area, volume, and center of massbanach.millersville.edu/~bob/math311/area/main.pdf · area, volume,...
TRANSCRIPT
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Area, Volume, and Center of MassMATH 311, Calculus III
J. Robert Buchanan
Department of Mathematics
Fall 2011
J. Robert Buchanan Area, Volume, and Center of Mass
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Introduction
Today we will apply the double integral to the problems offinding
the area of regions in the plane,the volume of a bounded region in three dimensions,the center of mass of a region in the plane.
J. Robert Buchanan Area, Volume, and Center of Mass
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Area
If R is a region in the xy -plane, the area A of R is
A =
∫∫R
1 dA
provided the value of the double integral is finite.
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (1 of 6)
Find the area of the triangular region with vertices at (0,2),(3,2), and (1,1).
0.0 0.5 1.0 1.5 2.0 2.5 3.0x
0.5
1.0
1.5
2.0
y
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 6)
The region R is bounded by the graphs of the lines x = 2− yand x = 2y − 1 for 1 ≤ y ≤ 2.
A =
∫∫R
1 dA
=
∫ 2
1
∫ 2y−1
2−y1 dx dy
=
∫ 2
1(2y − 1)− (2− y) dy
=
∫ 2
13y − 3 dy
=32
y2 − 3y∣∣∣∣21
= 6− 6−(
32− 3)
=32
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (3 of 6)
Find the area of the bounded region in the xy -plane betweeny =√
x and y = x2.
0.2 0.4 0.6 0.8 1.0x
0.2
0.4
0.6
0.8
1.0
y
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (4 of 6)
A =
∫∫R
1 dA
=
∫ 1
0
∫ √x
x21 dy dx
=
∫ 1
0
√x − x2 dx
=23
x3/2 − 13
x3∣∣∣∣10
=13
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (5 of 6)
Find the area of the bounded region in the xy -plane betweeny = x and x = y2 − y .
-1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5
-1
0
1
2
3
x
y
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (6 of 6)
A =
∫∫R
1 dA
=
∫ 2
0
∫ y
y2−y1 dx dy
=
∫ 2
0y − (y2 − y) dy
=
∫ 2
02y − y2 dy
= y2 − 13
y3∣∣∣∣20
= 4− 83
=43
J. Robert Buchanan Area, Volume, and Center of Mass
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Volume of a Solid Region
If f (x , y) ≥ 0 for all (x , y) in a region R, the volume of the solidbelow the surface z = f (x , y) and above the region R in thexy -plane is
V =
∫∫R
f (x , y) dA.
If f (x , y) ≥ g(x , y) for all (x , y) in a region R, the volume of thesolid below the surface z = f (x , y) and above the surfacez = g(x , y) within the the region R in the xy -plane is
V =
∫∫R(f (x , y)− g(x , y)) dA.
J. Robert Buchanan Area, Volume, and Center of Mass
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Volume of a Solid Region
If f (x , y) ≥ 0 for all (x , y) in a region R, the volume of the solidbelow the surface z = f (x , y) and above the region R in thexy -plane is
V =
∫∫R
f (x , y) dA.
If f (x , y) ≥ g(x , y) for all (x , y) in a region R, the volume of thesolid below the surface z = f (x , y) and above the surfacez = g(x , y) within the the region R in the xy -plane is
V =
∫∫R(f (x , y)− g(x , y)) dA.
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (1 of 4)
Find the volume of the solid region below the graph of z = ey2
and above the triangle in the xy -plane with vertices at (0,0),(1,1), (0,1).
0.0
0.5
1.0
x0.0
0.5
1.0
y1.0
1.5
2.0
2.5
z
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 4)
V =
∫∫R
f (x , y) dA
=
∫∫R
ey2dA
=
∫ 1
0
∫ y
0ey2
dx dy
=
∫ 1
0xey2
∣∣∣y0
dy
=
∫ 1
0yey2
dy
=12
ey2∣∣∣∣10
=12
e − 12
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (3 of 4)
Find the volume of the solid region above the graph ofz = 2x2 + y2 and below the graph of z = 8− x2 − 2y2 andwithin the rectangle R = {(x , y) |0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.
0.0
0.5
1.0
x0.0
0.5
1.0
y0
2
4
6
8
z
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (4 of 4)
V =
∫∫R
f (x , y) dA
=
∫ 1
0
∫ 1
0(8− x2 − 2y2)− (2x2 + y2) dx dy
=
∫ 1
0
∫ 1
0(8− 3x2 − 3y2) dx dy
=
∫ 1
08x − x3 − 3xy2
∣∣∣10
dy
=
∫ 1
0(8− 1− 3y2) dy
=
∫ 1
0(7− 3y2) dy
= 7y − y3∣∣∣10
= 6
J. Robert Buchanan Area, Volume, and Center of Mass
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Moments and Center of Mass
lamina: a thin sheet of material in the shape of a regionR ⊂ R2.
ρ(x , y): the density of the lamina at (x , y).m: mass of the lamina
m =
∫∫Rρ(x , y) dA.
Mx : moment with respect to the x-axis
Mx =
∫∫R
yρ(x , y) dA.
My : moment with respect to the y -axis
My =
∫∫R
xρ(x , y) dA.
(x , y) center of mass
(x , y) =
(My
m,Mx
m
)J. Robert Buchanan Area, Volume, and Center of Mass
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Example (1 of 2)
A lamina with density ρ(x , y) = y + 3 is bounded by the curvesx = y2 and x = 4. Find its mass, moments, and center of mass.
