Download presentation
Presentation is loading. Please wait.
Published byWill Stalker Modified over 9 years ago
1
Areas, Volumes, Work Sections 5.1, 5.2, 5.3, 5.4 Chapter 5. Applications of Integration
2
5.1 Area between curves
3
5.1 Area between two curves Two curves y = f(x), y = g(x) g(x) ≤ f(x), a ≤ x ≤ b R = { (x,y) | a ≤x ≤ b, g(x) ≤ y ≤ f(x) } x y a y = f(x) b x (x,y) What is the area of R ? First, assume f(x) ≥ 0 and g(x) ≥ 0 on [a,b] y = g(x)
4
5.1 Area between two curves Two curves y=f(x), y=g(x) g(x) ≤ f(x), a ≤ x ≤ b R = { (x,y) | a ≤x ≤ b, g(x) ≤ y ≤ f(x) } x y a y = f(x) b First, assume f(x) ≥ 0 and g(x) ≥ 0 on [a,b] y = g(x) Af Area = Af – Ag = Ag Af - Ag y = g(x)
5
General Case Two curves y=f(x), y=g(x) g(x) ≤ f(x), a ≤ x ≤ b R = { (x,y) | a ≤x ≤ b, g(x) ≤ y ≤ f(x) } x y a y = f(x) b y = g(x) Idea: shift up using transformations y = f(x)+ K and y = g(x) + K Shift does not change the areas
6
5.1 Area between two curves x y a y = f(x)+K b y = g(x) y = g(x) + K
7
Area between two curves x y a y = f(x) b y = g(x)
8
Intersections of graphs Often, a or b or both correspond to points where graphs y = f(x) and y = g(x) intersect To find intersection points, solve equation f(x) = g(x)
9
5.2 Volumes
10
Volumes of simple 3D objects L W hh r V = (L W) hV = (π r 2 ) h In both cases, V = A h = (Area of the base) (height)
11
Generalized Cylinder V = A h = (Area of the base) (height) h A
12
General 3D shape Cross section
13
Area of a cross section x x A(x)
14
Slicing =x n x1x1 x i-1 xixi x a b x0=x0=
15
i th slice x i-1 xixi x
16
i th slice – approximation by cylinder x i-1 xixi x
17
i th slice – approximation by cylinder x i-1 xixi x A(x i ) ΔxΔx V i ≈ A(x i ) Δx V(i th slice) ≈ V(i th cylinder) = Area of the base height = A(x i ) Δx ViVi
18
Total Volume Total volume = sum of volumes of all slices ≈ sum of volumes of all approximating cylinders
19
Volume as integral of areas of cross sections x x A(x) b a x0=x0=
20
Solids of revolution
21
Rotate a plane region around a line – axis of rotation
23
Volumes of solids of revolution using “washers”
24
“Washer” – region between two concentric circles r in r out Area of the “washer” = = A(outer disk) – A(inner disk) =π (r out ) 2 -π (r in ) 2 = π [ (r out ) 2 - (r in ) 2 ]
25
Cross sections are washers
26
Area of cross section x x b a r in (x) r out (x) A(x) = π [r 2 out (x) - r 2 in (x) ]
27
Volume x x b a r in (x) r out (x) A(x) = π [r 2 out (x) - r 2 in (x) ]
28
5.3 Volumes by cylindrical shells
29
Cylindrical Shell
30
Volume of Cylindrical Shell h R out R in V = V(outer cyllinder) – V(inner cyllinder)= = (π R out 2 ) h - (π R in 2 ) h = π(R out 2 - R in 2 ) h= = π(R out - R in ) (R out + R in ) h Let Δr = thickness = R out – R in Let r = average radius = (R out + R in )/2 V = (2 π r) h Δr
31
Volume of Cylindrical Shell V = (2 π r) h Δr = = circumference height thickness h r ΔrΔr
32
Solid of revolution
34
Approximate the region using rectangles
35
Rotating the rectangles, we get cylindrical shells =x n x1x1 x i-1 xixi x a b x0=x0= y h(x i ) ΔxΔx ΔxΔx r(x i )
36
Rotating the rectangles, we get cylindrical shells =x n x1x1 x i-1 xixi x a b x0=x0= y h(x i ) ΔxΔx ΔxΔx r(x i )
37
Volume of i-th shell =x n x1x1 x i-1 xixi x a b x0=x0= y h(x i ) ΔxΔx ΔxΔx r(x i ) V i = circumference height thickness = = 2 π r(x i ) h(x i )Δx
38
Approximation by shells V i = 2 π r(x i ) h(x i ) Δx
39
Volume by shells x x a b y h(x) r(x)
40
5.4 Work
41
Work done by constant force Work = Force Displacement
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.