Areas and double integrals. (Sect. 15.3) Areas of a region...

Areas and double integrals. (Sect. 15.3)

I Areas of a region on a plane.

I Average value of a function.

I More examples of double integrals.

Areas of a region on a plane

DefinitionThe area of a closed, bounded region R on a plane is given by

A =

∫∫R

dx dy .

Remark:

I To compute the area of a region R we integrate the functionf (x , y) = 1 on that region R.

I The area of a region R is computed as the volume of a3-dimensional region with base R and height equal to 1.


Example

Find the area of R = {(x , y) ∈ R2 : x ∈ [−1, 2], y ∈ [x2, x + 2]}.

Solution: We express the region R as anintegral Type I, integrating first onvertical directions:

A =

∫ 2

−1

∫ x+2

x2

dy dx .x21−1

4

2

y = x + 2

y = x2

y

A =

∫ 2

−1

(y∣∣∣x+2

x2

)dx =

∫ 2

−1

(x + 2− x2

)dx =

(x2

2+ 2x − x3

3

)∣∣∣2−1

.

A = 2− 1

2+ 4 + 2− 8

3− 1

3= 8− 1

2− 3 ⇒ A =

9

2. C


Example

Find the area of R = {(x , y) ∈ R2 : x ∈ [−1, 2], y ∈ [x2, x + 2]}integrating first along horizontal directions.

Solution: We express the region R as anintegral Type II, integrating first onhorizontal directions:

A =

∫∫R1

dx dy +

∫∫R2

dx dy . x = − y

21−1

4

2

y = x + 2

R1 x = y

R

y = xx = y − 2

y

x

2

2

A =

∫ 1

0

∫ √y

−√ydx dy +

∫ 4

1

∫ √y

y−2dx dy .

We must get the same result: A = 9/2.


Example

Find the area of R = {(x , y) ∈ R2 : x ∈ [−1, 2], y ∈ [x2, x + 2]}integrating first along horizontal directions.

Solution: Recall: A =

∫ 1

0

∫ √y

−√ydx dy +

∫ 4

1

∫ √y

y−2dx dy .

A =

∫ 1

02√

y dy +

∫ 4

1

(√y − y + 2

)dy

A = 2(2

3y3/2

)∣∣∣10+

(2

3y3/2 − y2

2+ 2y

)∣∣∣41

A =4

3+

16

3− 2

3− 8 +

1

2+ 8− 2 = 6− 3

2.

We conclude that A =9

2. C





Average value of a function

Review: The average of a single variable function.

DefinitionThe average of a function f : [a, b] → Ron the interval [a, b], denoted by f , isgiven by

f =1

(b − a)

∫ b

af (x) dx .

y

a b x

f(x)

f

DefinitionThe average of a function f : R ⊂ R2 → R on the region R witharea A(R), denoted by f , is given by

f =1

A(R)

∫∫R

f (x , y) dx dy .

Average value of a function

Example

Find the average of f (x , y) = xy on the regionR = {(x , y) ∈ R2 : x ∈ [0, 2], y ∈ [0, 3]}.

Solution: The area of the rectangle R is A(R) = 6.

We only need to compute I =

∫∫R

f (x , y) dx dy .

I =

∫ 2

0

∫ 3

0xy dy dx =

∫ 2

0x(y2

2

∣∣∣30

)dx =

∫ 2

0

9

2x dx .

I =9

2

(x2

2

∣∣∣20

)⇒ I = 9.

Since f = I/A(R) = 9/6, we get f = 3/2. C





More examples of double integrals

Example

Find the integral of ρ(x , y) = x + y in the triangle with boundariesy = 0, x = 1 and y = 2x .

Solution: We need to compute

M =

∫∫R

ρ(x , y) dxdy .

Remark: If ρ is the mass density, thenM is the total mass.

y

1

2

y = 0

x = 1

y = 2x

x

M =

∫ 1

0

∫ 2x

0(x + y) dy dx =

∫ 1

0

[x

(y∣∣∣2x

0

)+

(y2

2

∣∣∣2x

0

)]dx .

