Volumes 1 Computing Volumes by Adding Cross Sections Sample Problems Lecture Notes

Volumes 1
Lecture Notes
page 1
Computing Volumes by Adding Cross Sections
Sample Problems
1. Find the volume of the pyramid with a square base if its base has sides 10 meters long and its height is 12
meters.
2. A building’s base is a rectangle and a triangle as shown on the picture below. At each point, the vertical
cross section of the building is an equilateral triangle. Compute the volume of the building.
3. Find the volume of the solid whose base is a circle with radius 1 and each cross-section perpendicular to the
base is a square.
4. A wedge is cut out of a circular cylinder of radius 5 by two planes. One plane is perpendicular to the axis of
the cylinder. The other intersects the …rst at an angle of 30 along a diameter of the cylinder. Compute the
volume of the wedge.
c Hidegkuti, Powell, 2012
Last revised: October 11, 2012
Lecture Notes
Volumes 1
page 2
Practice Problems
1. a) Find the volume of the pyramid with a square base if its base has sides 20 meters long and its height is 15
meters.
b) We want to cut the pyramid by a plane parallel to its base. At what height should we cut if we wanted
to cut the pyramid into two parts of equal volume?
2. a) Use integration to compute the volume of a pyramid with a square base of sides L long and has a height
of H.
b) We want to cut the pyramid by a plane parallel to its base. At what height should we cut if we wanted
to cut the pyramid into two parts of equal volume?
3. Compute the volume of the building if its base is shown on the picture below and
a) cross-sections perpendicular to the base are squares.
b) cross-sections perpendicular to the base are equilateral triangles.
c) cross-sections perpendicular to the base are semi-circles.
d) cross-sections perpendicular to the base are isosceles right triangles where the hypotenuse lies on the base.
4. Consider a solid with a circular base of radius 1.
triangle. Compute the volume of the solid.
Cross-sections perpendicular to the base are equilateral
5. 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 …rst at an angle of 60 along a diameter of the cylinder. Compute the
volume of the wedge.
6. Use integration to compute the volume of the frustum of a pyramid shown on the picture below.
c Hidegkuti, Powell, 2012
Last revised: October 11, 2012
Volumes 1
Lecture Notes
page 3
Sample Problems - Answers
p
14 3
2.)
3
1.) 400
p
250 3
4.)
9
16
3.)
3
Practice Problems - Answers
1.) a)
2000 m3
2.) a)
1
HL2
3
b)
p
3.) a) 720 + 36 3
p
4 3
4.)
3
b) 15
1
r
3375
3: 0945 m from the ground
2
r !
3 1
H from the ground
2
3
p
b) 27 + 180 3
p
128 3
5.)
3
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c)
90 +
9p
3
2
p
d) 180 + 9 3
1
6.) V = h a2 + ab + b2
3
Last revised: October 11, 2012
Volumes 1
Lecture Notes
page 4
Sample Problems - Solutions
1. Find the volume of the pyramid with a square base if its base has sides 10 meters long and its height is 12
meters.
Solution:
Consider the picture above. We will slice the pyramid into thin slices perpendicular to the x axis. Each
slice has volume A (x) dx where A (x) is the area of the cross section. We will add the volume of these slices
ranging from x = 0 to x = 12. The volume is
V =
Z12
A (x) dx
0
We now need to …gure out A (x). The area of a square is s2 where s is the length of its side.
Consider the similar triangles P QR and P T U . Their heights are x and 12, correspondingly. Their bases are
s and 10; correspondingly. The ratio between these four quantites will give us s as a function of x.
s
10
=
x
12
=)
5
s= x
6
and so
Z12
25 x3
36 3
A (x) =
5
x
6
2
=
25 2
x
36
Thus
V =
Z12
A (x) dx =
0
c Hidegkuti, Powell, 2012
Z12
0
25
25 2
x dx =
36
36
0
x2 dx =
12
=
0
25
36
123
3
03
3
=
25
(576) = 400
36
Last revised: October 11, 2012
Volumes 1
Lecture Notes
page 5
2. A building’s base is a rectangle and a triangle as shown on the picture below. At each point, the vertical
cross section of the building is an equilateral triangle. Compute the volume of the building.
p !
p
1
3
3 2
Solution: If an equilateral triangle’s side is s, then its area is A = s
s =
s . Between 0 and 2;
2
2
4
the side of the triangle at x is x. Between 2 and 6, each triangle has sides 2.
V
=
Z6
A (x) dx =
0
=
Z6 p
3
[s (x)]2 dx =
4
0
p
2
3 x3
4 3
+
p
6
3x
=
0
2
Z2 p
3 2
x dx +
4
0
p
3 3
2
12
03 +
p
Z6 p
6
p Z2
p Z
3 2
3
2
2 dx =
x dx + 3 1 dx
4
4
2
3 (6
2) =
p
3
0
8
+4
12
=
2
p
14 3
3
3. Find the volume of the solid whose base is a circle with radius 1 and each cross-section perpendicular to the
base is a square.
2
2
Solution: Let us assume that the circle is centered at the origin;
p that is, its equation is x + y = 1. For each
value of x; the cross section is a square with sides s = 2y = 2 1 x2 . Tus the area of each cross section is
A (x) = s2 = 2
V =
Z1
1
A (x) dx =
Z1
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1
4 1
2
x
dx = 4
Z1
1
1
2
x
p
1
x2
dx = 8
Z1
0
2
1
=4 1
x2
x2 dx = 8 x
x3
3
1
=8 1
0
13
3
=
16
3
Last revised: October 11, 2012
Volumes 1
Lecture Notes
page 6
4. A wedge is cut out of a circular cylinder of radius 5 by
two planes. One plane is perpendicular to the axis of
the cylinder. The other intersects the …rst at an angle
of 30 along a diameter of the cylinder. Compute the
volume of the wedge.
Solution: We place the coordinate system so that the x axis is the intersection of the two planes and the
y axis also lies in the base of the wedge.
We slice the wedge into very thin slices along the x axis. The slices range from x = 5 to x = 5 but we
already see that the wedge is symmetrical to the y axis, so we can just add the volume of the slices from
x = 0 to x = 5 and double our result.
Each cross section has thickness dx and is in the shape of a right triangle with p
the right angle lying on the
circle. The angle lying on the x axis is 30 . The horizontal side is y where y = 25 x2 . The vertical side,
denoted by v is
p
p
v
3
3p
tan 30 =
=) v = y tan 30 =
y=
25 x2
y
3
3
The volume of a triangular slice is
0
10
p
1
1 @p
3p
B
dV = Adx = yvdx =
25 x2 A @
25
2
2 | {z }
| 3 {z
y
v
The total volume is then
V
=
=
=
Z
p
dV =
Z5 p
3
25
6
x2 dx = 2
5
3
53
25 5
25 0
3
3
p
p
2
3
250 3
125
=
3
3
9
Z5 p
0
03
3
1
p
3
C
25
x2 A dx =
6
}
x2 dx
p Z5
p
3
3
3
2
2
25 x dx = 2
25 x dx =
6
6
3
0
p
p
1
3
125
3
=
125
=
125 1
3
3
3
3
25x
x3
3
5
0
!
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Last revised: October 11, 2012