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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Homework

Read Section 13.2.Exercises: 1–17 odd, 23–31 odd, 37, 43, 45

J. Robert Buchanan Area, Volume, and Center of Mass