M =

∫ 1

0

[2x2 + 2x2

]dx = 4

x3

3

∣∣∣10

⇒ M =4

3. C


Example

Given the function ρ(x , y) = x + y , the number M computed inthe previous example, and the triangle with boundaries y = 0,x = 1 and y = 2x , find the numbers

r x =1

M

∫R

xρ(x , y) dy dx , r y =1

M

∫∫R

yρ(x , y) dy dx .

Remark: r = 〈r x , r y 〉 is the center of mass of the body.

Solution: Recall: M =4

3. We need to compute

r x =1

M

∫ 1

0

∫ 2x

0(x +y)x dydx =

3

4

∫ 1

0

[x2

(y∣∣∣2x

0

)+x

(y2

2

∣∣∣2x

0

)]dx

r x =3

4

∫ 1

0

[2x3 + 2x3

]dx =

3

4x4

∣∣∣10

⇒ r x =3

4.


Example

Given the function ρ(x , y) = x + y , the number M computed inthe previous example, and the triangle with boundaries y = 0,x = 1 and y = 2x , find the numbers

r x =1

M

∫R

xρ(x , y) dy dx , r y =1

M

∫∫R

yρ(x , y) dy dx .

Solution: Recall: M =4

3and r x =

3

4.

r y =1

M

∫ 1

0

∫ 2x

0(x + y)y dydx =

3

4

∫ 1

0

[x

(y2

2

∣∣∣2x

0

)+

(y3

3

∣∣∣2x

0

)]dx

r y =3

4

∫ 1

0

[2x3 +

8

3x3

]dx =

3

4

[2(x4

4

∣∣∣10

)+

8

3

(x4

4

∣∣∣10

)]r y =

3

4

[1

2+

2

3

]=

3

4

7

6⇒ r y =

7

8. C


DefinitionThe centroid of a region R in the plane is the vector c given by

c =1

A(R)

∫∫R〈x , y〉 dx dy , where A(R) =

∫∫R

dx dy .

Remark:

I The centroid of a region can be seen as the center of massvector of that region in the case that the mass density isconstant.

I When the mass density is constant, it cancels out from thenumerator and denominator of the center of mass.


Example

Find the centroid of the triangle inside y = 0, x = 1 and y = 2x .

Solution: The area of the triangle is

A(R) =

∫ 1

0

∫ 2x

0dy dx =

∫ 1

02x dx = x2

∣∣∣10

⇒ A(R) = 1.

Therefore, the centroid vector components are given by

cx =

∫ 1

0

∫ 2x

0x dy dx =

∫ 1

02x2 dx = 2

(x3

3

∣∣∣10

)⇒ cx =

2

3.

cy =

∫ 1

0

∫ 2x

0y dy dx =

∫ 1

0

(y2

2

∣∣∣2x

0

)dx =

∫ 1

02x2 dx = 2

(x3

3

∣∣∣10

)so cy =

2

3. We conclude, c =

2

3〈1, 1〉. C


Remark: The moment of inertia of an object is a measure of theresistance of the object to changes in its rotation along a particularaxis of rotation.

DefinitionThe moment of inertia about the x-axis and the y -axis of a regionR in the plane having mass density ρ : R ⊂ R2 → R are given by,respectively,

Ix =

∫∫R

y2ρ(x , y) dx dy , Iy =

∫∫R

x2ρ(x , y) dx dy .

If M denotes the total mass of the region, then the radii ofgyration about the x-axis and the y -axis are given by

Rx =√

Ix/M Ry =√

Iy/M.

The moment of inertia of an object.

Example

Find the moment of inertia and the radius of gyration about thex-axis of the triangle with boundaries y = 0, x = 1 and y = 2x ,and mass density ρ(x , y) = x + y .

Solution: The moment of inertia Ix is given by

Ix =

∫ 1

0

∫ 2x

0x2(x + y) dy dx =

∫ 1

0

[x3

(y∣∣∣2x

0

)+ x2

(y2

2

∣∣∣2x

0

)]dx

Ix =

∫ 1

04x4 dx = 4

(x5

5

∣∣∣10

)⇒ Ix =

4

5.

Since the mass of the region is M = 4/3, the radius of gyration

along the x-axis is Rx =√

Ix/M =√

45

34 , that is, Rx =

√35 . C

Areas and double integrals. (Sect. 15.3) Areas of a region...

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