0 1 2 3 4
-2
-1
0
1
2
x
y
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
m =
∫ 2
−2
∫ 4
y2(y + 3) dx dy
=
∫ 2
−2(12 + 4y − 3y2 − y3) dy = 32
Mx =
∫ 2
−2
∫ 4
y2y(y + 3) dx dy
=
∫ 2
−2(12y + 4y2 − 3y3 − y4) dy =
12815
My =
∫ 2
−2
∫ 4
y2x(y + 3) dx dy
=
∫ 2
−2
(24 + 8y − 3
2y4 − 1
2y5)
dy =384
5
(x , y) =
(My
m,Mx
m
)=
(125,
415
)
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
m =
∫ 2
−2
∫ 4
y2(y + 3) dx dy =
∫ 2
−2(12 + 4y − 3y2 − y3) dy = 32
Mx =
∫ 2
−2
∫ 4
y2y(y + 3) dx dy
=
∫ 2
−2(12y + 4y2 − 3y3 − y4) dy =
12815
My =
∫ 2
−2
∫ 4
y2x(y + 3) dx dy
=
∫ 2
−2
(24 + 8y − 3
2y4 − 1
2y5)
dy =384
5
(x , y) =
(My
m,Mx
m
)=
(125,
415
)
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
m =
∫ 2
−2
∫ 4
y2(y + 3) dx dy =
∫ 2
−2(12 + 4y − 3y2 − y3) dy = 32
Mx =
∫ 2
−2
∫ 4
y2y(y + 3) dx dy
=
∫ 2
−2(12y + 4y2 − 3y3 − y4) dy =
12815
My =
∫ 2
−2
∫ 4
y2x(y + 3) dx dy
=
∫ 2
−2
(24 + 8y − 3
2y4 − 1
2y5)
dy =384
5
(x , y) =
(My
m,Mx
m
)=
(125,
415
)
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
m =
∫ 2
−2
∫ 4
y2(y + 3) dx dy =
∫ 2
−2(12 + 4y − 3y2 − y3) dy = 32
Mx =
∫ 2
−2
∫ 4
y2y(y + 3) dx dy
=
∫ 2
−2(12y + 4y2 − 3y3 − y4) dy =
12815
My =
∫ 2
−2
∫ 4
y2x(y + 3) dx dy
=
∫ 2
−2
(24 + 8y − 3
2y4 − 1
2y5)
dy =384
5
(x , y) =
(My
m,Mx
m
)=
(125,
415
)
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
m =
∫ 2
−2
∫ 4
y2(y + 3) dx dy =
∫ 2
−2(12 + 4y − 3y2 − y3) dy = 32
Mx =
∫ 2
−2
∫ 4
y2y(y + 3) dx dy
=
∫ 2
−2(12y + 4y2 − 3y3 − y4) dy =
12815
My =
∫ 2
−2
∫ 4
y2x(y + 3) dx dy
=
∫ 2
−2
(24 + 8y − 3
2y4 − 1
2y5)
dy =384
5
(x , y) =
(My
m,Mx
m
)=
(125,
415
)J. Robert Buchanan Area, Volume, and Center of Mass
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Moment of Inertia
Inertia: tendency of an object to resist a change in motion.Ix : moment of inertia about the x-axis
Ix =
∫∫R
y2ρ(x , y) dA.
Iy : moment of inertia about the y -axis
Iy =
∫∫R
x2ρ(x , y) dA.
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (1 of 2)
A lamina with density ρ(x , y) = y + 3 is bounded by the curvesx = y2 and x = 4. Find its moments of inertia.
0 1 2 3 4
-2
-1
0
1
2
x
y
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
Ix =
∫ 2
−2
∫ 4
y2y2(y + 3) dx dy
=
∫ 2
−212y2 + 4y3 − 3y4 − y5 dy
= 4y3 + y4 − 35
y5 − 16
y6∣∣∣∣2−2
=1285
Iy =
∫ 2
−2
∫ 4
y2x2(y + 3) dx dy
=
∫ 2
−264 +
643
y − y6 − 13
y7 dy
= 64y +323
y2 − 17
y7 − 124
y8∣∣∣∣2−2
=1536
7
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
Ix =
∫ 2
−2
∫ 4
y2y2(y + 3) dx dy
=
∫ 2
−212y2 + 4y3 − 3y4 − y5 dy
= 4y3 + y4 − 35
y5 − 16
y6∣∣∣∣2−2
=128
5
Iy =
∫ 2
−2
∫ 4
y2x2(y + 3) dx dy
=
∫ 2
−264 +
643
y − y6 − 13
y7 dy
= 64y +323
y2 − 17
y7 − 124
y8∣∣∣∣2−2
=1536
7
J. Robert Buchanan Area, Volume, and Center of Mass
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Example (2 of 2)
Ix =
∫ 2
−2
∫ 4
y2y2(y + 3) dx dy
=
∫ 2
−212y2 + 4y3 − 3y4 − y5 dy
= 4y3 + y4 − 35
y5 − 16
y6∣∣∣∣2−2
=128
5
Iy =
∫ 2
−2
∫ 4
y2x2(y + 3) dx dy
=
∫ 2
−264 +
643
y − y6 − 13
y7 dy
= 64y +323
y2 − 17
y7 − 124
y8∣∣∣∣2−2
=1536
7
J. Robert Buchanan Area, Volume, and Center of Mass
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Homework
Read Section 13.2.Exercises: 1–17 odd, 23–31 odd, 37, 43, 45
J. Robert Buchanan Area, Volume, and Center of Mass