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Volumes: Slabs, Washers, ShellsCalculus II
Josh Engwer
TTU
27 January 2014
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 1 / 64
Volumes (PART I)
VOLUMES PART I:
SOLIDS WITH KNOWN CROSS SECTIONS
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 2 / 64
Solids with Known Cross Sections (Slabs)
(Thickness of kth H-Slab
)=(
Width of kth generating H-Rect)
= ∆yk(Thickness of kth V-Slab
)=(
Width of kth generating V-Rect)
= ∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 3 / 64
Solids with Known Cross Sections (Demo)(DEMO) SOLIDS WITH KNOWN CROSS SECTION (Click below):
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 4 / 64
Volumes of Slabs
Proposition
Volume of Slab =(
Area of Cross Section)×(
Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 5 / 64
Volume of Solid with Known Cross Section (CS)
EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Let solid S have base R and equilateral triangular cross sections ⊥ to x-axis.
Setup integral(s) to compute Volume(S).
⊥ ≡ ”perpendicular”
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 6 / 64
Volume of Solid with Known Cross Section (CS)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R (label BP’s & BC’s in terms of x)Notice region R is V-simple.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 7 / 64
Volume of Solid with Known Cross Section (CS)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R. (remove unnecessary clutter)Notice region R is V-simple.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 8 / 64
Volume of Solid with Known Cross Section (CS)
Key Element: V-Slab (CS = Equilateral Triangle) generated by V-Rect
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 9 / 64
Volume of Solid with Known Cross Section (CS)
kth V-Slab on R:Thickness = (Width of generating V-Rect)Side Length = (Height of generating V-Rect)Volume =
√3
4 × (Side Length)2 × (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 10 / 64
Volume of Solid with Known Cross Section (CS)
kth V-Slab on R:Thickness = ∆xk
Side Length = 2x∗k − 4−x∗k
Volume =√
34
[2x∗k − 4−x∗k
]2∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 11 / 64
Volume of Solid with Known Cross Section (CS)
kth V-Slab on R:Thickness = ∆xk
Side Length = 2x∗k − 4−x∗k
Volume =√
34
[2x∗k − 4−x∗k
]2∆xk
Riemann Sum: Volume(S) ≈ V∗N =N∑
k=1
√3
4
[2x∗k − 4−x∗k
]2∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 12 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S) ≈ V∗N =
N∑k=1
√3
4
[2x∗k − 4−x∗k
]2∆xk
Integral: Volume(S) = limN→∞
V∗N =
∫ largest x-coord. in R
smallest x-coord. in R
√3
4(2x − 4−x)2
dx
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 13 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S) ≈ V∗N =
N∑k=1
√3
4
[2x∗k − 4−x∗k
]2∆xk
Integral: Volume(S) =
∫ 2
0
√3
4(2x − 4−x)2
dx ≈ 1.7954
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 14 / 64
Volume of Solid with Known Cross Section (CS)
EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Let solid S have base R and equilateral triangular cross sections ⊥ to y-axis.
Setup integral(s) to compute Volume(S).
