1. For each point (x,y,z) in R3, the cylindrical coordinates (r,,z) are defined by the polar coordinates r and  (for x and y) together with z.
Example Find the cylindrical coordinates for each of the following:
(x , y , z) = (6 , 63 , 8)
(x , y , z) = (6 , –63 , 0)
(x , y , z) = (–6 3 , –6 , –23) .
(r ,  , z) = (12 , /3 , 8)
(r ,  , z) = (12 , 5/3 , 0)
(r ,  , z) = (12 , 7/6 , –23)

2. Example Find the rectangular (Cartesian) coordinates for each of
the following:
(r ,  , z) = (20 , /2 , 4)
(r ,  , z) = (20 , /4 , 4)
(r ,  , z) = (15 , 2/3 , –16)
(r ,  , z) = (6 , 4/3 , 0)
(r ,  , z) = (0 ,  , –3)
(x , y , z) = (0 , 20 , 4)
(x , y , z) = (102 , 102 , 4)
(x , y , z) = (–7.5 , 7.53 , –16)
(x , y , z) = (–3 , –33 , 0)
(x , y , z) = (0 , 0 , –3)

3. We can generalize the change of variables to integrals involving 3 (or any number of ) variables. If T(u,v,w) = (x(u,v,w), y(u,v,w), z(u,v,w)) is a transformation mapping the region W in R3 described by rectangular uvw coordinates to the region W* in R3 described by rectangular xyz coordinates, then
f(x,y,z) dx dy dz = W*
(x,y,z)
f(x(u,v,w),y(u,v,w),z(u,v,w)) ——— du dv dw
(u,v,w) W
x x x
— — —
u v w
(x,y,z) y y y
——— = det — — —
(u,v,w) u v w
z z z
where, of course,

4. Using what we already know about polar coordinates, we have
f(x,y,z) dx dy dz =
f(r cos  , r sin  , z) r dr d dz W* W
where W* and W are the same region described respectively in terms of x, y, and z and in terms of r, , and z. (See also page 399 of the text.)
Example Integrate the function f(x,y,z) = xyz over the region W where x, y, and z are all positive and between the cone z2 = x2 + y2 and the sphere x2 + y2 + z2 = 100. z y
The region W can be described by
  < , r  ,  z 
/2
50 r
 100 – r2 x
/2
50
(100 – r2)1/2
xyz dx dy dz =
(r cos )(r sin )zr dz dr d = W r

5. /2
50
(100 – r2)1/2
r3 z cos  sin  dz dr d =
/2
50
(100 – r2)1/2 r
r3 z2 cos  sin 
—————— dr d = 2
/2
50
z = r
100r3 – 2r5
————— cos  sin  dr d = 2
/2
50
/2
75r4 – r6
———— cos  sin  d = 6
31250
——— cos  sin  d = 3
15625
——— 3
r = 0

6. Example Find the volume inside the sphere of radius a defined by
x2 + y2 + z2 = a2 .
Let W be the region inside the sphere which can be described as
  < , r  ,  z  2 a
– a2 – r2
 a2 – r2 2 a
(a2 – r2)1/2 2 a
(a2 – r2)1/2
dx dy dz =
r dz dr d =
rz dr d = W
z = –(a2 – r2)1/2
–(a2 – r2)1/2 2 a 2 a 2
– 2(a2 – r2)3/2
————— d = 3
2a3
— d = 3
2r(a2 – r2)1/2 dr d =
r = 0
4a3 —— 3

7. For each point (x,y,z) in R3, the spherical coordinates (,,) are defined by
x =  sin  cos  , y =  sin  sin  , z =  cos  , where
 =  x2 + y2 + z2 is the length of the vector (x,y,z) ,
 = the angle that the vector (x,y,z) makes with the positive z axis,
 = the angle that the vector (x,y,0) makes with the positive x axis .
We have that 0   ,
0     , and
0   < 2 .
Also, note that  sin  = r =  x2 + y2 .

