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450
❙❙❙❙
CHAPTER 6
APPLICATIONS OF INTEGRATION
SOLUTION
Figure 11 shows a horizontal crosssection. It is a washer with inner radius
and outer radius
, so the crosssectional area is
The volume is
We now ﬁnd the volumes of three solids that are
not
solids of revolution.
EXAMPLE 7
Figure 12 shows a solid with a circular base of radius 1. Parallel crosssections
perpendicular to the base are equilateral triangles. Find the volume of the solid.
Let’s take the circle to be
. The solid, its base, and a typical cross
section at a distance
from the origin are shown in Figure 13.
FIGURE 13
yy
60°
60°
B
A
C
œ„3y
(c) A crosssection
x
y
0
y
x
A
B(x, y)
y=œ„„„„„„
1≈
(b) Its base
(a) The solid
0
x
y
A
B
C
1
_1
x
x
2
±
y
2
±
1
FIGURE 11
x
0
x=œ„
y
x=y
x=_1
y
y
1
1+y
1+œ„
y
±
²
±
4
y
3
²
2
3
³
y
2
2
³
y
3
3
³
0
1
±
2
±
y
1
0
(2
s
y
³
y
³
y
2
)
dy
±
y
1
0
[(1
±
s
y
)
2
³
´
1
±
y
µ
2
]
dy
V
±
y
1
0
A
´
y
µ
dy
±
(1
±
s
y
)
2
³
´
1
±
y
µ
2
A
´
y
µ
±
´
outer radius
µ
2
³
´
inner radius
µ
2
1
±
s
y
1
±
y
FIGURE 12
Computergenerated picture
of the solid in Example 7
y
x
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lies on the circle, we have
and so the base of the triangle
is
. Since the triangle is equilateral, we see from Figure 13(c) that its
height is
. The crosssectional area is therefore
and the volume of the solid is
EXAMPLE 8
Find the volume of a pyramid whose base is a square with side
and whose
height is .
SOLUTION
We place the origin
at the vertex of the pyramid and the axis along its cen
tral axis as in Figure 14. Any plane
that passes through
and is perpendicular to the
axis intersects the pyramid in a square with side of length , say. We can express
in
terms of
by observing from the similar triangles in Figure 15 that
and so
. [Another method is to observe that the line
has slope
and
so its equation is
.] Thus, the crosssectional area is
The pyramid lies between
and
, so its volume is
NOTE
■
We didn’t need to place the vertex of the pyramid at the origin in Example 8. We
did so merely to make the equations simple. If, instead, we had placed the center of the
base at the origin and the vertex on the positive axis, as in Figure 16, you can verify that
y
±
L
2
h
2
x
3
3
±
0
h
±
L
2
h
3
V
±
y
h
0
A
²
x
³
dx
±
y
h
0
L
2
h
2
x
2
x
±
h
x
±
0
x
y
O
xh
FIGURE 14
s
x
h
L
x
y
O
P
FIGURE 15
A
²
x
³
±
s
2
±
L
2
h
2
x
2
y
±
Lx
´²
2
h
³
L
´²
2
h
³
OP
s
±
´
h
x
h
±
s
´
2
L
´
2
±
s
L
x
s
s
x
x
P
x
x
O
h
L
±
2
y
1
0
s
3
²
1
±
x
2
³
±
2
s
3
µ
x
±
x
3
3
±
0
1
±
4
s
3
3
V
±
y
1
±
1
A
²
x
³
±
y
1
±
1
s
3
²
1
±
x
2
³
A
²
x
³
±
1
2
±
2
s
1
±
x
2
±
s
3
s
1
±
x
2
±
s
3
²
1
±
x
2
³
s
3
y
±
s
3
s
1
±
x
2
¶
AB
¶
±
2
s
1
±
x
2
ABC
y
±
s
1
±
x
2
B
SECTION 6.2
VOLUMES
❙❙❙❙
451
x
y
h
0
y
FIGURE 16
Resources / Module 7
/ Volumes
/ Start of Mystery of the
Topless Pyramid
452
❙❙❙❙
CHAPTER 6
APPLICATIONS OF INTEGRATION
we would have obtained the integral
EXAMPLE 9
A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane
is perpendicular to the axis of the cylinder. The other intersects the ﬁrst at an angle of
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 Fall '07
 Temkin
 Calculus

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