:
x + y/2 = 1

So, if you let the inner integral run from x=0 to x=1, the y limits at each x will be from y=0 to y=2(1-x), and the z limits will be from 0 to 3(1-x-y/2).

Piece of pie, huh?

Answer 2

You first need to work out the equation for the plane passing through the vertices (1,0,0), (0,2,0) and (0,0,3) [see for example Anton: Calculus with Analytic Geometry, Section on Planes in 3-space]. This plane has equation 6x+3y+2z=6, or solving for z,
z=3-3/2y-3x.
This gives the upper limit for your z-integration with the lower limit being z=0.

Now project the region E down on to the positive x-y quadrant. This region is a triangle with vertices (0,0), (1,0) and (0,2). The bounding line has equation y=-2x+2, which gives the upper limit for the y-integration, the lower limit being y=0.

Finally, the x-integration goes from x=0 to x=1.