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?

Question

asked Jun 16, 2022 219k views

1 Answer

Matt Pileggi · Answer 1 · 2022-06-20T16:13:31+0000

Answer:

$\mathbf{ \int^2_0 (x^2)/(4). dx = (2)/(3)}$

Explanation:

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

$V = \int^h_o (L^2 *y^2)/(h^2). dy \ or \ V = \int^h_o (L^2 *x^2)/(h^2). dx$

Given that:

height h = 2; &

side (l) = 1

Then;

$V = \int^2_0 ((1)^2 *x^2)/((2)^2). dx$

$V = \int^2_0 (x^2)/(4). dx$

$V =(1)/(4) \Big [ (x^3)/(3) \Big ] ^2_0$

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

$V =(1)/(4) \Big [ (8)/(3) -0\Big ]$

$\mathbf{V = (2)/(3)}$