example of Riemann triple integral

example of Riemann triple integral

Determine the volume of the solid in

ℝ

3

by the part of the surface

(

x

2

+

y

2

+

z

2

)

3

=

3

⁢

a

3

⁢

x

⁢

y

⁢

z

being in the

first octant

(

a

>

0

).

Since

x

2

+

y

2

+

z

2

is the squared

distance

of the point

(

x

,

y

,

z

)

from the origin, the solid is apparently defined by

D

:=

{

(

x

,

y

,

z

)

∈

ℝ

3

⋮

x

≧

0

,

y

≧

0

,

z

≧

0

,

(

x

2

+

y

2

+

z

2

)

3

≦

3

a

3

x

y

z

}

.

By the definition

𝐦𝐞𝐚𝐬

⁢

(

D

)

:=

∫

χ

D

⁢

(

v

)

⁢

𝑑

v

in the parent entry (

http://planetmath.org/RiemannMultipleIntegral

), the volume in the question
is

V

=

∫

D

1

⁢

𝑑

v

=

∭

D

𝑑

x

⁢

𝑑

y

⁢

𝑑

z

.

(1)

For calculating the

integral

(1) we express it by the (geographic)

spherical coordinates

through

{

x

=

r

⁢

cos

⁡

φ

⁢

cos

⁡

λ

y

=

r

⁢

cos

⁡

φ

⁢

sin

⁡

λ

z

=

r

⁢

sin

⁡

φ

where the

latitude

angle

φ

of the

position vector

r

→

is measured from the

x

⁢

y

-plane (not as the

colatitude

ϕ

from the positive

z

-axis);

λ

is the

longitude

.  For the

change of coordinates

, we need the Jacobian determinant

∂

⁡

(

x

,

y

,

z

)

∂

⁡

(

r

,

φ

,

λ

)

=

|

∂

⁡

x

∂

⁡

r

∂

⁡

y

∂

⁡

r

∂

⁡

z

∂

⁡

r

∂

⁡

x

∂

⁡

φ

∂

⁡

y

∂

⁡

φ

∂

⁡

z

∂

⁡

φ

∂

⁡

x

∂

⁡

λ

∂

⁡

y

∂

⁡

λ

∂

⁡

z

∂

⁡

λ

|

=

|

cos

⁡

φ

⁢

cos

⁡

λ

cos

⁡

φ

⁢

sin

⁡

λ

sin

⁡

φ

-

r

⁢

sin

⁡

φ

⁢

cos

⁡

λ

-

r

⁢

sin

⁡

φ

⁢

sin

⁡

λ

r

⁢

cos

⁡

φ

-

r

⁢

cos

⁡

φ

⁢

sin

⁡

λ

r

⁢

cos

⁡

φ

⁢

cos

⁡

λ

0

|

,

which is simplified to

r

2

⁢

cos

⁡

φ

.  The equation of the surface attains the form

r

6

=

3

⁢

a

3

⁢

r

3

⁢

cos

2

⁡

φ

⁢

sin

⁡

φ

⁢

cos

⁡

λ

⁢

sin

⁡

λ

,

or

r

=

3

⁢

a

3

⁢

cos

2

⁡

φ

⁢

sin

⁡

φ

⁢

cos

⁡

λ

⁢

sin

⁡

λ

3

:=

r

⁢

(

φ

,

λ

)

.

In the solid, we have

0

≦

r

≦

r

⁢

(

φ

,

λ

)

and

r

=

0

if only if

⁢

cos

2

⁡

φ

⁢

sin

⁡

φ

⁢

cos

⁡

λ

⁢

sin

⁡

λ

=

0

.

Thus we can write

V

=

∫

0

π

2

∫

0

π

2

∫

0

r

⁢

(

φ

,

λ

)

r

2

⁢

cos

⁡

φ

⁢

d

⁢

φ

⁢

d

⁢

λ

⁢

d

⁢

r

=

1

3

⁢

∫

0

π

2

∫

0

π

2

(

/

r

=

0

r

⁢

(

φ

,

λ

)

⁡

r

3

)

⁢

cos

⁡

φ

⁢

d

⁢

φ

⁢

d

⁢

λ

,

getting then

V

=

a

3

⁢

∫

0

π

2

(

cos

3

⁡

φ

)

⁢

(

-

sin

⁡

φ

)

⁢

𝑑

φ

⋅

∫

0

π

2

(

cos

⁡

λ

)

⁢

(

-

sin

⁡

λ

)

⁢

𝑑

λ

=

a

3

⁢

/

φ

=

0

π

2

⁡

cos

4

⁡

φ

4

⋅

/

λ

=

0

π

2

⁡

cos

2

⁡

λ

2

=

a

3

8

.

Remark.

The general for variable changing in a triple integral is

∭

D

f

⁢

(

x

,

y

,

z

)

⁢

𝑑

x

⁢

𝑑

y

⁢

𝑑

z

=

∭

Δ

f

⁢

(

x

⁢

(

ξ

,

η

,

ζ

)

,

y

⁢

(

ξ

,

η

,

ζ

)

,

z

⁢

(

ξ

,

η

,

ζ

)

)

⁢

|

∂

⁡

(

x

,

y

,

z

)

∂

⁡

(

ξ

,

η

,

ζ

)

|

⁢

𝑑

ξ

⁢

𝑑

η

⁢

𝑑

ζ

.

Title

example of Riemann triple integral

Canonical name

ExampleOfRiemannTripleIntegral

Date of creation

2013-03-22 19:10:59

Last modified on

2013-03-22 19:10:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Example

Classification

msc 26A42

Synonym

volume as triple integral

Related topic

Volume2

Related topic

VolumeAsIntegral

Related topic

SubstitutionNotation

Related topic

ChangeOfVariablesInIntegralOnMathbbRn

Related topic

ExampleOfRiemannDoubleIntegral

Generated on Fri Feb 9 19:53:10 2018 by

LaTeXML