Example Find the cylindrical coordinates for each of the following:

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 — — — u v w 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  0 /2 50 r  100 – r2 x /2 50 (100 – r2)1/2 xyzdx dy dz = (r cos )(r sin )zrdz dr d = W 0 0 r

5. /2 50 (100 – r2)1/2 r3z cos  sin dz dr d = /2 50 (100 – r2)1/2 0 0 r r3z2cos  sin  —————— dr d = 2 /2 50 0 0 z = r 100r3– 2r5 ————— cos  sin dr d = 2 0 0 /2 50 /2 75r4– r6 ———— cos  sin d = 6 31250 ——— cos  sin d = 3 15625 ——— 3 0 r = 0 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  0 2 a – a2– r2  a2– r2 2 a (a2– r2)1/2 2 a (a2– r2)1/2 dx dy dz = rdz dr d = rz dr d = W z = –(a2– r2)1/2 0 0 –(a2– r2)1/2 0 0 2 a 2 a 2 – 2(a2– r2)3/2 ————— d = 3 2a3 — d = 3 2r(a2– r2)1/2 dr d = 0 0 0 r = 0 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  0 –  sin  | – 2sin  | = 2sin 

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 ) 2sin 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 0 2 0   2 a dx dy dz = 2sin d d d = W 0 0 0 a   2  2 2 a3sin  ————— d 3 3sin  ——— d d = 3 a3sin  ——— d d = 3 0 0 0 0 0  = 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 0 /4 xyz dx dy dz = W /4 /2 6 (sincos)(sinsin)(cos) 2sind d d = 0 /4 0

17. /4 /2 6 5 sin3 cos  sin  cos d d d = 0 /4 0 /4 /2 7776 sin3 cos  sin  cos d d = 0 /4 /4 /4 /2 1944 sin3 cos d = 3888 sin3 cos  sin2 d =  = /4 0 0 /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 0 2 0 /3 dx dy dz = W /3 2 5 /3 2 5 2 sin d d d = 3 sin  ———d d = 3 0 0 0 0 0  = 0

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