1 / 19

Example Find the cylindrical coordinates for each of the following:

For each point ( x,y,z ) in R 3 , 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)

chad
Télécharger la présentation

Example Find the cylindrical coordinates for each of the following:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related