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Unformatted text preview: Chap . 10 Vector Integral Calculus . Integral Theorems Sec . 10 . 1 Line Integrals Problem Set 10 . 1 . Page 425 1 . Line integral in the plane . This generalizes a definite integral of calculus. Instead of integrating along the x-axis you now integrate over a curve C , a portion of a parabola, in the xy-plane. Integrals of the form (3) have various applications, for instance, in connection with work done by a force in a displacement. The right side of (3) shows how such an integral is converted to a definite integral with t as the variable of integration. The conversion is done by using the representation of the path of integration C . The problem parallels Example 1 in the text. The parabola C can be represented by r u t U u ¡ t , 5 t 2 ¢ , in components, x u t , y u 5 t 2 , (I) where t varies from t u 0 (the initial point of C on the x-axis) to t u 2 (the terminal point of C ). The given function is a vector function F u ¡ y 3 , x 3 ¢ . (II) F defines a vector field in the xy-plane. At each point u x , y U it gives a certain vector, which you could draw as a little arrow. In particular, at each point of C the vector function F gives a vector. You can obtain these vectors simply by substituting x and y from (I) into (II). This gives F u r u t UU u ¡ 125 t 6 , t 3 ¢ . (III) This is now a vector function of t defined on the parabola C . Now comes an important point to observe. You do not integrate F itself, but you integrate the dot product of F in (III) and the tangent vector r U u t U of C . This dot product F u r U can be “visualized” because it is the component of F in the direction of the tangent of C (times the factor | r U u t U |), as you can see from (11) in Sec. 9.2 with F playing the role of a and r U playing the role of b . (Note that if t is the arc length s of C , then r U is a unit vector, so that that factor equals 1 and you get exactly that tangential projection.) Think this over before you go on calculating. Differentiation with respect to t gives the tangent vector r U u t U u ¡ 1, 10 t ¢ . Hence the dot product is F u r u t UU u r U u t U u 125 t 6 u 1 ¡ t 3 u 10 t u 125 t 6 ¡ 10 t 4 . Integration with respect to t (the parameter in the parametric representation of the path of integration C ) from t u 0 to t u 2 gives U 2 F u r u t UU u r U u t U dt u U 2 u 125 t 6 ¡ 10 t 4 U dt u 125 t 7 7 ¡ 10 t 5 5 2 u 125 7 u 2 7 ¡ 2 u 2 5 u 2349.71 rounded to 2350 on p. A27 in Appendix 2 of the book. 7 . Line integral ( 3 ) in space . Line integrals in space are handled by the same method as line integrals in the plane. Quite generally, it is a great advantage of vector methods that methods in space and in the plane are very similar in most cases. Integrals (3), as just discussed, are suggested by work. 98 Linear Algebra, Vector Calculus Part B In the present problem, the path of integration C is a portion of a helix (see Example 4 on p. 391) C : r u t U u ¡ cos t , sin t , t ¢ with t varying from 0 to 4 u . It lies on the circular cylinder of radius....
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
- Spring '12