Higher_Mathematics_Part_3

y = x2

1

– 1

1

x

– 1

x = y2

Fig. 5.10

Therefore, we can apply here (5.20)

1

x

1

∫∫(x + 2y)dxdy = ∫ dx ∫ (x + 2 y)dy = ∫(xy + y2 )

x2x dx =

D

0 x2

0

1

5

2

4

5

x2

x

x

x

2

1

1

1

9

= ∫(x x + x − x3 − x4 )dx = (

+

−

−

)

10

=

+

−

−

=

.

0

5

2

4

5

5

2

4

5

20

2

I = ∫∫(x + 2y)dxdy

Example 2. We take the double integral

over a domain

y = 2 − x and y = x2 .

D

D bounded by the lines

We have x2 = 2 − x,

then x2 + x − 2 = 0. We obtain x

= −2, x = 1.

1

2

y = 2 – x

y

y = x2

4

1

–2 –1 O 1

y

Fig. 5.11

1

2− x

1

2− x

I = ∫∫ (x + 2y)dxdy = ∫ dx ∫

(x + 2y)dy = ∫ dx(xy + y2 )

=

1

D

−2

x2

1

−2

x2

(4 − 2x − x3 − x4 )dx =

= ∫

(x(2 − x) + (2 − x)2 − x3 − x4 )dx = ∫

−2

−2

4x

− x2

x4

x5

1

1

1

−

32

=

−

−

= 4 − 1−

−

−8 − 4 − 4 +

= 12,15.

4

5

−2

4 5

5

domain of integration: ∫∫ f (x,

y)dxdy

over the domain D bounded by the lines

D

y = 2x, x + y = 3, y =

0 .

Solution. Examine D. The straight lines

y = 2x

and

y = 3 − x

meet at point

A(1, 2). Domain D is a triangle (Fig. 5.12).

1. This domain is regular. So, y

varies from 0 to 2, and ϕ1 (x) = 2x and

ϕ2 (x) = 3 − x.

Any straight line y = const, (0 ≤ y ≤ 2)

meets the boundary of

the region at not more than two points.

Therefore, we can apply here

∫∫ f (x,

2

3− x

y)dxdy = ∫ dy ∫ f (x,

y)dy.

D

0

2x

y

y

3

y = 2x

3

А

А

2

2

y = 3 – x

О

В

x

О

В

x

1

3

1

3

а

b

Fig. 5.12

∫∫ f (x,

1

2 x

3

3− x

y)dxdy = ∫ dx ∫

f (x, y)dy + ∫ dx ∫ f (x, y)dy.

D

0

0

1

0

2

5− x2

2

y

2

5− x2

Therefore ∫∫ f (x,

y)dxdy = ∫ dx ∫

(x + 2 y)dy = ∫(xy + 2

)

dx =

2

D

0

x

0

x

2

2

2

x

2

x

2

2

43

= ∫(x(5 − x2 ) + (5

− x2 )2 −

−

)dx = ∫(x4 − x3 −

x2

+

5x + 25)dx =

0

2

4

0

4

x5

x4

43x3

5x2

2

25

86

=

−

−

+

+ 25x

=

− 4 −

+ 10

+ 50 ≈ 34.

4

12

2

5

3

5

0

different lines, then we shall divide this domain into two parts by line

y = 1.

Defined from the equations

y =

x

and

y = 5 − x2

a variable y through

x, we

shall receive

2

1

2 y

5

5− y

∫∫ f (x, y)dxdy = ∫ dy ∫ (x + 2 y)dx + ∫ dy ∫ (x + 2 y)dx =

D

0

0

1

0

1 x2

2 y

5 x2

5− y

= ∫

+ 2yx

dy + ∫

+ 2 yx

dy ≈ −21.56.

2

2

0

0

0

0

Example 5. Calculate the integral

∫∫(6x − 3y)dxdy, where D :{x + y = 1,

Let’s construct the domain

D* :

{

}

in uv-plane (Fig. 5.15).

u = 1, u = 2, v = 1, v = 3

x =

1

(u + v)

3x = u + v

3

x + y = u

1

2x − y = v

2x − y = v

y =

(2u − v)

3

∂x

Now we find partial derivatives of the first order with respect to u and v,

=

1

,

∂x =

1

,

∂y

=

2

,

∂y

= −

1

. Then Jacobian equals

∂v

∂u 3

∂v 3

∂u 3

3

I (u,v) =

∂x

∂x

1

1

= −

1

−

2

1

≠ 0 .

