Q 70. - Calculus

Question

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

Step-by-Step Solution

Answer

Use type I integral to prove the above result.

Step 1: 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.

Step 2: 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

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