Application of Cylindrical and Spherical coordinate system in double-triple integration

APPLICATION OF CYLINDRICAL ANDAPPLICATION OF CYLINDRICAL AND
SPHERICAL COORDINATE SYSTEM INSPHERICAL COORDINATE SYSTEM IN
DOUBLE-TRIPLE INTEGRATIONDOUBLE-TRIPLE INTEGRATION
DR. Sonendra Kumar Gupta
(Associate Professor)
Department of Basic Science
(Engineering Mathematics)

CONTENTSCONTENTS
►Jacobian Transformation function
►Use in Double integration.
►Use in Triple integration.
►Polar Coordinate system
►Use of polar coordinate in Double integral
►Solved example in polar coordinate
►Cylindrical Coordinate system
►Use of Cylindrical Coordinate in Triple Coordinate
►Spherical Coordinate System
►Limits of Spherical Coordinate System
►Problem on Spherical Coordinate System
►Problem on Cylindrical Coordinate System

JACOBIAN FUNCTIONJACOBIAN FUNCTION
Jacobian function is a transformation function , which convert a function into one
plane to another plane.
Y
X
v
u
P(x, y) P(u, v)
( ) ( ), ,P x y P u v⇒
It is denoted by J(u, v) and defined as
( )
( )
( )
,
,
,
x x
x y u v
J u v
y yu v
u v
∂ ∂
∂ ∂ ∂
= =
∂ ∂∂
∂ ∂
Carl Gustav Jacob
10 December 1804 – 18 February 1851
German mathematician

USE IN DOUBLE INTEGRATIONUSE IN DOUBLE INTEGRATION
Double integral
1 2
( , ) ( , )
R R
f x y dx dy f u v J du dv=
∫∫ ∫∫
Where
( )
( )
( )
,
,
,
x x
x y u v
J u v
y yu v
u v
∂ ∂
∂ ∂ ∂
= =
∂ ∂∂
∂ ∂

USE IN TRIPLE INTEGRATIONUSE IN TRIPLE INTEGRATION
Triple integral
1 2
( , , ) ( , , )
V V
f x y z dx dy dz f u v w J du dv dw=
∫∫∫ ∫∫∫
Where
( )
( )
( )
, ,
, ,
, ,
x x x
u v w
x y z y y y
J u v w
u v w u v w
z z z
u v w
∂ ∂ ∂
∂ ∂ ∂
∂ ∂ ∂ ∂
= =
∂ ∂ ∂ ∂
∂ ∂ ∂
∂ ∂ ∂

POLAR COORDINATE SYSTEMPOLAR COORDINATE SYSTEM
( ) ( ), ,P x y P r θ⇒
x
y
2 2
r x y= +
X
Y
cosx r= θ
θ
siny r= θ
( ) ( ), ,P x y P r= θ
Cartesian to Polar Coordinate
cos , sinx r y r= θ = θ
Therefore, we have
2 2
tan
y
r x y and
x
= + θ =
Y
X
θ
r
P(x, y) P(r, θ)

Use of polar coordinate in Double integralUse of polar coordinate in Double integral
1 2
( , ) ( cos , sin )
R R
f x y dx dy f r r J dr d= θ θ θ
∫∫ ∫∫
( )
( )
( )
( )2 2
cos sin,
,
sin cos,
cos sin
x x
rx y r
and J r
y y rr
r
r r
∂ ∂
θ − θ∂ ∂ ∂θ
θ = = =
∂ ∂ θ θ∂ θ
∂ ∂θ
= θ+ θ =
cos cos , sin
sin sin , cos
x x
Here x r r
r
y y
y r r
r
∂ ∂
= θ ⇒ = θ = − θ
∂ ∂θ
∂ ∂
= θ ⇒ = θ = θ
∂ ∂θ
1 2
( , ) ( cos , sin )
R R
f x y dx dy f r r r dr d= θ θ θ
∫∫ ∫∫∴

