x2 =a2 y2 =b2 z2 =c2 1. Ans: 1 2 abc 4 9.57. in VS .NET

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x2 =a2 y2 =b2 z2 =c2 1. Ans: 1 2 abc 4 9.57.
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If r is the region x2 xy y2 @ 1, prove that
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x u cos v sin , y u sin v cos and choose so as to eliminate the xy term in the integrand. Then let u a cos , v b sin  where a and b are appropriately chosen.] x x 9.58. Prove that
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x u n 1 F u du for n 1; 2; 3; . . . (see Problem 9.22).
Line Integrals, Surface Integrals, and Integral Theorems
Construction of mathematical models of physical phenomena requires functional domains of greater complexity than the previously employed line segments and plane regions. This section makes progress in meeting that need by enriching integral theory with the introduction of segments of curves and portions of surfaces as domains. Thus, single integrals as functions de ned on curve segments take on new meaning and are then called line integrals. Stokes s theorem exhibits a striking relation between the line integral of a function on a closed curve and the double integral of the surface portion that is enclosed. The divergence theorem relates the triple integral of a function on a three-dimensional region of space to its double integral on the bounding surface. The elegant language of vectors best describes these concepts; therefore, it would be useful to reread the introduction to 7, where the importance of vectors is emphasized. (The integral theorems also are expressed in coordinate form.)
LINE INTEGRALS The objective of this section is to geometrically view the domain of a vector or scalar function as a segment of a curve. Since the curve is de ned on an interval of real numbers, it is possible to refer the function to this primitive domain, but to do so would suppress much geometric insight. A curve, C, in three-dimensional space may be represented by parametric equations: x f1 t ; y f2 t ; z f3 t ; or in vector notation: x r t where r t xi yj zk (see Fig. 10-1). 229
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a@t@b
1 2
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
[CHAP. 10
Fig. 10-1
For this discussion it is assumed that r is continuously di erentiable. While (as we are doing) it is convenient to refer the Euclidean space to a rectangular Cartesian coordinate system, it is not necessary. (For example, cylindrical and spherical coordinates sometimes are more useful.) In fact, one of the objectives of the vector language is to free us from any particular frame of reference. Then, a vector A x t ; y t ; z t or a scalar, , is pictured on the domain C, which according to the parametric representation, is referred to the real number interval a @ t @ b. The Integral A dr 3
of a vector eld A de ned on a curve segment C is called a line integral. representation A1 dx A2 dy A3 dz obtained by expanding the dot product. The scalar and vector integrals t dt lim
C n!1 n X k 1 n X k 1
The integrand has the
k ; k ; k tk A k ; k ; k t k
4 5
A t dt lim
can be interpreted as line integrals; however, they do not play a major role [except for the fact that the scalar integral (3) takes the form (4)]. The following three basic ways are used to evaluate the line integral (3): 1. The parametric equations are used to express the integrand through the parameter t. Then A dr
t2
dr dt dt
If the curve C is a plane curve (for example, in the xy plane) and has one of the representations y f x or x g y , then the two integrals that arise are evaluated with respect to x or y, whichever is more convenient.
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
If the integrand is a perfect di erential, then it may be evaluated through knowledge of the end points (that is, without reference to any particular joining curve). (See the section on independence of path on Page 232; also see Page 237.)
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