⊥ ≡ ”perpendicular”
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 15 / 64
Volume of Solid with Known Cross Section (CS)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R (label BP’s & BC’s in terms of y)Notice region R is NOT H-simple, so subdivide into subregions R1,R2.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 16 / 64
Volume of Solid with Known Cross Section (CS)
Key Element: H-Slab (CS = Equilateral Triangle) generated by H-Rect
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 17 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R1:Thickness = (Width of generating H-Rect)Side Length = (Length of generating H-Rect)Volume =
√3
4 × (Side Length)2 × (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 18 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R1:Thickness = ∆yk
Side Length = 2− (− log4 y∗k )
Volume =√
34 (2 + log4 y∗k )
2∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 19 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R1:Thickness = ∆yk
Side Length = 2− (− log4 y∗k )
Volume =√
34 (2 + log4 y∗k )
2∆yk
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
√3
4(2 + log4 y∗k )
2∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 20 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
√3
4(2 + log4 y∗k )
2∆yk
Integral: Volume(S1) = limN→∞
V∗N =
∫ largest y-coord. in R1
smallest y-coord. in R1
√3
4(2 + log4 y)
2 dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 21 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
√3
4(2 + log4 y∗k )
2∆yk
Integral: Volume(S1) =
∫ 1
1/16
√3
4(2 + log4 y)
2 dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 22 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R2:Thickness = (Width of generating H-Rect)Side Length = (Length of generating H-Rect)Volume =
√3
4 × (Side Length)2 × (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 23 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R2:Thickness = ∆yk
Side Length = 2− log2 y∗kVolume =
√3
4 (2− log2 y∗k )2
∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 24 / 64
Volume of Solid with Known Cross Section (CS)
kth H-Slab on R2:Thickness = ∆yk
Side Length = 2− log2 y∗kVolume =
√3
4 (2− log2 y∗k )2
∆yk
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
√3
4(2− log2 y∗k )
2∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 25 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
√3
4(2− log2 y∗k )
2∆yk
Integral: Volume(S2) = limN→∞
V∗N =
∫ largest y-coord. in R2
smallest y-coord. in R2
√3
4(2− log2 y)
2 dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 26 / 64
Volume of Solid with Known Cross Section (CS)
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
√3
4(2− log2 y∗k )
2∆yk
Integral: Volume(S2) =
∫ 4
1
√3
4(2− log2 y)
2 dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 27 / 64
Volume of Solid with Known Cross Section (CS)
Region R = R1 ∪ R2 =⇒ Solid S = S1 ∪ S2
∴ Volume(S) = Volume(S1) + Volume(S2)
=∫ 1
1/16
√3
4(2 + log4 y)
2 dy +
∫ 4
1
√3
4(2− log2 y)
2 dy ≈ 2.0818
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 28 / 64
Volumes (PART II)
VOLUMES PART II:
SOLIDS OF REVOLUTION
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 29 / 64
Solids of Revolution
Axis of Revolution: x-axis
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 30 / 64
Solids of Revolution
Axis of Revolution: y-axis
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 31 / 64
Solids of Revolution (V-Washers)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 32 / 64
Solids of Revolution (H-Shells)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 33 / 64
Solids of Revolution (V-Shells)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 34 / 64
Solids of Revolution (H-Washers)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 35 / 64
Volume of Washers
Proposition
Volume of Washer = π×[(
Outer Radius)2−(
Inner Radius)2]×(
Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 36 / 64
Volume of V-Shells
Proposition
Volume of kth V-Shell = 2π ×(
Radius)×(
Length)×(
Thickness)
Radius involves the tag y∗k Thickness = ∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 37 / 64
Volume of H-Shells
Proposition
Volume of kth H-Shell = 2π ×(
Radius)×(
Height)×(
Thickness)
Radius involves the tag x∗k Thickness = ∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 38 / 64
Volume of Solid of Revolution (using Washers)
EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Let S be the solid formed by revolving region R about x-axis using washers.
Setup integral(s) to compute Volume(S).
⊥ ≡ ”perpendicular”
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 39 / 64
Volume of Solid of Revolution (using Washers)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R (label BP’s & BC’s in terms of x)Notice region R is V-simple.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 40 / 64
Volume of Solid of Revolution (using Washers)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R. (remove unnecessary clutter)Sketch & label the axis of revolution (dashed line).