Solve the integral
2
2 2
2 20 0
x x x
dx dy
x y
−
+
∫ ∫ by changing to polar coordinate system.
Solution: Given
2
2 2
2 20 0
x x x
I dx dy
x y
−
=
+
∫ ∫
The region of integration bounded by following limits,
(i). y = 0 i.e. X-axis
( ) ( )2 22 2 2
( ). 2 2 0 . . 1 0 1ii y x x x y x i e x y= − ⇒ + − = − + − =
equation of circle with centre (1, 0) and radius 1.
(iii). x = 0 i.e., Y-axis.
(iv). x = 2 i.e. straight line parallel to Y-axis.
Using polar coordinate system,
Putting x = r cos θ and y = r sinθ, So that 2 2 2
r x y= +
1
tan
y
x
−  
θ =  ÷
 
and
…. (1)
Use of Polar Coordinate in Double integralUse of Polar Coordinate in Double integral
Y
X
0x =
( )2, 0A
2x =
2 2
2 0x y x+ − =
g1 1
( )1, 0C( )0, 0O

Y
X
0x =
( )2, 0A ( )2, 0A
2x =
2 2
2 0x y x+ − =
g1 1
( )1, 0C( )0, 0O
Polar Coordinate System
Limits in polar form:
(i) Since, 2 2
2 0x y x+ − =
2
2 cos 0r r⇒ − θ =
( )2cos 0r r⇒ − θ =
0, 2cosr r⇒ = = θ
(ii). At y = 0, then 1 0
tan 0
x
−  
θ = = ÷
 
and at x = 0, then ( )1 1
tan tan
0 2
y− − π 
θ = = ∞ = ÷
 
Thus limits are,
0 2cos , 0
2
r and dx dy r dr d
π
≤ ≤ θ ≤ θ ≤ = θ

2
2 2 /2 2cos
2 20 0 0 0
cosx x x r
dx dy r dr d
rx y
− π θ θ 
= θ ÷
 +
∫ ∫ ∫ ∫
/2 2cos
0 0
cos r dr d
π θ 
= θ θ 
 ∫ ∫
2cos2 2/2 /2
0 0
0
4cos
cos cos 0
2 2
r
d d
θ
π π   θ
= θ θ = θ − θ   
      
∫ ∫
/2
0 3
0
0 1 3 1
2 2
2 sin cos 2
0 3 2
2
2
d
π
 + +    
Γ Γ ÷  ÷ 
    = θ θ θ =
+ +  Γ ÷   
∫
( )
1
2
1.1 42
3 15 3
2 22
 
Γ Γ ÷ π = = =
  × × πΓ ÷
 
2
2 2 /2 2cos
2 20 0 0 0
cos 4
3
x x x r
dx dy r dr d
rx y
− π θ θ
= θ =
+
∫ ∫ ∫ ∫
From equation (1), we get

CYLINDRICAL AND
SPHERICAL COORDINATES
Integration

CYLINDRICAL COORDINATES.CYLINDRICAL COORDINATES.
φr
Recall polar representations in the plane
( ) ( ), , , ,P x y z P r zφ⇒

REPRESENTING 3D POINTS INREPRESENTING 3D POINTS IN
φr
Cylindrical coordinates just adds a z-coordinate to the polar coordinates (r, φ).

φr
φr
(r, φ, z)

CONVERTING BETWEENCONVERTING BETWEEN
RECTANGULAR ANDRECTANGULAR AND
CYLINDRICAL COORDINATESCYLINDRICAL COORDINATES
cos
sin
x r
y r
z z
φ
φ
=
=
=
φr
φr
(r, φ,z)
Rectangular to Cylindrical
2 2 2
tan( )
r x y
y
x
z h
φ
= +
=
=
Cylindrical to rectangularNo real surprises here!