∂u

∂v

=

3

3

= −

∂y

∂y

2

−

1

9

9

3

∂u

∂v

3

3

V

2x – y = 1

v

2

2x – y = 3

v = 3

1

3

O

1

2

3

X

v = 1

– 1

1

x + y = 2

– 2

u

O

1

2

– 3

x + y = 1

Fig. 5.14

Fig. 5.15

114

We obtain

I (ρ, ϕ)

=

1

.

3

Now the given integral is equal to

∫∫(6x − 3y)dxdy = ∫∫[6

1

(u + v)

−

3

1

(2u − v)]

1

dudv =

3

3

D

D*

3

1

2

3

2

v

2

2

9

1

=

∫∫3vdudv =∫ du∫vdv = ∫

13

du = ∫(

−

)du = 4.

3

2

2

2

*

1

1

1

1

D

Example 6. Take the

∫∫

1

dxdy

where D :

{

x2 + y2 ≤ 1, x ≥ 0, y ≥ 0

x2 + y2

}

D

x = ρ cos ϕ and y = ρ sin ϕ,

then

D

*

=

≤ ρ ≤ 1; 0

≤ ϕ ≤

π

and our integrable

0

2

function f (x, y) =

1

=

1

=

1

,

then the given

integral can be

ρ2

ρ

x2 + y2

1

O

1

x

Fig. 5.16

1

π

π

π

π

π .

∫∫

ρdρdϕ = ∫02 dϕ∫01 dρ = ∫2 ρ

10 dϕ = ∫2 1dϕ = ϕ

02 =

2

D* ρ

0

0

Example 7. Find the area S if D: x = y2 − 2 y, x − y = 0.

Solution. We write the given equation of

x = y2 − 2 y

in canonical form,

y = x

3

A

1

–1 O

1 2

3

x

–1

x = y2 – 2y

Hence

Fig. 5.17

y

3

y

3

3

S = ∫∫ dxdy = ∫ dy

∫

dx = ∫ x

dy = ∫( y − ( y2 − 2y))dy =

D

0

y2 − 2 y

0

y2 − 2 y

0

3

3y

2

y

3

9

= ∫(3y − y2 )dy = (

−

)

30

=

.

0

2

3

2

Example 8. Find the area S if D: x2 + y2

= 4x,

y = x, y = 0.

y

y = x

C

ρ = 4cosφ

D

O

π/4

B

x

A(2; 0)

ρ2 cos2 ϕ + ρ2 sin2 ϕ = 4ρ cosρ,

ρ2

= 4ρ cosρ,

ρ = 4cos ϕ .

Straight lines

y = 0 and y = x in the polar components have the form ϕ = 0

and

ϕ = π

correspondently. Hence, polar components ϕ and ρ of the points which

4

π ,

belong to the domain, are changed within the limits: 0 ≤ ϕ ≤

0 ≤ ρ ≤ 4cos ϕ.

We have

4

π

4cos ϕ

π

ρ2

4cos ϕ dϕ =

S = ∫∫ dxdy = ∫∫ρdρdϕ = ∫4 dϕ ∫

ρ dρ = ∫4

D

D

0

0

0

2

0

π

π

1

sin 2ϕ

π

π

1+ cos 2ϕ

1

4

4

2

4

= 8∫cos

ϕdϕ = 8∫

dϕ = 4

ϕ +

= 4

+

= π + 2.

2

2

0

0

2

0

4

2

2

2

y

2

V = ∫∫(2 − y)dxdy = ∫

dx ∫ (2 − y)dy =

∫

(2y −

)

2x2 dx =

D

− 2

x2

− 2

2

2

2

x4

2x3

x5

2

32 2

= ∫ (2 − 2x

+

)dx =

2x −

+

2 =

.

2

3

10

−

15

− 2

O

y

2

2 3

y