Finding Volumes. In General: Vertical Cut:Horizontal Cut:
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Transcript of Finding Volumes. In General: Vertical Cut:Horizontal Cut:
![Page 1: Finding Volumes. In General: Vertical Cut:Horizontal Cut:](https://reader036.fdocuments.net/reader036/viewer/2022081506/56649d0a5503460f949dcd05/html5/thumbnails/1.jpg)
Finding Volumes
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In General:
Vertical Cut: Horizontal Cut:
A =topfunction
⎛⎝⎜
⎞⎠⎟−
bottomfunction
⎛⎝⎜
⎞⎠⎟dx
a
b
∫ a≤b
A =rightfunction
⎛⎝⎜
⎞⎠⎟−
leftfunction
⎛⎝⎜
⎞⎠⎟dy
c
d
∫ c≤d
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Find the area of the region bounded by
Bounds? In terms of y: [-2,1]
Points: (0,-2), (3,1)
Right Function?
Left Function?
Area?
x =2 + y
x =4 −y2
[(4 −y2
−2
1
∫ )−(2 + y)]dy
x + y2 =4 and x−y=2
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Volume & Definite Integrals
We used definite integrals to find areas by slicing the region and adding up the areas of the slices.
We will use definite integrals to compute volume in a similar way, by slicing the solid and adding up the volumes of the slices.
For Example………………
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Blobs in SpaceVolume of a blob:
Cross sectional area at height h: A(h)
Volume =
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ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.
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Volumes:
We will be given a “boundary” for the base of the shape which will be used to find a length.
We will use that length to find the area of a figure generated from the slice .
The dy or dx will be used to represent the thickness.
The volumes from the slices will be added together to get the total volume of the figure.
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
Bounds?
Top Function?
Bottom Function?
[-1,1]
y = 1−x2
y =− 1−x2
Length? 1−x2 − − 1−x2( )
=2 1 − x2
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
We use this length to find the area of the square.
Length? =2 1 − x2
Area? 2 1−x2( )
2
4 1−x2( )
4 1−x2( )dx−1
1
∫
Volume?
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
What does this shape look like?
4 1−x2( )dx−1
1
∫Volume?
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a circle with diameter in the plane.
x2 + y2 ≤1
Length? =2 1 − x2
Area?
π 1 − x2( )
2
π 1− x2( )
π 1− x2( )dx−1
1
∫Volume?
2 1−x2
2Radius:
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Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles.
x2 + y2 ≤1
Length? =2 1 − x2
Area?1
22 1−x2
( ) 2 1−x2( )
Volume? 2 1−x2( )dx0
1
∫
Bounds? [0,1]
Visual?
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The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.
y =2 sinx
Bounds?
Top Function?
Bottom Function?
[0,π]
y =2 sinx
y =0
Length? 2 sin x
Area of an equilateral triangle?
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Area of an Equilateral Triangle?
S
S
S
S/2
S
S
Sqrt(3)*S/2
S/2
Area = (1/2)b*h
=1
2⎛⎝⎜
⎞⎠⎟S( )
3
2S
⎛
⎝⎜⎞
⎠⎟=
3
4S2
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The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.
y =2 sinx
Bounds?
Top Function?
Bottom Function?
[0,π]
y =2 sinx
y =0
Length? 2 sin x
Area of an equilateral triangle?3
4(2 sin x )2
3
4(2 sin x )2dx
0
π
∫Volume?
3
4(S)2
⎛
⎝⎜⎞
⎠⎟
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.
x2 + y2 ≤1
We used this length to find the area of the square whose side was in the plane….
Length? =2 1 − x2
Area with the length representing the diagonal?
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Area of Square whose diagonal is in the plane?
D
SS
S2 +S2 =D2
2S2 =D2
S2 =D2
2⇒ S =
D
2
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.
x2 + y2 ≤1
Length of Diagonal? =2 1− x2
Length of Side?
2 1−x2
2= 2 1 − x2
2(1−x2 )Area?
Volume? 2 1−x2( )dx−1
1
∫
(S =D2)