By Donald D. Fitts
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Example text
Divide the surface into n infinitesimal surface elements of areas dst, . . is shown in Fig. 3-4. Let , \ 2,. . , An be the values of A(xj>,z ds2, . . , dsn, respectively. We define jdsas a vector of magnitude dsj with a direction normal to the surface at the point in question. If the unit vector normal to the surface at that point is n/, then we have dsj = n j d s j (3-56) 46 VECTOR CALCULUS Since the surface elements are infinitesimally small, a dsj is defined for each point on the surface S and, therefore, a vector field ds is determined.
3-72) to obtain / F ( 0 7 20 + 7 0 - 7 0 ) dv = fs (0 7 0 )-d s (3-73) We next subtract Eq. (3-73) from Eq. (3-72) and thereby obtain the desired result: f v (0 7 20 ~ 0 7 20) dv = fs (0 7 0 - 0 7 0 ) • ds Another useful integral relationship may be obtained by setting 0 = 1 in Eq. (3-74). In this case, V0 and 7 20 vanish and Eq. (3-74) reduces to $v V2
3 4 4 ) that A is irrotational. Thus, a necessary condition for a vector LINE INTEGRAL 43 field to be represented as the gradient of a scalar field is that the vector field be irrotational. It can be shown through the use of Stokes’ theorem (to be discussed in Sec. 3-12) that any irrotational vector field can be expressed as the gradient of a scalar field. Thus, the condition V X A = 0 is both necessary and sufficient for the relation A = V0 to be valid. The scalar function