shown in Fig. 1. These vectors define a parallelogram (Fig.1) lying in the tangent plane toSatP. The area of this parallelogram is[recall(10)in Section 14.2] q,dA=||RuduRvdv||=||RuRv||dudv.Asdu and dv tend to zero this plane area element lies closer and closer toSso that it seems reasonable to define the area of the curved surface Sby the double integral where Ris the region in the u,v plane that corresponds to the surface Sin3-space. Figure 1. Area element aA. the area element onS.To obtain a computational version of(4), cross Ru=xuhat(i)+yuhat(j)+zuhat(k) withRv=xvhat(i)+yvhat(j)+zvhat(k). The norm of the resulting vector is the square root of thesum of the squares of its components. Carrying out these steps, we obtaindA=EG-F22dudvE=xu2+yu2+zu2F=xuxv+yuyv+zuzvG=xv2+yv2+zv2Notice that we may attach a geometrical significance toF for, by inspection,we see that F=Ru*Rv. Thus, ifR(u,v)is such that Fis identically zero, thenthat condition implies that Ru and Rv are perpendicular to each other at each pointonS.In that event, we