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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?

Sagot :

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Answer:

[tex]\mathbf{ \int^2_0 \dfrac{x^2}{4}. dx = \dfrac{2}{3}}[/tex]

Step-by-step explanation:

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

[tex]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[/tex]

Given that:

height h = 2; &

side (l) = 1

Then;

[tex]V = \int^2_0 \dfrac{(1)^2 *x^2}{(2)^2}. dx[/tex]

[tex]V = \int^2_0 \dfrac{x^2}{4}. dx[/tex]

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

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

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

[tex]\mathbf{V = \dfrac{2}{3}}[/tex]

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