Question

Set up and evaluate an integral for the volume of a right pyramid whose height is 2 and whose base is a square with a side of 1?

Answer 1

Answer:

$\mathbf{ \int^2_0 \dfrac{x^2}{4}. dx = \dfrac{2}{3}}$

Step-by-step explanation:

The volume of right pyramid whose base is a square is expressed by the formula:

$V = \int^h_o \dfrac{L^2 *y^2}{h^2}. dy \ or \ V = \int^h_o \dfrac{L^2 *x^2}{h^2}. dx$

Given that:

height h = 2; &

side (l) = 1

Then;

$V = \int^2_0 \dfrac{(1)^2 *x^2}{(2)^2}. dx$

$V = \int^2_0 \dfrac{x^2}{4}. dx$

$V =\dfrac{1}{4} \Big [ \dfrac{x^3}{3} \Big ] ^2_0$

$V =\dfrac{1}{4} \Big [ \dfrac{(2)^3}{3} -0\Big ]$

$V =\dfrac{1}{4} \Big [ \dfrac{8}{3} -0\Big ]$

$\mathbf{V = \dfrac{2}{3}}$