Vectors Calaulus 1.24

Triple Integrals in Cylindrical and Spherical Coordinates Cylindrical Coordinates        When we were working with double integrals, we saw that it was often easier to convert to polar coordinates.  For triple integrals we have been introduced to three coordinate systems.  The rectangular coordinate system (x,y,z) is the system that we are used to.  The other two systems, cylindrical coordinates (r,q,z) and spherical coordinates  (r,q,f) are the topic of this discussion.  Recall that cylindrical coordinates are most appropriate when the expression         x2 + y2  occurs.  The construction is just an extension of polar coordinates.          x  =  r cos q        y  =  r sin q        z  =  z Since triple integration can be looked at as iterated integration we have         This leads us the the following theorem Theorem:  Integration With Cylindrical  Coordinates Let f(x,y,z) be a continuous function on a solid Q.  Then            Example Find the moment of inertia about the z-axis of the solid that lies below the paraboloid         z  =  25 - x2 - y2  inside the cylinder         x2 + y2  =  4 above the xy-plane, and has density function         r(x,y,z)  =  x2 + y2 + 6z             Solution     By the moment of inertia formula, we have         The region, being inside of a cylinder is ripe for cylindrical coordinates.  We get         Spherical Coordinates Another coordinate system that often comes into use is the spherical coordinate system.  To review, the transformations are          x  =  r cosq sinf        y  =  r sinq sinf        z  =  r cosf   In the next section we will show that          dzdydx  =  r2 sinf drdfdq   This leads us to  Theorem:  Integration With Spherical Coordinates Let f(x,y,z) be a continuous function on a solid Q.  Then            Example Find the volume of solid that lies inside the sphere         x2 + y2 + z2  =  2 and outside of the cone         z2  =  x2 + y2              Solution We convert to spherical coordinates.  The sphere becomes         r  =  To convert the cone, we add z2 to both sides of the equation         2z2  =  x2 + y2 +z2  Now convert to         2r2cos2f  =  r2  Canceling the r2 and solving for f we get         f  =  cos-1(1/)  =  p/4  or 7p/4   In spherical coordinates (since the coordinates are p periodic)         7p/4  =  3p/4 To find the volume we compute                   Evaluating this integral should be routine at this point and is equal to                    8p      V  =                                      3