INTEGRATION ELEMENTS:INTEGRATION ELEMENTS:
RECTANGULARRECTANGULAR
COORDINATESCOORDINATES
We know that in a Riemann Sum approximation for a
triple integral, a summand
This computes the function value at some point in the little
“sub-cube” and multiplies it by the volume of the little
cube of length , width , and height .
* * *
( , , ) .i i i i i if x y z x y z∆ ∆ ∆
* * *
function value volume of the small
at a sampling point cube
( , , ) .i i i i i if x y z x y z∆ ∆ ∆
14243 14243
ix∆ iy∆ iz∆

CYLINDRICALCYLINDRICAL
What happens when we consider small changes
in the cylindrical coordinates r, φ, and z?
, , andr z∆ ∆φ ∆
We no longer get a
cube, and (similarly to
the 2D case with polar
coordinates) this affects
integration.

, , andr z∆ ∆φ ∆
Start with our previous
picture of cylindrical
coordinates:
φr
φr

, , andr z∆ ∆φ ∆
Start with our previous
picture of cylindrical
coordinates:
Expand the radius by a
small amount:
r∆
r
r+∆r
r

r+∆r
r
This leaves us with a thin cylindrical shell of inner radius r
and outer radius r +∆ r.
r
r+∆r

INTEGRATION ELEMENTS:
CYLINDRICAL
COORDINATES
Now we consider the angle φ.
We want to increase it by a small amount ∆φ.
φ
φ
∆φ

φ
This give us a “wedge.”
Combining this with the cylindrical shell created by the
change in r, we get

This give us a “wedge.”
Intersecting this wedge with the cylindrical shell created by
the change in r, we get

Finally , we look at a small vertical change ∆ z .

INTEGRATION IN CYLINDRICALINTEGRATION IN CYLINDRICAL
dA r dr d≈ φ
We need to find the volume of this little solid. As in polar coordinates,
we have the area of a horizontal cross section is. .

INTEGRATION IN CYLINDRICALINTEGRATION IN CYLINDRICAL
COORDINATES.COORDINATES.
dV r dr d dz≈ φ
We need to find the volume of this little solid.
Since the volume is just the base times the height. . .
So . . .
( , , )
S
f r z r dr d dzφ φ∫∫∫

Use of Cylindrical Coordinate in Triple integralUse of Cylindrical Coordinate in Triple integral
1 2
( , , ) ( cos , sin , )
V V
f x y z dx dy dz f r r z J dr d dz= φ φ φ
∫∫∫ ∫∫∫
cos cos , sin , 0
sin sin , cos , 0
0, 0, 1
x x x
Here x r r
r z
y y y
y r r
r z
z z z
z z
r z
∂ ∂ ∂
= φ ⇒ = φ =− φ =
∂ ∂φ ∂
∂ ∂ ∂
= φ ⇒ = φ = φ =
∂ ∂φ ∂
∂ ∂ ∂
= ⇒ = = =
∂ ∂φ ∂
( )
( )
( )
cos sin 0
, ,
, , sin cos 0
, ,
0 0 1
x x x
r z
r
x y z y y y
J r z r r
r z r z
z z z
r z
∂ ∂ ∂
∂ ∂φ ∂
φ − φ
∂ ∂ ∂ ∂
φ = = = φ φ =
∂ φ ∂ ∂φ ∂
∂ ∂ ∂
∂ ∂φ ∂
and
1 2
( , , ) ( cos , sin , )
V V
f x y z dx dy dz f r r z r dr d dz∴ = φ φ φ
∫∫∫ ∫∫∫

