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I am calculating volume of body that is defined by $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ to do this I have two possible ways:

  1. $$\int\limits_0^2\int\limits_0^{2-x}\int\limits_0^{2-x-y}1\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}dx-\int\limits_1^2\int\limits_1^{2-x}\int\limits_1^{2-x-y}1\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x=\frac{2}{3}$$ where the first integral is volume without $z=1$ and second integral is the part over $z=1$.

  2. I expect that $z = 0, x+y+z=2, x = 0, y = 0$ is half of cube $2\times2\times2$. Volume of that cube is $8$ so volume of half is $4$. The part over $z=1$ is half of cube $1\times1\times1$. Volume of this cube is 1 and volume of half is $\frac{1}{2}$. So volume I trying to find is $4-\frac{1}{2}=\frac{7}{2}$

At least one of (1), (2) must by wrong. What is right volume of area defined by $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ ?

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It seems to me the part with the halves of cubes is wrong. The objects are tetrahedra. –  mne__povezlo Mar 18 '13 at 13:55

3 Answers 3

up vote 5 down vote accepted

It is a bit hard to me if I want to tell you the proof, but I made a plot in which you can read the limits for $x,y,z$ and solve the triple integral. In fact, I rotated the coordinate axes such that the $y$ axes is perpendicular.

enter image description here

And so you just need to evaluate the following integral: $$\int_0^1\int_0^{2-z}\int_0^{2-x-z}dydxdz=7/6$$

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If you look at a cross-section of the volume in a plane $z=\text{constant}$, then the region is a right triangle of area

$$\frac{1}{2} (2-z)^2$$

The volume is then

$$\int_0^1 dz \: \frac{1}{2} (2-z)^2 = \left [ \frac{1}{6} (2-z)^3\right ]_1^0 = \frac{7}{6}$$

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both are wrong, if you subtract the part over z=1 you still have to subtract the part that is below z=1! (for x and y between 1 and 2). you should calculate the correct integral which is: $\int\limits_0^1\mathrm{d}x\int\limits_0^{1-x}\mathrm{d}y\int\limits_0^1\mathrm{d}z+\int\limits_0^2\mathrm{d}x\int\limits_{1-x}^{2-x}\mathrm{d}y\int\limits_0^{2-x-y}\mathrm{d}z\ $ (the first integral is the flat part at z=1 (x and y between 0 an 1, actually, between the origin an y=1-x), the second part is the planar surface that goes down from z=1 to z=0, for x and y between 1 and 2, actually between y=1-x and y=2-x)(see drawing by BabakS). The result should be the same as Ron Gordon's very elegant solution. Regarding the second method, the plane does not cut the cube in half (the shape of the top "half" has four triangles as sides, the bottom part has 3 triangles and one square.

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