The tetrahedron having its four vertices at the origin, and the points on the three
axes where respectively x = 1, y = 2, and z = 2. Use the order dz dy dx.
Please explain how you did it. I'm trying to understand the process by which one goes about doing this for most problems.
First make sure you are visualising the tetrahedron against three co-ordinate axes.
Then, how would you use calculus to find the area of the xy face? Probably, thinking, integrating stuff collected as you travel along the x axis from 0 to 1...
$\displaystyle \int_0^1\ dx$
... but integrating what stuff? Every vertical strip going from wherever you are on the x axis at any moment up to the line y = 2 - 2x...
$\displaystyle \int_0^1\ 2 - 2x\ dx$
But you could think of this as taking, at every x value, a detour from the x axis in the y direction, and integrating all the ones collected at every point on this y-direction detour...
$\displaystyle \int_0^1\ \int_0^{2 - 2x}\ 1\ dy\ dx$
Notice this is equivalent to the previous formula.
Now, the volume. For every point on the y-direction detour, you want to take a z-direction detour from z = 0 up to the plane that goes through the lines z = 2 - 2x in the xz plane and z = 2 - y in the yz plane. Which is the plane z = 2 - y - 2x.
$\displaystyle \int_0^1\ \int_0^{2 - 2x}\ \int_0^{2 - y - 2x}\ 1\ dz\ dy\ dx$
Notice you could now take in another detour (e.g. in the mass direction).
Then you begin to unwrap, by taking the equivalent form,
$\displaystyle \int_0^1\ \int_0^{2 - 2x}\ 2 - y - 2x\ dy\ dx$
(because 2 - y - 2x is the definite integral of 1 with respect to z going from limits zero up to 2 - y - 2x.)
And so on.
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Originally Posted by kkar 
The volume.