Problem on Cylindrical Coordinate systemProblem on Cylindrical Coordinate system
Evaluate
over the region bounded by the paraboloid x2
+ y2
= 3z and the plane z = 3
( )2 2
x y dxdydz∫ ∫ ∫ +
Solution: Given ( )2 2
I x y dxdy dz∫ ∫ ∫= +
Using the cylindrical coordinate system,
x = r cos θ , y = r sin θ and z = z, where 2 2 2
, tan
y
r x y and
x
= + =θ
(i). Now, the equation of paraboloid,
2 2 2
3 3x y z r z+ = ⇒ =
(ii). Draw the elementary volume AB parallel to z-axis in the bounded region, which
starts from the paraboloid r2
= 3z and terminates on the plane z = 3
∴ Limits of z :
2
3
3
r
z to z= =
(iii). Projection of the region in rθ–plane in the curve of intersection of the paraboloid
r2
= 3z and plane z = 3, which is obtained as r2
= 9 i.e. r = 3, a circle with centre at the
origin and radius 3.
(iv). Draw the elementary radius vector OA′ in the region (circle r = 3) in XY(rθ) plane,
which starts from the origin and terminates on the circle r = 1
… (1)

Limits in Cylindrical Coordinate form are
Limits of z :
Limits of r :
Limits of θ :
2
3
3
r
z to z= =
r = 0 to r = 3
θ = 0 to θ = 2π
and dxdydz r dr d dz= θ
Z
Y
X
3z =
Cylindrical Coordinate
0r =
3r =
2 2
3x y z+ =
2 2
9x y+ =
From equation (1), we have
( ) ( )2
2 3 32 2 2
0 0
3
rr
z
x y dxdydz r r dr d dz= =
=
∫ ∫ ∫ ∫ ∫ ∫+ = π
θ θ
2
2 3 33
0 0 /31r z rr dz dr d= = =∫ ∫ ∫=   
π
θ θ

[ ] 2
2
32 3 2 33 3
0 0 0 0/3
3
3
r rr
r
r z dr d r dr d= = = =∫ ∫ ∫ ∫
 
= = − 
 
π π
θ θθ θ
[ ]
35 4 6
22 3 3
0 0 0
0
3
1 3
3 4 18
r r r
d r dr=∫ ∫
   
= − = × − ÷  
   
ππ
θ θ θ
( )
5 6
53 3 9 6
2 0 2 3
4 18 36
−   
= − × − = × ×     
π π
5 3 81
2 3
36 2
= × × =
π
π
( ) ( )2
2 3 32 2 2
0 0
3
81
2
rr
z
x y dxdydz r r dr d dz= =
=
∫ ∫ ∫ ∫ ∫ ∫+ = =π
θ
π
θ

SPHERICAL COORDINATESSPHERICAL COORDINATES
Spherical Coordinates are the 3D analog of polar
representations in the plane.
We divide 3-dimensional space into
1. a set of concentric spheres centered at the origin.
2. rays emanating outward from the origin
( ) ( ), , , ,P x y z P r φ θ⇒

(x,y,z) We start with a point (x, y, z)
given in rectangular coordinates.
Then, measuring its distance r
from the origin, we locate it on a
sphere of radius r centered at the
origin.
Next, we have to find a way to
describe its location on the sphere.
r

We use a method similar to the method
used to measure latitude and longitude on
the surface of the Earth.
We find the great circle that goes through
the “north pole,” the “south pole,” and the
point.

SPHERICAL COORDINATES SYSTEMSPHERICAL COORDINATES SYSTEM
θ
We measure the latitude or polar
angle starting at the north pole in the
plane given by the great circle.
This angle is called θ. The range of
this angle is
Note:
all angles are measured in
radians, as always.
0 .≤ ≤θ π

We use a method similar to the
method used to measure latitude and
longitude on the surface of the Earth.
Next, we draw a horizontal circle on
the sphere that passes through the
point.

and “drop it down” onto the xy-plane.

We measure the latitude or azimuthal
angle on the latitude circle, starting at the
positive x-axis and rotating
toward the positive y-axis.
The range of the angle is
Angle is called φ.
0 2 .≤ <φ π
Note that this is the same angle as the φ in cylindrical coordinates!

