From “Calculus, Single Variable” by Hughes-Hallett, Gleason, McCallum et. al.
Copyright 2005 by John Wiley & Sons, Inc.
This material is used by permission of John Wiley & Sons, Inc.
SUGGESTED PROBLEMS
In Exercises 1-10, the region is rotated around the x-axis. Find the volume.
3.
Bounded by y = 4 − x 2 , y = 0 , x = − 2 , x = 0 .
A parabola, upside down, shifted up by 4. We take only the part to the left of x = 0 and above the axes. In the second picture, we see that area spun around the x -axis.
x
If we take vertical slices of this object, we’ll get circles. The circle at point x will have radius r = 4 − x 2 . Thus the volume of the object can be approximated by
Riemann Sum =
X
( πr
2
)∆ x =
X
π (4 − x
2
)
2
∆ x
Converting that sum to integral format,
1

Exact Volume =
=
Z
0
−
2
Z
0
π (4 − x
2
)
2
π
−
2
(16 − 8 x 2
8 x 3
= π 16 x −
3
+ dx
+ x
5 x
5
4 ) dx
0
−
2
= π 0 − (16)( − 2) −
(8)( − 8)
3
=
256
π
15
+
− 32
5
5.
Bounded by y = e x
, y = 0 , x = − 1 , x = 1 .
−1 1 x
If we take vertical slices of this object, we’ll get circles. The circle at point x will have radius r = e x . Thus the volume of the object can be approximated by
Riemann sum =
X
πr
2
∆ x = π
X
( e x
)
2
∆ x = π
X e
2 x
∆ x
2

Using integration to compute the exact volume,
Exact Volume = π
Z
1
−
1 e 2 x e
2 x
1 dx
= π
2
−
1
=
π e
2
− e −
2
2
≈ 11 .
39 units cubed
10.
Bounded by y = e 3 x
, y = e x
, x = 0 , x = 1 .
20
10
0 e
(
3x
) e x
1 x
Cutting vertically through this object gives disc shapes. The cross-sectional area of a disk is
πr 2 out
− πr 2 in the area of the outer circle minus the area of the inner circle. See the example on page
376 for a nicer looking diagram.
3

The outer radius is given by r out
= e 3 x
, while the inner radius is r in
= e x
. This gives us
Riemann sum =
X
( πr 2 out
− πr 2 in
)∆ x = π
X
( e 3 x
) 2
− ( e x
) 2 ∆ x = π
X
( e 6 x
− e 2 x
)∆ x
The corresponding integral gives us the exact area, given that we’re adding up these slices from x = 0 to x = 1:
Exact Volume = π
Z
1
( e 6 x
0
− e 2 x
) dx
= π e 6 x
6
− e 2 x
2
1
0
= π ( e 6
6
− e 2
2
) − ( e 0
6
− e 0
)
2
= π e 6
6
− e 2
2
+
1
3
≈ 200 .
7 units cubed
35.
The design of boats is based on Archimedes Principle, which states that the buoyant force on an object in water is equal to the weight of the water displaced. Suppose you want to build a sailboat whose hull is parabolic with cross section y = ax 2 , where a is a constant.
Your boat will have length L and its maximum draft (the maximum vertical depth of any point of the boat beneath the water line) will be H . See Figure 8.28. Every cubic meter of water weighs 10,000 newtons. What is the maximum possible weight for your boat and cargo?
Figure 8.28
The first difficulty in this question is the wording. It sounds like a is something you want to optimize, but really it is a fixed constant, and you are just trying to find the volume of the boat.
With that in mind, computing the volume becomes another example of slicing up an object to find an appropriate volume integral. For this object, the easiest way to slice it would be parallel to the parabolic shape (perpendicular to L ), so that every slice is an identical parabola.
Volume ≈ P (Parabola Area) · ∆ x for x = 0 to x = L .
The area of the parabola is a 1D integration problem:
4

