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

`\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)}`