FINALLY, A POINT INFINALLY, A POINT IN
SPHERICAL COORDINATES!SPHERICAL COORDINATES!
(r ,φ ,θ)
Our designated point on the sphere is
indicated by the three spherical
coordinates (r, φ, θ ) … (radial distance,
azimuthal angle, polar angle).
Please note that this notation is not at all
standard and varies from author to author
and discipline to discipline. (In
particular, physicists often use θ to refer
to the azimuthal angle and φ refer to the
polar angle.)
r

RECTANGULAR AND SPHERICALRECTANGULAR AND SPHERICAL
θ
(x,y,z)
z
r
ρ
First note that if ρ is the usual cylindrical
coordinate for (x,y,z)
we have a right triangle with
•acute angle θ,
•hypotenuse r, and
•legs ρ and z.
It follows that
.
sin( ) sin
.
cos( ) cos
.
.
tan( ) tan
Per
r
Hyp r
Base z
z r
Hyp r
Per
z
Base z
ρ
θ ρ θ
θ θ
ρ
θ ρ θ
= = ⇒ =
= = ⇒ =
= = ⇒ =
What happens if
θ is not acute?

RECTANGULAR AND SPHERICALRECTANGULAR AND SPHERICAL
θ
(x,y,z)
z
r
ρ
Spherical to rectangular in
φ
ρ
φ
ρ
cos( )x ρ φ=
sin( )y ρ φ=
( ). cos( )i x = ρ φ
sin( )cos( )x r⇒ = θ φ
( ). sin( )ii y = ρ φ
sin( )sin( )y r⇒ = θ φ
( ) cos( )iii z r= θ

CONVERTING FROMCONVERTING FROM
SPHERICAL TO RECTANGULARSPHERICAL TO RECTANGULAR
θ
(x,y,z)
z
r
ρ
Rectangular to Spherical
φ
ρ
2 2 2
r x y z= + +
tan( )
y
x
=φ
2 2
tan( )
x y
z z
+
= =
ρ
θ
2 2
x y= +ρ

LIMITS OF SPHERICAL COORDINATELIMITS OF SPHERICAL COORDINATE
SYSTEMSYSTEM
If the region of integration is a sphere x2
+ y2
+z2
= a2
with centre at (0, 0, 0) and
radius a, then limits of r, φ, θ are
1. Positive octant of a sphere
: 0
: 0 / 2
: 0 / 2
r r to r a
to
to
= =
θ θ= θ=π
φ φ= φ=π
2. Hemisphere (Above XY-plane, i.e. z > 0)
: 0
: 0 / 2
: 0 2
r r to r a
to
to
= =
θ θ= θ=π
φ φ= φ= π

Use of Spherical coordinate in Triple integralUse of Spherical coordinate in Triple integral
1 2
( , , ) ( sin cos , sin sin , cos )
V V
f x y z dxdy dz f r r r J dr d d= θ φ θ φ θ φ θ
∫∫∫ ∫∫∫
sin cos sin cos , sin sin , cos cos
x x x
x r r r
r
∂ ∂ ∂
= θ φ ⇒ = θ φ = − θ φ = θ φ
∂ ∂φ ∂θ
sin sin sin sin , sin cos , cos sin
y y y
y r r r
r
∂ ∂ ∂
= θ φ ⇒ = θ φ = θ φ = θ
∂ ∂φ ∂θ
cos cos , 0, sin
z z z
z r r
r
∂ ∂ ∂
= θ ⇒ = θ = = − θ
∂ ∂φ ∂θ
∴
and
( )
( )
( )
2
sin cos sin sin cos cos
, ,
, , sin sin sin cos cos sin sin
, ,
cos 0 sin
x x x
r
r r
x y z y y y
J x y z r r r
r r
r
z z z
r
∂ ∂ ∂
∂ ∂φ ∂θ
θ φ − θ φ θ φ
∂ ∂ ∂ ∂
= = = θ φ θ φ θ φ = θ
∂ φ θ ∂ ∂φ ∂θ
θ − θ
∂ ∂ ∂
∂ ∂φ ∂θ
1 2
2
( , , ) ( sin cos , sin sin , cos ) sin
V V
f x y z dxdy dz f r r r r dr d d∴ = θ φ θ φ θ θ φ θ
∫∫∫ ∫∫∫

