Integrals in cylindrical, spherical coordinates (Sect. 15 ...

1 Integrals in cylindrical, spherical coordinates (Sect. )IIntegration in spherical : Cylindrical coordinates in integral in spherical coordinates in coordinatesof a pointP R3is the ordered triple(r, ,z)defined by the :Cylindrical coordinates are just polar coordinates on theplanez= 0 together with the vertical (Cartesian-cylindrical transformations)The Cartesian coordinates of a point P= (r, ,z)are given byx=rcos( ),y=rsin( ), andz= cylindrical coordinates of a point P= (x,y,z)in the first andfourth quadrant arer= x2+y2, = arctan(y/x), andz= in cylindrical, spherical coordinates (Sect.)

2 IIntegration in spherical : Cylindrical coordinates in integral in spherical coordinates inR3 DefinitionThespherical coordinatesof a pointP R3is the ordered triple( , , )defined by the (Cartesian- spherical transformations)The Cartesian coordinates of P= ( , , )in the first quadrant aregiven byx= sin( ) cos( ),y= sin( ) sin( ), andz= cos( ).The spherical coordinates of P= (x,y,z)in the first quadrant are = x2+y2+z2, = arctan(yx), and = arctan( x2+y2z). spherical coordinates inR3 ExampleUse spherical coordinates to express region between the spherex2+y2+z2= 1 and the conez= x2+ :(x= sin( ) cos( ),y= sin( ) sin( ),z= cos( ).)

3 2y1/ 2xx + y = 1/222zz = 1 x y222z = x + yThe top surface is the sphere = bottom surface is the cone: cos( ) = 2sin2( )cos( ) = sin( ),so the cone is = :R={( , , ) : [0,2 ], [0, 4], [0,1]}. Integrals in cylindrical, spherical coordinates (Sect. )IIntegration in spherical : Cylindrical coordinates in integral in spherical integral in spherical coordinatesTheoremIf the function f:R R3 Ris continuous, then the tripleintegral of function f in the region R can be expressed in sphericalcoordinates as follows, Rf dv= Rf( , , ) 2sin( )d d d .Remark:ISpherical coordinates are useful when the integration regionRis described in a simple way using spherical the extra factor 2sin( )on the right-hand integral in spherical coordinatesExampleFind the volume of a sphere of :Sphere:S={ [0,2 ], [0, ], [0,R]}.

4 V= 2 0 0 R0 2sin( )d d d ,V=[ 2 0d ][ 0sin( )d ][ R0 2d ],V= 2 [ cos( ) 0]R33,V= 2 [ cos( ) + cos(0)]R33;hence:V=43 integral in spherical coordinatesExampleUse spherical coordinates to find the volume below the spherex2+y2+z2= 1 and above the conez= x2+ :R={( , , ) : [0,2 ], [0, 4], [0,1]}.The calculation is simple, the region is a simple section of a 2 0 /40 10 2sin( )d d d ,V=[ 2 0d ][ /40sin( )d ][ 10 2d ],V= 2 [ cos( ) /40]( 33 10),V= 2 [ 22+ 1]13 V= 3(2 2).CTriple integral in spherical coordinatesExampleFind the integral off(x,y,z) =e(x2+y2+z2)3/2in the regionR={x>0,y>0,z>0,x2+y2+z261}using :R={ [0, 2], [0, 2], [0,1]}.

5 Hence,I= Rf dv= /20 /20 10e 3 2sin( )d d d ,I=[ /20d ] [ /20sin( )d ] [ 10e 3 2d ].Use substitution:u= 3, hencedu= 3 2d , soI= 2[ cos( ) 20] 10eu3du Rf dv= 6(e 1).CTriple integral in spherical coordinatesExampleChange to spherical coordinates and compute the integralI= 2 2 4 x20 4 x2 y20y x2+y2+z2dz dy :(x= sin( ) cos( ),y= sin( ) sin( ),z= cos( ).)ILimits inx:|x|62;ILimits iny: 06y6 4 x2,sothe positive side of the diskx2+ inz:06z6 4 x2 y2,so apositive quarter of the ballx2+y2+ integral in spherical coordinatesExampleChange to spherical coordinates and compute the integralI= 2 2 4 x20 4 x2 y20y x2+y2+z2dz dy :(x= sin( ) cos( ),y= sin( ) sin( ),z= cos( ).)

6 2zxy22 ILimits in : [0. ];ILimits in : [0, /2];ILimits in : [0,2].IThe function to integrate is:f= 2sin( ) sin( ).I= 0 /20 20 2sin( ) sin( )( 2sin( ))d d d .Triple integral in spherical coordinatesExampleChange to spherical coordinates and compute the integralI= 2 2 4 x20 4 x2 y20y x2+y2+z2dz dy :I= 0 /20 20 2sin( ) sin( )( 2sin( ))d d d .I=[ 0sin( )d ][ /20sin2( )d ][ 20 4d ],I=( cos( ) 0)[ /2012(1 cos(2 ))d ]( 55 20),I= 212[( 2 0) 12(sin(2 ) /20)]255 I=24 integral in spherical coordinatesExampleCompute the integralI= 2 0 /30 2sec( )3 2sin( )d d d .Solution:Recall: sec( ) = 1/cos( ).

7 I= 2 /30( 3 2sec( ))sin( )d ,I= 2 /30(23 1cos3( ))sin( )d In the second term substitute:u= cos( ),du= sin( )d .I= 2 [23( cos( ) /30)+ 1/21duu3].Triple integral in spherical coordinatesExampleCompute the integralI= 2 0 /30 2sec( )3 2sin( )d d d .Solution:I= 2 [23( cos( ) /30)+ 1/21duu3].I= 2 [23( 12+ 1) 11/2u 3du]= 2 [4 (u 2 2 11/2)],I= 2 [4 +12(u 2 11/2)]= 2 [4 +12(1 22)]= 2 [82 32]We conclude:I= 5 .CTriple integral in spherical coordinates (Sect. )ExampleUse spherical coordinates to find the volume of the region outsidethe sphere = 2 cos( ) and inside the half sphere = 2 with [0, /2].

8 Solution:First sketch the integration = 2 cos( ) is a sphere,since 2= 2 cos( ) x2+y2+z2= 2zx2+y2+ (z 1)2= = 2 is a sphere radius 2and [0, /2] sayswe only considerthe upper half of the = 2xrho = 2 cos ( 0 )Triple integral in spherical coordinates (Sect. )ExampleUse spherical coordinates to find the volume of the region outsidethe sphere = 2 cos( ) and inside the sphere = 2 with [0, /2].Solution:2y2z12rho = 2xrho = 2 cos ( 0 )V= 2 0 /20 22 cos( ) 2sin( )d d d .V= 2 /20( 33 22 cos( ))sin( )d V=2 3 /20[8 sin( ) 8 cos3( ) sin( )]d .V=16 3[( cos( ) /20) /20cos3( ) sin( )d ].

9 Triple integral in spherical coordinates (Sect. )ExampleUse spherical coordinates to find the volume of the region outsidethe sphere = 2 cos( ) and inside the sphere = 2 with [0, /2].Solution:V=16 3[( cos( ) /20) /20cos3( ) sin( )d ].Introduce the substitution:u= cos( ),du= sin( )d .V=16 3[1 + 01u3du]=16 3[1 +(u44 01)]=16 3(1 14).V=16 334 V= 4 .C