H
H a
0
H a x
The area of this parabola can be computed by taking vertical slices. Each slice will be
( H − ax 2 ) high: the difference between the horizontal line y = H and the parabola, y = ax 2 . The width of each slice will be ∆ x . This area is given by the integral below:
Parabola Area =
=
Z
√
H/a
−
√
H/a
( H − ax
2
) dx
Hx − ax
3
3
√
H/a
−
√
H/a
= H r
H a
− a
3
H a
3 / 2 !
− − H r
H a
− a
3
= 2
H 3 / 2
√ a
−
1
3
H 3 / 2
√ a
!
=
4
3
H 3 / 2
√ a
− H a
3 / 2 !
If this is the area of the cross-section of the boat for its whole length, then the total volume of the boat is (area) · (length):
Volume =
4
3
H
√
3 / 2 a
!
· L
If L and H are in meters,
Buoyancy force =
40 , 000 LH
3
√ a
3 / 2 newtons
5

This is the maximum weight supportable by the boat. Converting to a mass, the boat
1 can hold m = (Buoyancy force), where g is the acceleration due to gravity, 9.8
m/s 2 .
g
QUIZ PREPARATION QUESTIONS
34.
A 100 cm long gutter is made of three strips of metal, each 5 cm wide; Figure 8.27 shows a cross-section.
(a) Find the volume of water in the gutter when the depth is h cm.
(b) What is the maximum value of h ?
(c) What is the maximum volume of water that the gutter can hold?
(d) If the gutter is filled with half the maximum volume of water, is the depth larger or smaller than half of the answer to part (b)? Explain how you can answer without any calculation.
(e) Find the depth of the water when the gutter contains half the maximum possible volume.
Figure 8.27
(a) There are several approaches possible to solving this problem. We show two possibilities below.
Solution 1 - Using Integration
If we take horizontal slices of the gutter, of thickness ∆ h , we will get rectangular slabs. These slabs will be 100 cm long, and w ( h ) wide. To determine the width of a given slice, we look at the details of the trapezoid: h
3
5 w
5 h h
3
5
The volume of any slice will be w · 100 · ∆ h . From the diagram, w = 5 + 2 h/
√
3, so our Riemann sum for the volume is
6

Riemann sum =
X
100 5 + 2 h
√
3
∆ h .
Converting to an integral form for the exact computation, we want to use h as the upper limit of integration. This means we should use a different variably inside the integral, say y :
Volume up to h =
Z h
0
100 5 + 2 y
√
3
= 100 5 y + 2
2 y 2
√
3
= 100 5 h + h
√
2
3 h
0 dy
Solution 2 - Without Integration
If the gutter has the same trapezoidal profile for its whole 100 cm length, then its maximum volume will be (Area of trapezoid) × (100 cm).
You can use the 60 o angle to find the ratio of the sides in the triangles. See the diagram below.
5
5
5 3
2
5
2
5
When the height of the water in the gutter is at h , we have the following dimensions:
5 h h
3
2h
3
7

Cross-section area = Rectangle + 2Triangle
= w × h + 2
1
2 b × h
= 5 h +
= 5 h + h
√
3 h h 2
√
3
Total Volume = 100 cm × (Cross-section area)
= 100 5 h + h 2
√
3
5
√
3
(b) From the second diagram above, the maximum value of h is
2
≈ 4 .
33. At that height, the gutter is filled so that the side length of 5 is completely used.
mula for volume, and setting h to its largest possible value,
5
√
3
2
Max volume = 100 5 h +
√
3 h 2
√
3
= 100
25
+
2 for h
25
√
· 3
3
!
=
5
√
3
2
≈ 3247 .
6 cm 3
(d) If the gutter currently holds half of its maximum volume , then the depth will be above half the maximum depth. This is because the gutter starts narrow at the bottom. If one unit of volume were added, it would be spread around a smaller area, and so would be deeper. As the gutter fills, the cross-section gets larger, meaning that the same volume could be spread thinner across a wider area.
(e) We are given that V = equation
1
2 max V , or V ≈ 1623. We want to solve for height in the
8

1623 = 100 5 h + h 2
√
3
16 .
23 = 5 h + h 2
√
3 h 2
√
3
+ 5 h − 16 .
23 = 0
Quadratic formula: h =
− 5 ± q
25 − 4(
2 /
√
3
1
√
3
)(16 .
23)
≈ 2 .
52 , − 11 .
2 h = 2 .
52 makes sense in the problem, as it is a little over half the maximum depth of ≈ 4 .
3. Thus, with half of its maximum volume, the gutter will be filled to approximately 2.5 cm depth of water.
9