example of Riemann double integral

example of Riemann double integral

Let us determine the value of the

double integral

I

:=

∬

D

d

⁢

x

⁢

d

⁢

y

(

1

+

x

2

+

y

2

)

2

(1)

where

D

is the

triangle

by the lines

x

=

0

,

y

=

0

and

x

+

y

=

1

.

Since the triangle is defined by the

inequalities

0

≦

x

≦

1

,

0

≦

y

≦

1

-

x

,
one can write

I

=

∫

0

1

∫

0

1

-

x

d

⁢

x

⁢

d

⁢

y

(

1

+

x

2

+

y

2

)

2

=

∫

0

1

d

⁢

x

(

1

+

x

2

)

2

⁢

∫

0

1

-

x

d

⁢

y

[

1

+

(

y

1

+

x

2

)

2

]

2

=

∫

0

1

1

(

1

+

x

2

)

2

⋅

1

+

x

2

2

⁢

/

y

=

0

1

-

x

⁡

(

arctan

⁡

y

1

+

x

2

+

y

1

+

x

2

1

+

y

2

1

+

x

2

)

⁡

d

⁢

x

=

∫

0

1

(

1

2

⁢

(

1

+

x

2

)

-

3

2

⁢

arctan

⁡

1

-

x

1

+

x

2

+

1

-

x

(

1

-

x

+

x

2

)

⁢

(

1

+

x

2

)

)

⁢

𝑑

x

.

The last

expression

seems quite difficult to calculate to a

closed form

…

Some appropriate substitution (

http://planetmath.org/ChangeOfVariablesInIntegralOnMathbbRn

)

x

:=

x

⁢

(

u

,

v

)

,

y

:=

y

⁢

(

u

,

v

)

directly to the form (1) could offer a better is

∬

D

f

⁢

(

x

,

y

)

⁢

𝑑

x

⁢

𝑑

y

=

∬

Δ

f

⁢

(

x

⁢

(

u

,

v

)

,

y

⁢

(

u

,

v

)

)

⁢

|

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

|

⁢

𝑑

u

⁢

𝑑

v

.

(2)

What kind a change of variables would be good?  One idea were to use some “natural substitution”, i.e. such one that would give constant limits (

http://planetmath.org/DefiniteIntegral

).  For example, the equations

x

+

y

:=

u

,

y

x

:=

v

,

map the triangular domain (

http://planetmath.org/Domain2

)

D

to the “

rectangle

”

Δ

:

0

≦

u

≦

1

,

0

≦

v

<

∞

.

Then we need the

Jacobian

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

=

u

+

v

2

(

v

+

1

)

3

.

By (2), we have

I

=

∫

0

1

∫

0

∞

(

v

+

1

)

4

u

2

+

2

⁢

v

2

+

2

⁢

v

+

1

⋅

u

+

v

2

(

v

+

1

)

3

⁢

𝑑

u

⁢

𝑑

v

=

∫

0

∞

(

v

+

1

)

⁢

𝑑

v

⁢

∫

0

1

u

+

v

2

u

2

+

2

⁢

v

2

+

2

⁢

v

+

1

⁢

𝑑

u

.

But here after integrating with respect to

u

, one obtains a difficult single

integral

.  Thus, when the , the

integrand

may become awkward.

A second idea would be to try to make the integrand simpler.  For this end, the transition to the

polar coordinates

x

:=

r

⁢

cos

⁡

φ

,

y

:=

r

⁢

sin

⁡

φ

in (1) is more suitable.  We have

∂

⁡

(

x

,

y

)

∂

⁡

(

r

,

φ

)

=

|

cos

⁡

φ

-

r

⁢

sin

⁡

φ

sin

⁡

φ

r

⁢

cos

⁡

φ

|

≡

r

.

The

Pythagorean theorem

gives the equation

r

2

=

x

2

+

y

2

=

(

r

⁢

cos

⁡

φ

)

2

+

(

1

-

r

⁢

cos

⁡

φ

)

2

,  i.e.

r

2

⁢

cos

⁡

2

⁢

φ

-

2

⁢

r

⁢

cos

⁡

φ

+

1

=

0

,

from which we get the

upper limit

r

=

2

⁢

cos

⁡

φ

±

4

⁢

cos

2

⁡

φ

-

4

⁢

cos

⁡

2

⁢

φ

2

⁢

cos

⁡

2

⁢

φ

=

cos

⁡

φ

±

sin

⁡

φ

cos

2

⁡

φ

-

sin

2

⁡

φ

;

this is

1

cos

⁡

φ

+

sin

⁡

φ

, since the “+” alternative can be excluded by choosing e.g.

φ

=

π

2

.  Thus

Δ

:

0

≦

φ

≦

π

2

,

0

≦

r

≦

1

cos

⁡

φ

+

sin

⁡

φ

and

I

=

1

2

⁢

∫

0

π

2

∫

0

1

cos

⁡

φ

+

sin

⁡

φ

2

⁢

r

⁢

d

⁢

r

(

1

+

r

2

)

2

⁢

𝑑

φ

=

1

2

⁢

∫

0

π

2

d

⁢

φ

2

+

sin

⁡

2

⁢

φ

.

Here, the

http://planetmath.org/node/9380

Weierstrass substitution

tan

⁡

φ

:=

t

easily yields the final result

I

=

2

⁢

π

⁢

3

9

.

(3)

Title

example of Riemann double integral

Canonical

name

ExampleOfRiemannDoubleIntegral

Date of creation

2013-03-22 19:12:22

Last modified on

2013-03-22 19:12:22

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Example

Classification

msc 26A42

Classification

msc 28-00

Related topic

SubstitutionNotation

Related topic

ChangeOfVariablesInIntegralOnMathbbRn

Related topic

ExampleOfRiemannTripleIntegral

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