fiziks 
                                  Institute for NET/JRF, GATE, IIT-JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics 
                                                                                                       
                                                                                   4(b). Surface Integrals 
                              A surface integral is an expression of the form                                                        z
                                                               Ada                                                                                          da
                                                             
                                                              S
                              where  A is again some vector function, and da  is 
                              an  infinitesimal  patch  of  area,  with  direction                                                                                         y
                              perpendicular  to  the  surface(as  shown  in  figure).  x
                              There are, of course, two directions perpendicular to any surface, so the sign of a surface 
                              integral is intrinsically ambiguous. If the surface is closed then “outward” is positive, but 
                              for open surfaces it’s arbitrary. 
                               If  A  describes  the  flow  of  a  fluid  (mass  per  unit  area  per  unit  time),  then                                           Ada 
                                                                                                                                                                 
                              represents the total mass per unit time passing through the surface-hence the alternative 
                              name, “flux.” 
                              Ordinarily, the value of a surface integral depends on the particular surface chosen, but 
                              there is a special class of vector functions for which it is independent of the surface, and 
                              is determined entirely by the boundary line.  
                               
                               
                               
                               
                               
                               
                               
                               
                               
                               
                               
                               
                               
                               
                                        H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 
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                                                                                       fiziks 
                                  Institute for NET/JRF, GATE, IIT-JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics 
                                                                                                        
                               Example:  
                                                                                                           ˆ                ˆ          2        ˆ
                               Calculate  the  surface  integral  of  A  2xzx x  2y  yz 3z over  five  sides 
                               (excluding the bottom) of the cubical box (side 2) as shown in figure. Let “upward and 
                               outward” be the positive direction, as indicated by the arrows. 
                                
                               Solution:  
                               Taking the sides one at a time: 
                                                                                                                                     z          (v)          (ii)
                                                                 ˆ
                               (i)    x  2, da  dydzx, Ada  2xzdydz  4zdydz,                                                   2
                               so      Ada 4 2dy 2zdz 16. 
                                                            
                                                        0      0                                                       (iv)        (i)                        (iii)
                                                                        ˆ                                                                                2                 y
                               (ii)       x  0, da  dydzx, Ada  2xzdydz  0, 
                                                                                                                                 2
                               so Ada0.                                                                            x
                                   
                                                                    ˆ                                                               2                 2
                               (iii)  y  2, da  dxdz y, Ada  x  2dxdz, so  Ada                                               x  2dx         dz 12. 
                                                                                                                                                   
                                                                                                                                   0                  0
                                                                      ˆ                                                                   2                  2
                               (iv)  y  0, da  dxdz y, Ada  x  2dxdz, so  Ada                                                   x  2dx         dz  12. 
                                                                                                                                                         
                                                                                                                                          0                 0
                                                                  ˆ                       2                                                         2       2
                               (v)  z  2, da  dxdy z, Ada  yz 3dxdy  ydxdy, so Ada                                                          dx      ydy  4 
                                                                                                                                                        
                                                                                                                                                   0       0
                               Evidently the total flux is             
                                                                      Ada1601212420 
                                                                   
                                                                     surface
                                
                                         H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 
                                                                      Phone: 011-26865455/+91-9871145498 
                                                      Website: www.physicsbyfiziks.com  | Email: fiziks.physics@gmail.com  
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