Q 69. - Calculus

Question

Let $a,b,c$ be positive real numbers. Prove that the volume of the pyramid with vertices $(0,0,0), (a,0,0), (0,b,0), (0,0,c)$ is $\frac{1}{6}abc.$

Step-by-Step Solution

Answer

It can be proved by solving ${\iint }_{\Omega }f(x,y)dydx$.

Step 1: Given Information

A pyramid has vertices $(0,0,0),(a, 0,0),(0, b, 0),(0,0, c)$

$a,b,c$ are real positive numbers.

We need to prove that volume is $\frac{1}{6}abc$

Step 2: Solving for type I integral

The equation of line lying in cartesian plane is

$y-0=\frac{b-0}{0-a}(x-a)$

$y=-\frac{b}{a}x+b$

Region of integration is bounded by lines $y=-\frac{b}{a}x+b, x=0, y=0$ 

For type I integral $0\le x\le a, 0\le y\le -\frac{b}{a}x+b$

Step 3: Evaluating Volume

Volume is given by

${\iint }_{\Omega }f(x,y)dydx={\int }_0^a{\int }_0^{-\frac{b}{a}x+b}f(x,y)dydx$

Function of pyramid by equation of plane is given by

$f(x,y)=z=c-\frac{c}{a}x-\frac{c}{b}y$

It gives

${\int }_0^a{\int }_0^{-\frac{b}{a}}f+b(x,y)dydx={\int }_0^a{\int }_0^{-\frac{b}{a}}\left(c-\frac{c}{a}x-\frac{c}{b}y\right)dydx$

Step 4: Simplification

Evaluation of double integral gives

${\int }_0^a{\int }_0^{-\frac{b}{a}}\left(c-\frac{c}{a}x-\frac{c}{b}y\right)dydx={\int }_0^a{\left[cy-\frac{c}{a}xy-\frac{c}{2b}{y}^2\right]}_0^{\frac{-b}{a}x+b}dx$

$={\int }_0^a\left[c\left(-\frac{b}{a}x+b\right)-\frac{c}{a}x\left(-\frac{b}{a}x+b\right)-\frac{c}{2b}{\left(-\frac{b}{a}x+b\right)}^2\right]dx$

$={\int }_0^a\left(-\frac{bc}{a}x+\frac{bc}{2a^2}{x}^2+\frac{1}{2}bc\right)dx$

It implies

${\int }_0^a{\int }_0^{-\frac{b}{a}}\left(c+\frac{c}{a}x-\frac{c}{b}y\right)dydx={\left[-\frac{bc}{2a}{x}^2+\frac{bc}{6a^2}{x}^3+\frac{1}{2}bcx\right]}_0^a$

$=-\frac{1}{2}abc+\frac{1}{6}abc+\frac{1}{2}abc$

$=\frac{1}{6}abc$

Hence volume is $\frac{1}{6}abc$