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# SSM-Ch10 - Chap 10 Sec 10.1 Vector Integral Calculus...

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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 ue0a2 t ue0a3 u003d ue0a4 t , 5 t 2 ue0a5 , in components, x u003d t , y u003d 5 t 2 , (I) where t varies from t u003d 0 (the initial point of C on the x -axis) to t u003d 2 (the terminal point of C ). The given function is a vector function F u003d ue0a4 y 3 , x 3 ue0a5 . (II) F defines a vector field in the xy -plane. At each point ue0a2 x , y ue0a3 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 ue0a2 r ue0a2 t ue0a3ue0a3 u003d ue0a4 125 t 6 , t 3 ue0a5 . (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 u2032 ue0a2 t ue0a3 of C . This dot product F u2219 r u2032 can be “visualized” because it is the component of F in the direction of the tangent of C (times the factor | r u2032 ue0a2 t ue0a3 |), as you can see from (11) in Sec. 9.2 with F playing the role of a and r u2032 playing the role of b . (Note that if t is the arc length s of C , then r u2032 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 u2032 ue0a2 t ue0a3 u003d ue0a4 1, 10 t ue0a5 . Hence the dot product is F ue0a2 r ue0a2 t ue0a3ue0a3 u2219 r u2032 ue0a2 t ue0a3 u003d 125 t 6 u2219 1 u002b t 3 u2219 10 t u003d 125 t 6 u002b 10 t 4 . Integration with respect to t (the parameter in the parametric representation of the path of integration C ) from t u003d 0 to t u003d 2 gives u222b 0 2 F ue0a2 r ue0a2 t ue0a3ue0a3 u2219 r u2032 ue0a2 t ue0a3 dt u003d u222b 0 2 ue0a2 125 t 6 u002b 10 t 4 ue0a3 dt u003d 125 t 7 7 u002b 10 t 5 5 0 2 u003d 125 7 u2219 2 7 u002b 2 u2219 2 5 u003d 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.

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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 ue0a2 t ue0a3 u003d ue0a4 cos t , sin t , t ue0a5 with t
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