8. Example Find the spherical coordinates for each of the following:
(x , y , z) = (3 , –1 , 0)
(x , y , z) = (3 , –1 , 2)
(x , y , z) = (–3 , –1 , –2)
(x , y , z) = (0 , 0 , 10)
(x , y , z) = (0 , 0 , –10)
(x , y , z) = (0 , 0 , 0).
( ,  , ) = (2 , 11/6 , /2)
( ,  , ) = (22 , 11/6 , /4)
( ,  , ) = (22 , 7/6 , 3/4)
( ,  , ) = (10 ,  , 0)
( ,  , ) = (10 ,  , )
( ,  , ) = (0 ,  , )

9. Example Find the rectangular (Cartesian) coordinates for each of
the following:
( ,  , ) = (4 , /4 , /4)
( ,  , ) = (4 , 3/4 , 3/4)
( ,  , ) = (5 ,  , )
( ,  , ) = (2 ,  , 0) .
(x , y , z) = (2 , 2 , 22)
(x , y , z) = (–2 , 2 , –22)
(x , y , z) = (0 , 0 , –5)
(x , y , z) = (0 , 0 , 2)

10. Example Express each of the following surfaces in R3 in cylindrical coordinates and in spherical coordinates:
xyz = 1
x2 + y2 – z2 = 1
r2z sin cos = 1
3 sin2 cos sin cos = 1
r2 – z2 = 1
2 – 22cos2 = 1

11. Example Express each of the following surfaces in R3 in rectangular (Cartesian) coordinates, and describe the surface:
r = 9
 = 1
sin  = 0.5
cos  = 0.6
x2 + y2 = 81 This is a circular cylinder.
x2 + y2 + z2 = 1
This is a sphere of radius 1 centered at the origin.
3x2 + 3y2 – z2 = 0 for z  0
This is the “top” half of a cone.
y = 4x/3 or y = – 4x/3 for x  0
These are two half-planes.

12. If W* and W are the same region described respectively in terms of x, y, and z and in terms of , , and , then
f(x,y,z) dx dy dz = W*
(x,y,z)
———
(,,)
f( sin  cos  ,  sin  sin  ,  cos ) d d d = W
We need the Jacobian determinant.
f( sin  cos  ,  sin  sin  ,  cos ) d d d . W
x =  sin  cos 
y =  sin  sin  
z =  cos 
(x,y,z)
——— =
(,,)

13. x x x
— — —
  
y y y
det — — — =
z z z
x =  sin  cos 
y =  sin  sin  
z =  cos 
(x,y,z)
——— =
(,,)
sin  cos 
–  sin  sin 
 cos  cos 
det
sin  sin 
 sin  cos 
 cos  sin  =
cos 
–  sin 
| – 2 sin  | =
2 sin 

14. If W* and W are the same region described respectively in terms of x, y, and z and in terms of , , and , then
f(x,y,z) dx dy dz = W*
(x,y,z)
———
(,,)
f( sin  cos  ,  sin  sin  ,  cos ) d d d = W
f( sin  cos  ,  sin  sin  ,  cos ) 2 sin  d d d . W

15. Example Find the volume inside the sphere of radius a defined by
x2 + y2 + z2 = a2 .
Let W be the region inside the sphere which can be described as
  ,   < ,    a 2   2 a
dx dy dz =
2 sin  d d d = W a   2  2
3 sin 
——— d d = 3
a3 sin 
——— d d = 3
2 a3 sin 
————— d 3
 = 0
4a3 —— 3 =

16. Example Integrate the function f(x,y,z) = xyz over the region between the cone z2 = x2 + y2 and the sphere x2 + y2 + z2 = 36 where x, y, and z are all positive and x < y.
Let W be the region of integration which can be described as
  ,    ,    6
/4
/2
/4
xyz dx dy dz = W
/4
/2 6
(sincos)(sinsin)(cos) 2sin d d d =
/4

17. /4
/2 6
5 sin3 cos  sin  cos  d d d =
/4
/4
/2
7776 sin3 cos  sin  cos  d d =
/4
/4
/4
/2
1944 sin3 cos  d =
3888 sin3 cos  sin2 d =
 = /4
/4
486 sin4 =
121.5
 = 0

18. Example Find the volume of the “ice cream cone” above the xy plane described by the cone 3z2 = x2 + y2 and the sphere x2 + y2 + z2 = 25.
Let W be the region of integration which can be described as
  ,   < ,    5 2
/3
dx dy dz = W
/3 2 5
/3 2 5
2 sin  d d d =
3 sin 
——— d d = 3
 = 0

19. /3 2
/3
125sin 
——— d d = 3
250 sin 
———— d = 3
/3
– 250 cos 
————— = 3
– 125 250
——— + —— =
 = 0
125 —— 3