EXAMPLE 8 Find the volume of a pyramid whose base is a square with side L and whose height is h. SOLUTION We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in the figure. Any plane P, that passes through x and is perpendicular to the x-axis intersects the pyramid in square with side of length s. We can express s in terms of x by using similar triangles. s/2 and so s = Video Example ) Tutorial [Another method is to observe that the line OP has slope L/(2h) and so its equation is y = Online Textbook Lx/(2h).] Thus the cross-sectional area is A(x) = s' = The pyramid lies between x = 0 and x = , so its volume is L? A (2) dr = V = %3= h?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 68E
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EXAMPLE 8
Find the volume of a pyramid whose base is a square with side L and whose height is h.
SOLUTION
We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in the
figure. Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square
h
with side of length s. We can express s in terms of x by using similar triangles.
s/2
h
L/2
L
and so s =
Video Example
Tutorial
. [Another method is to observe that the line OP has slope L/(2h) and so its equation is y =
Online Textbook
Lx/(2h).] Thus the cross-sectional area is
A(x) = s² :
The pyramid lies between x = 0 and x =
so its volume is
h
ch
L2
V:
A (x) dr
16
=
h2
Transcribed Image Text:EXAMPLE 8 Find the volume of a pyramid whose base is a square with side L and whose height is h. SOLUTION We place the origin O at the vertex of the pyramid and the x-axis along its central axis as in the figure. Any plane Px that passes through x and is perpendicular to the x-axis intersects the pyramid in a square h with side of length s. We can express s in terms of x by using similar triangles. s/2 h L/2 L and so s = Video Example Tutorial . [Another method is to observe that the line OP has slope L/(2h) and so its equation is y = Online Textbook Lx/(2h).] Thus the cross-sectional area is A(x) = s² : The pyramid lies between x = 0 and x = so its volume is h ch L2 V: A (x) dr 16 = h2
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