The pyramid has the base on the xy plane; vertices $(\pm 1,0,0), (0,\pm 1,0),(0,0,1)$

So basically, with my integration limits I thought I was calculating the volume of $\frac14$ of the pyramid when I actually calculated $\frac12$, I need help figuring out why.

So I took the xy positive quadrant, sketched it and integrated with limits: 0 to (-z+1) dx then 0 to (-z+1) dy, then 0 to 1 dz

Why does this represent a half of the pyramid?

  • $\begingroup$ Seems to understand ($0$ to $1$ dz) that you considered just one face, while, on the first quadrant, the pyramid has two faces. You shall work on the first octant. $\endgroup$
    – G Cab
    Jun 18 '16 at 15:19
  • $\begingroup$ @GCab i meant to say that i considered and sketched the part of the pyramid in x>0, y>0, z>0 $\endgroup$
    – user348617
    Jun 18 '16 at 15:31
  • $\begingroup$ yes, but there is a hedge along $x=y$, did you consider that ? can you show your integral formula ? $\endgroup$
    – G Cab
    Jun 18 '16 at 16:26
  • $\begingroup$ Why do people set these wretched questions to which the answer is obvious without any integration? $\endgroup$
    – almagest
    Jun 18 '16 at 17:15
  • $\begingroup$ @almagest practicing multiple integration... $\endgroup$
    – user348617
    Jun 18 '16 at 19:25

Sorry, I realize now that the base of your pyramid is rhombic, not square (vertices at ..): so my previous comment does not apply.
Well, being the base rhombic, its section at $z$=const. is also rhombic, that is, in the $1$st quadrant, a triangle with sides $(1-z)$ and diagonal $x+y=1-z$. So its area is half than if you integrate $x$ and $y$ from $0$ to $1-z$.


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