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 41 / 64
Volume of Solid of Revolution (using Washers)
Key Element: V-Washer (generated by V-Rect)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 42 / 64
Volume of Solid of Revolution (using Washers)
kth V-Washer on R:Thickness = (Width of generating V-Rect)Outer Radius = (Distance from farther BC to Axis of Revolution)Inner Radius = (Distance from closer BC to Axis of Revolution)
Volume = π ×[(Outer Radius)
2 − (Inner Radius)2]× (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 43 / 64
Volume of Solid of Revolution (using Washers)
kth V-Washer on R:
Thickness = ∆xk
Outer Radius = 2x∗k − 0Inner Radius = 4−x∗k − 0
Volume = π[(
2x∗k)2 −
(4−x∗k
)2]
∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 44 / 64
Volume of Solid of Revolution (using Washers)
kth V-Washer on R:
Thickness = ∆xk
Outer Radius = 2x∗k − 0Inner Radius = 4−x∗k − 0Volume = π
[22x∗k − 4−2x∗k
]∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 45 / 64
Volume of Solid of Revolution (using Washers)
kth V-Washer on R:
Thickness = ∆xk
Outer Radius = 2x∗k − 0Inner Radius = 4−x∗k − 0Volume = π
[4x∗k − 16−x∗k
]∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 46 / 64
Volume of Solid of Revolution (using Washers)
kth V-Washer on R:
Thickness = ∆xk
Outer Radius = 2x∗k − 0Inner Radius = 4−x∗k − 0Volume = π
[4x∗k − 16−x∗k
]∆xk
Riemann Sum: Volume(S) ≈ V∗N =
N∑k=1
π[4x∗k − 16−x∗k
]∆xk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 47 / 64
Volume of Solid of Revolution (using Washers)
Riemann Sum: Volume(S) ≈ V∗N =
N∑k=1
π[4x∗k − 16−x∗k
]∆xk
Integral: Volume(S) = limN→∞
V∗N =
∫ largest x-coord. in R
smallest x-coord. in Rπ(4x − 16−x) dx
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 48 / 64
Volume of Solid of Revolution (using Washers)
Riemann Sum: Volume(S) ≈ V∗N =
N∑k=1
π[4x∗k − 16−x∗k
]∆xk
Integral: Volume(S) =
∫ 2
0π(4x − 16−x) dx ≈ 32.864
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 49 / 64
Volume of Solid of Revolution (using Shells)
EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Let S be the solid formed by revolving region R about line y = 5 using shells.
Setup integral(s) to compute Volume(S).
⊥ ≡ ”perpendicular”
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 50 / 64
Volume of Solid of Revolution (using Shells)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Sketch & characterize region R (label BP’s & BC’s in terms of y)Notice region R is NOT H-simple, so subdivide into subregions R1,R2.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 51 / 64
Volume of Solid of Revolution (using Shells)EXAMPLE: Let region R be bounded by curves y = 2x, y = 4−x, x = 2.
Key Element: V-Shell (generated by H-Rect)Sketch & label the axis of revolution y = 5.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 52 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R1:Thickness = (Width of generating H-Rect)Radius = (Positive Distance from H-Rect to Axis of Revolution)Length = (Length of generating H-Rect)Volume = 2π × (Radius)× (Length)× (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 53 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R1:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− (− log4 y∗k )Volume = 2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 54 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R1:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− (− log4 y∗k )Volume = 2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 55 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R1:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− (− log4 y∗k )Volume = 2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Integral: Volume(S1) = limN→∞
V∗N =
∫ largest y-coord. in R1
smallest y-coord. in R1
2π (5− y) (2 + log4 y) dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 56 / 64
Volume of Solid of Revolution (using Shells)
Riemann Sum: Volume(S1) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2 + log4 y∗k ) ∆yk
Integral: Volume(S1) =
∫ 1
1/162π (5− y) (2 + log4 y) dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 57 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R2:Thickness = (Width of generating H-Rect)Radius = (Positive Distance from H-Rect to Axis of Revolution)Length = (Length of generating H-Rect)Volume = 2π × (Radius)× (Length)× (Thickness)
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 58 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R2:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− log2 y∗kVolume = 2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 59 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R2:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− log2 y∗kVolume = 2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 60 / 64
Volume of Solid of Revolution (using Shells)
kth V-Shell on R2:
Thickness = ∆yk
Radius = 5− y∗kLength = 2− log2 y∗kVolume = 2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Integral: Volume(S2) = limN→∞
V∗N =
∫ largest y-coord. in R2
smallest y-coord. in R2
2π (5− y) (2− log2 y) dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 61 / 64
Volume of Solid of Revolution (using Shells)
Riemann Sum: Volume(S2) ≈ V∗N =
N∑k=1
2π (5− y∗k ) (2− log2 y∗k ) ∆yk
Integral: Volume(S2) =
∫ 4
12π (5− y) (2− log2 y) dy
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 62 / 64
Volume of Solid of Revolution (using Shells)
Region R = R1 ∪ R2 =⇒ Solid S = S1 ∪ S2
∴ Volume(S) = Volume(S1) + Volume(S2)
=∫ 1
1/162π (5− y) (2 + log4 y) dy +
∫ 4
12π (5− y) (2− log2 y) dy ≈ 81.8613
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 63 / 64
Fin
Fin.
Josh Engwer (TTU) Volumes: Slabs, Washers, Shells 27 January 2014 64 / 64