Problem on Spherical Coordinate systemProblem on Spherical Coordinate system
Evaluate
2
2 2
1 1 1
0 0 2 2 2
x
x y
dxdydz
x y z
−
+∫ ∫ ∫
+ +
Solution:
2
2 2
1 1 1
0 0 2 2 2
x
x y
dxdydz
I
x y z
−
+∫ ∫ ∫=
+ +
It is difficult to integrate in Cartesian form, so we can solve by spherical coordinate
system and using the following steps.
Given the limits of integration is
Limits of x : 0 1x to x= =
Limits of y : 2 2 2
0 1 1y to y x x y= = − ⇒ + =
Limits of z : 2 2
1z x y to z= + = i.e., equation of cone 2 2 2
,z x y= + 1z =
Clearly the region of integration is bounded by the cone 2 2 2
,z x y= +
and the cylinder x2
+ y2
= 1, plane z = 1, in the positive octant of the plane as shown in
figure.
Now, using spherical coordinate system
x = r sinθ cosφ, y = r sin θ sinφ, z = r cos θ and r2
= x2
+ y2
+ z2
… (1)

2 2
2
tan , tan , sin
x yy
and dxdydz r dr d d
x z
+
= = =φ θ θ θ φ
(i). z = r cosθ
1 = r cosθ i.e., r = secθ
∴ 0 ≤ r ≤ secθ
Limits in Spherical Coordinate form
2 2 2
tan 1
4
x y z
z z
+
= = = ⇒ =
π
θ θ(ii).
at y = 0 ⇒
1 0
tan 0
x
−  
= = ÷
 
φ
and at x = 0 ⇒
1
tan
0 2
y−  
= = ÷
 
π
φ
(iii).
∴ Limits in spherical coordinate system are
Limits of r : r = 0 to r = secθ
Limits of θ : θ = 0 to θ = π/4
Limits of φ : φ = 0 to φ = π/2
From equation (1), we have
z
X
Y
φ
θ r
2 2
1x y+ =
2 2 2
z x y= +
O

2
2 2
1 1 1 / 2 / 4 sec 2
0 0 0 0 02 2 2 2
1
sinx
x y r
dxdydz
r dr d d
x y z r
−
+ = = =∫ ∫ ∫ ∫ ∫ ∫=
+ +
π π θ
φ θ θ θ φ
/ 2 / 4 sec
0 0 0 sinr r dr d d= = =∫ ∫ ∫= π π θ
φ θ θ θ φ
/ 2 / 4 sec
0 0 0sin r r dr d d= = =∫ ∫ ∫=   
π π θ
φ θ θ θ φ
sec2
/ 2 / 4
0 0
0
sin
2
r
d d= =∫ ∫
 
=  
 
θ
π π
φ θ θ θ φ
/ 2 / 4 2
0 0
1
sin sec
2
d d= =∫ ∫= π π
φ θ θ θ θ φ
/ 2 / 4
0 0
1
sec tan
2
d d= =∫ ∫=   
π π
φ θ θ θ θ φ
[ ]
/ 4/ 2 / 2
0 00
1 1
sec sec 1
2 2 4
d d= =∫ ∫
 
= = −  
ππ π
φ φ
π
θ φ φ
[ ] ( )/ 2
0
1
2 1 2 1
2 4
 = × − = − 
π π
φ
( )
2
2 2
1 1 1 / 2 / 4 sec 2
0 0 0 0 02 2 2 2
1
sin 2 1
4
x
x y r
dxdydz
r dr d d
x y z r
−
+ = = =∫ ∫ ∫ ∫ ∫ ∫= = −
+ +
π π θ
φ θ
π
θ θ φ

Application of Cylindrical and Spherical coordinate system in double-triple integration

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Application of Cylindrical and Spherical coordinate system in double-triple integration