math 211, calculus ii - millersville university of...
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Volume: Slicing, Disks, and WashersMATH 211, Calculus II
J. Robert Buchanan
Department of Mathematics
Spring 2018
Background
Today’s discussion will center on using the definite integral tofind the volume of a three-dimensional object.
Recall: for a rectangular box
V = l w h = (length× width)︸ ︷︷ ︸cross-sectional area
×height
and for a cylinder
V = πr2h = π(radius)2︸ ︷︷ ︸cross-sectional area
×height
General IdeaSuppose a three-dimensional object lies along the x-axiscovering the interval [a,b] and the cross-sectional area is acontinuous function of x , call it A(x).
Riemann Sum Approach
Let ∆x =b − a
nfor n ∈ N and define xi = a + i∆x for
i = 1,2, . . . ,n.
V ≈n∑
i=1
A(xi)∆x
V = limn→∞
n∑i=1
A(xi)∆x
=
∫ b
aA(x) dx
ExampleFind the volume of a right circular cone whose base radius is 3and whose height is 7.
Solution
The cross sectional area of the cone at 0 ≤ x ≤ 7 isA(x) = π(3x/7)2.
V =
∫ 7
0π(3x/7)2 dx
=9π49
∫ 7
0x2 dx
=
[3π49
x3]x=7
x=0= 21π.
ExampleA solid object has as its base the circular region defined byx2 + y2 = 1. Every cross section of the object perpendicular tothe x-axis is a triangle whose base vertices are on the circleand whose height equals the length of the base. Find thevolume of this object.
Solution
The cross sectional area of the solid at −1 ≤ x ≤ 1 is
A(x) =12
(2√
1− x2)2.
V =
∫ 1
−1
12
(2√
1− x2)2 dx
= 2∫ 1
−1(1− x2) dx
= 4∫ 1
0(1− x2) dx
=
[4(
x − 13
x3)]x=1
x=0
=83.
ExampleA solid object has as its base the ellipse defined byx2
25+
y2
36= 1. Every cross section of the object perpendicular
to the x-axis is a square whose base vertices are on the ellipse.Find the volume of this object.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.
Method of Disks
Suppose f (x) ≥ 0 and continuous on [a,b] and the regionbounded beneath the graph of y = f (x), the x-axis, and thelines x = a and x = b is revolved around the x-axis.
Every cross section is a circular disk of radius r = f (x), thus thevolume of the solid of revolution is
V =
∫ b
aπ[f (x)]2︸ ︷︷ ︸
cross-sectional area
dx .
ExampleFind the volume of the solid of revolution generated wheny = 2− 1
2x is revolved around the x-axis while 0 ≤ x ≤ 4.
Method of Disks
Suppose f (x) ≥ 0 and continuous on [a,b] and the regionbounded beneath the graph of y = f (x), the x-axis, and thelines x = a and x = b is revolved around the x-axis.
Every cross section is a circular disk of radius r = f (x), thus thevolume of the solid of revolution is
V =
∫ b
aπ[f (x)]2︸ ︷︷ ︸
cross-sectional area
dx .
ExampleFind the volume of the solid of revolution generated wheny = 2− 1
2x is revolved around the x-axis while 0 ≤ x ≤ 4.
Solution
V = π
∫ 4
0
(2− 1
2x)2
dx
= π
∫ 4
0
(4− 2x +
14
x2)
dx
=
[π
(4x − x2 +
112
x3)]x=4
x=0
=16π
3
y as Independent VariableIf x = g(y) and g(y) ≥ 0 for c ≤ y ≤ d then the volume of thesolid of revolution generated by revolving the region boundedbetween the graph of x = g(y), the y -axis, and the lines y = cand y = d is given by
V =
∫ d
cπ[g(y)]2︸ ︷︷ ︸
cross-sectional area
dy .
x
c
d
y
x=g(y)
ExampleFind the volume of the solid of revolution generated wheny =√
x for 1 ≤ y ≤ 4 is revolved around the y -axis.
