Question

Let $a,c$be positive real numbers. Prove that the volume of the pyramid with vertices$(-a,a,0),(a,-a,0),(-a,-a,0),\text{and}(0,0,c)\text{is}\frac{4}{3}{a}^{2}c$.

Short answer

Use type I integral to prove the above result.

Step-by-step solution

Given Information

The vertices of pyramid are $(a,a,0),(-a,a,0),(-a,-a,0),(0,0,c)$

$a,c$are real positive numbers.

Using Concept of Type I Integral

The region of integration $\Omega$is bounded by $y=-a,y=a,x=a,x=-a$

For type I integral, $-a\le x\le a,-a\le y\le a$

Since the volume of pyramid with vertices $(0,0,0),(a,0,0),(0,b,0),(0,0,c)$is $\frac{1}{6}abc$

If pyramid has square base as $b\equiv 4a\text{and}a\equiv 2a$

Hence, volume is

$\frac{1}{6}\left(2a\right)\left(4a\right)c=\frac{4}{3}{a}^{2}c$