Integrals in cylindrical, spherical coordinates (Sect. 15 ...
Integrals in cylindrical, spherical coordinates (Sect. 15.7)
Integration in spherical coordinates.
Review: Cylindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Cylindrical coordinates in space.
z
Definition
P
The cylindrical coordinates of a point
z
P R3 is the ordered triple (r , , z)
defined by the picture.
r
y
x
0
Remark: Cylindrical coordinates are just polar coordinates on the
plane z = 0 together with the vertical coordinate z.
Theorem (Cartesian-cylindrical transformations)
The Cartesian coordinates of a point P = (r , , z) are given by x = r cos(), y = r sin(), and z = z.
The cylindrical coordinates of a point P = (x, y , z) in the first and fourth quadrant are r = x2 + y 2, = arctan(y /x), and z = z.
Integrals in cylindrical, spherical coordinates (Sect. 15.7)
Integration in spherical coordinates.
Review: Cylindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Spherical coordinates in R3
Definition
z 0
The spherical coordinates of a point
rho
P R3 is the ordered triple (, , )
y
defined by the picture.
0
x
Theorem (Cartesian-spherical transformations)
The Cartesian coordinates of P = (, , ) in the first quadrant are given by x = sin() cos(), y = sin() sin(), and z = cos().
The spherical coordinates of P = (x, y , z) in the first quadrant are
= x2 + y 2 + z2, = arctan y , and = arctan
x2 + y2 .
x
z
Spherical coordinates in R3
Example
Use spherical coordinates to express region between the sphere x2 + y 2 + z2 = 1 and the cone z = x2 + y 2.
Solution: (x = sin() cos(), y = sin() sin(), z = cos().)
z z = 1- x2 - y2
The top surface is the sphere = 1. The bottom surface is the cone:
z = x 2+ y 2
cos() = 2 sin2() cos() = sin(),
1/ 2 y
x
x2+ y 2= 1/2
so the cone is = . 4
Hence: R = (, , ) : [0, 2], 0, , [0, 1] .
4
Integrals in cylindrical, spherical coordinates (Sect. 15.7)
Integration in spherical coordinates.
Review: Cylindrical coordinates. Spherical coordinates in space. Triple integral in spherical coordinates.
Triple integral in spherical coordinates
Theorem
If the function f : R R3 R is continuous, then the triple integral of function f in the region R can be expressed in spherical
coordinates as follows,
f dv =
R
f (, , ) 2 sin() d d d.
R
Remark:
Spherical coordinates are useful when the integration region R is described in a simple way using spherical coordinates. Notice the extra factor 2 sin() on the right-hand side.
Triple integral in spherical coordinates
Example
Find the volume of a sphere of radius R.
Solution: Sphere: S = { [0, 2], [0, ], [0, R]}.
2 R
V=
2 sin() d d d,
0 00
2
V=
d
0
sin() d
0
R
2 d ,
0
R3
V = 2 - cos()
,
03
R3 V = 2 - cos() + cos(0) ;
3
hence: V = 4 R3. 3
Triple integral in spherical coordinates
Example
Use spherical coordinates to find the volume below the sphere x2 + y 2 + z2 = 1 and above the cone z = x2 + y 2.
Solution: R = (, , ) : [0, 2], 0, , [0, 1] .
4 The calculation is simple, the region is a simple section of a sphere.
2 /4 1
V=
2 sin() d d d,
00
0
2
V=
d
/4
sin() d
1
2 d ,
0
0
0
/4 3 1
V = 2 - cos()
,
0
30
21
V = 2 - + 1
V = (2 - 2).
2
3
3
Triple integral in spherical coordinates
Example
Find the integral of f (x , y , z) = e(x2+y2+z2)3/2 in the region R = {x 0, y 0, z 0, x2 + y 2 + z2 1} using spherical
coordinates.
Solution: R = 0, , 0, , [0, 1] . Hence,
2
2
/2 /2 1
I=
f dv =
e32 sin() d d d ,
R
0
0
0
/2
I=
d
0
/2
sin() d
0
1
e3 2 d .
0
Use substitution: u = 3, hence du = 32 d, so
I=
- cos() 2
1 eu du
2
0 03
f dv = (e - 1).
R
6
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