Solution
Since y =√
x then x = y2 and the volume is calculated asfollows.
V = π
∫ 4
1(y2)2 dy
= π
∫ 4
1y4 dy
=[π
5y5]y=4
y=1
=π(4)5
5− π(1)5
5
=1024π
5− π
5=
1023π5
Method of WashersSometimes a solid region has a “hole” inside it.
x
y
Question: how can we find the volume of the solid of revolutionwith a cavity?
Method of Washers Formula
If f (x) and g(x) are continuous on [a,b] and 0 ≤ f (x) ≤ g(x) on[a,b] the volume of the solid of revolution generated byrevolving the region bounded between the graphs of y = f (x),y = g(x), and the lines x = a and x = b is given by
V =
∫ b
aπ( [g(x)]2 − [f (x)]2︸ ︷︷ ︸
(outer radius)2−(inner radius)2
) dx
Example
Find the volume of the solid of revolution generated byrevolving the region bounded by y = 2− x2/2, y = 1− x2/4,x = −1, and x = 1 around the x-axis.
SolutionSince 2− x2/2 ≥ 1− x2/4 for −1 ≤ x ≤ 1 then
V = π
∫ 1
−1
[(2− x2
2
)2
−(
1− x2
4
)2]dx
= π
∫ 1
−1
[(4− 2x2 +
x4
4
)−(
1− x2
2+
x4
16
)]dx
= π
∫ 1
−1
[3− 3
2x2 +
316
x4]
dx
= 2π∫ 1
0
[3− 3
2x2 +
316
x4]
dx
=
[2π(
3x − 12
x3 +380
x5)]x=1
x=0
V = 2π(
3− 12
+3
80
)=
203π40
.
Example
Find the volume of the solid of revolution generated when theregion bounded by the curves y =
√x and y = x2 is revolved
around the y -axis.
x
y
Solution
Since y =√
x ⇐⇒ y2 = x and y = x2 ⇐⇒ √y = x and√
y ≥ y2 for 0 ≤ y ≤ 1 then
V = π
∫ 1
0
[(√
y)2 − (y2)2]
dy
= π
∫ 1
0
[y − y4
]dy
=
[π
(y2
2− y5
5
)]y=1
y=0
V = π
(12− 1
5
)=
3π10.
Revolving Around Different LinesWe can modify our previous formulas to handle the cases inwhich we revolve a region around horizontal or vertical line notintersecting the region.
The key is to think about the radius of revolution in each case.
x
y
axis of revolution
inner radius
outer radius
Example
Find the volume of the solid of revolution generated when theregion bounded between y =
√x , y = 0 and x = 4 is revolved
around the line y = 2.
Solution
The outer radius is ro = 2− 0 = 2 while the inner radius isri = 2−
√x .
V = π
∫ 4
0
[(2)2 − (2−
√x)2]
dx
= π
∫ 4
0
[4− (4− 4
√x + x)
]dx
= π
∫ 4
0
[4x1/2 − x
]dx
=
[π
(83
x3/2 − 12
x2)]x=4
x=0
V = π
(643− 8)
=40π
3.
Example
Find the volume of the solid of revolution generated when theregion bounded between y =
√x and y = x2 is revolved
around the line x = 2.
0.5 1.0 1.5 2.0 2.5x
0.2
0.4
0.6
0.8
1.0
y
axis
Solution
The curve furthest from the axis is y =√
x ⇐⇒ y2 = x . Thecurve closest is y = x2 ⇐⇒ √
y = x .
V = π
∫ 1
0
[(2− y2)2 − (2−
√y)2]
dy
= π
∫ 1
0
[(4− 4y2 + y4)− (4− 4
√y + y)
]dy
= π
∫ 1
0
[4y1/2 − y − 4y2 + y4
]dy
=
[π
(83
y3/2 − 12
y2 − 43
y3 +15
y5)]y=1
y=0
V = π
(83− 1
2− 4
3+
15
)=
31π30
.
Homework
I Read Section 6.2I Exercises: see D2L/WebAssign