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Set up (do not evalutate) a triple integral that represents the volume in the first octant below the plane 2x+3y+z=6. Use the integration order dx dy dz. So the answer is (limits of integration): for x: 0 to 3-3/2y-1/2z for y: 0 to 2-1/3z for z: 0 to 6 how did they get these limits of integration? Thanks

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Set up (do not evalutate) a triple integral that represents the volume in the first octant below the plane

2x+3y+z=6. Use the integration order dx dy dz.

So the answer is (limits of integration):

for x: 0 to 3-3/2y-1/2z

for y: 0 to 2-1/3z

for z: 0 to 6

how did they get these limits of integration? Thanks

 
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