#1
Jun2903, 04:45 PM

P: n/a

Compute triple integral f(x,y)dV given f(x,y,z)=2x+3y. T is the tetrahedron bounded by the coordinate planes and first octant part of the plane with equation 2x + 3y + z = 6.
how do i solve for this, can someone get me started halfway, plz? Dx [g)] 



#2
Jun3003, 04:22 AM

P: 17

I dont know if you have solved this problem already, but here goes:
The first step you will want to do is draw two diagrams, one of the solid, and one projection of it on the xy plane. Then solve the equation for z (or x or y whichever is going to produce the easiest integral outcome). In this case I would go for the z solution: z=2x+3y6 Then determine the axis intersections in order to find the integral limits. Set up your triple integral, solve.[6)] 



#3
Jun3003, 08:40 AM

Math
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P: 38,877

The "innermost" integral will be with respect to z (dz) and will have limits z= 0 to z= 6 2x 3y.
"projecting" down to the x,y plane (z= 0) gives the line where the plane 2x+ 3y+ z= 6 crosses the x,y plane: 2x+ 3y= 6. We need to integrate over the triangle with edges x=0, y=0, 2x+ 3y= 6. Solving for y, y= 2 (2/3)x. The second integral will be with respect to y (dy) and will have limits y= 0 and y= 2(2/3)x. Finally, project to the xaxis itself. The line 2x+3y= 6 crosses the xaxis when y= 0: 2x= 6 or x= 3. The final integral will be with respect to x (dx) and will have limits x= 0, x= 3. The triple integral you want is integral (x=0 to 3) integral (y=0 t0 2(2/3)x) integral (z= 0 to 6 2x+3y)(2x+3y)dzdydx. 


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