(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the volume between the plane x+y+z = 1 and the xy-plane, for x+y[itex]\leq[/itex]2, x[itex]\geq[/itex]0, y[itex]\geq[/itex]0.

3. The attempt at a solution

First, the plane is above the xy-plane for y < 1-x and below the xy-plane for y > 1-x, so we'll need two integrals. This is how I set them up.

[itex]\int[/itex][itex]^{1}_{0}[/itex][itex]\int[/itex][itex]^{1-x}_{0}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx - [itex]\int[/itex][itex]^{2}_{0}[/itex][itex]\int[/itex][itex]^{2-x}_{1-x}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx

This gives me an answer of 7/6, but my book has 1 as the answer. I'm assuming my setup is wrong, since I checked the evaluation of the integrals in wolfram alpha. But I can't see what I did wrong. The innermost integrals give z-value of the given plane. The middle integrals sum those stacks from 0 to 1-x above the xy-plane and from 1-x to 2-x below the xy-plane. And then the outermost integrals sum those slices from the y axis to the boundary at 0=1-x and 0=2-x respectively.

Help would be much appreciated.

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# Homework Help: Volume with triple integral

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