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 a square with side of length s. We can express s in terms of x by using similar triangles. s/2 L/2 ar %3D 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 V = A (x) dr = h

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 68E
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y
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 Py 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
L/2
h
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)
= s :
The pyramid lies between x = 0 and x =
so its volume is
h
L2
| A (x) dr
V:
h?
h2
II
Transcribed Image Text:y 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 Py 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 L/2 h 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) = s : The pyramid lies between x = 0 and x = so its volume is h L2 | A (x) dr V: h? h2 II
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