# triple integral square pyramid

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?

• 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. Jun 18 '16 at 15:19
• @GCab i meant to say that i considered and sketched the part of the pyramid in x>0, y>0, z>0 Jun 18 '16 at 15:31
• yes, but there is a hedge along $x=y$, did you consider that ? can you show your integral formula ? Jun 18 '16 at 16:26
• Why do people set these wretched questions to which the answer is obvious without any integration? Jun 18 '16 at 17:15
• @almagest practicing multiple integration... Jun 18 '16 at 19:25

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 $1-z$.