# Triple Integral of Square Based Pyramid given vertices

Use a triple integral to find the volume of the pyramid P whose base is the square with vertices (1,0,0), (0,1,0), (−1,0,0), and (0,−1,0) and whose top vertex is (0,0,1).

I've been given the problem above however I'm not sure how to go about answering it, I understand how to triple integrate given a function however I'm quite stumped how to answer this.

Any help would be appreciated!

• It is bad enough being required to use calculus to solve a simple geometry problem, but requiring a triple integral is ridiculous! Apr 11, 2016 at 12:21
• I know right! I get each individually however combining them has got me completely confused Apr 11, 2016 at 12:24
• This is an exercise in setting up the integral. A good choice of the outer variable of integration will make the two inner integrals easy to state. Note that limits of the inner integrals can depend on the variable used in the outer integral. Apr 11, 2016 at 12:26

The "function" which you are integrating is in this case plainly $$f(x,y,z) = 1$$

Important are the integration borders. First, formulate the borders for $z$ depending on $x$ and $y$. For the lower border you just get $0$, which is correspondent to the base area of the pyramid. The upper border, however, is a function of $x$ and $y$, which also depends on the quadrant you are looking at. For the quadrant where $x\geq 0, y\geq 0$ the upper integration border is $-x-y+1$, which is just inclinations of $x$ and $y$ ($-1$ each) and the $z$-axis intercept.

The other quadrants have upper borders deviating from this one in sign combination, but you can make use of the symmetry that the pyramid volume is exactly quartered by the $xy$-crossing. So you just set up the integral for one quadrant and multiply by $4$ after integration.

The $y$ lower border is again $0$, the upper border is a function of $x$ namely $1-x$. The borders for $x$ are plainly $0$ and $1$.

So we have the borders and are ready to set up our integral.

$$V=4\cdot\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{-x-y+1}dz\,dy\,dx$$

First integration over $z$ yields

$$V=4\int_{0}^{1}\int_{0}^{1-x}\left[z\right]_{0}^{-x-y+1}dy\,dx =4\int_{0}^{1}\int_{0}^{1-x}(-x-y+1)dy\,dx$$

And so on... You will find $V=2/3$