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## Calculus IV

by: Felicita Lockman

25

0

2

# Calculus IV MATH 2242

Felicita Lockman
UNCC
GPA 3.68

Michael Fairchild

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COURSE
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Michael Fairchild
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Felicita Lockman on Sunday October 25, 2015. The Class Notes belongs to MATH 2242 at University of North Carolina - Charlotte taught by Michael Fairchild in Fall. Since its upload, it has received 25 views. For similar materials see /class/228910/math-2242-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.

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Date Created: 10/25/15
Marsden 7215 Evalute the line integral 2zyz dm 22 dy zzy dz 0 where C is an oriented simple curve connecting 111 to 1 2 4 SOLUTION One hint is that the problem does not tell you what path to integrate along The wording of the problem seems to suggest that no matter what path you pick you get the same answer Question What kind of line integrals have this property Answer Line integrals of gradient elds see Theorem 73 on p440 Thus according to Theorem 73 if you can nd a scalar function 9 such that Vg Qxyz x22 y then 124 2xyzdx szzdy zzy dz Vg ds g124 7 g111 0 111 Its not hard to see that if gxyz xzyz then Vg 2syzx22x2y Thus the answer is 917274 9171718 1 7 This is probably the method the authors wanted you to use for this problem See note below for advice on how to nd such a function 9 Another solution is to explicitly parameterize a simple path connecting 111 to 124 A line segment connecting the two points is the simplest such path Using the techniques of chapter 1 we see that Ct 111 t013 11 t1 3t for 0 S t S 1 does the job Then c t 013 and using the formula f0 F ds Fct c t dt we get 1 2xyzdxz22dyz2ydz 28t6t213t1t013dt C 0 113t31tdt 1 6t4dt 0 I think the rst method is far superior since it does not require an explicit parameter ization or calculation of a messy integral But to use the rst method we need to nd a scalar eld 9 whose gradient is F What if you cant just see77 what 9 should be How do you systematically nd one In this case F Qxyz x22x2y The vector equation F Vg implies 69 69 69 2xzx22x2 iii lt y y Mawz The rst component of this equation says 2zyz Multiplying each side of this equa tion by dm and integrating with respect to m gives gxyz xzyz yz where our constant77 of integration is any function of y and 2 but not How do we nd yz 1 Well we have two other conditions to satsify namely 22 375 and y Using our temporary function gy z xzyz My 2 the rst of these implies 22 22 thus 37 0 so evidently b doesn7t depend on y after all That is b may only depend on 2 But doing the same trick for 2 that we did for y we nd that b doesn7t depend on 2 either Thus b is at most an honest to goodness constant like 3 or 7139 or H7 let7s call that constant a Thus we have gxyz xzyz a We may as well take 04 0 Why Because any choice for 04 doesn7t affect the relationship F Vg since 04 being a constant disappears when we take the derivatives involved in V9 IMPORTANT NOTE Its not always possible to nd a scalar function 9 whose gradient is F When you can F is called a gradient eld or a conservative eld It turns out that if V gtlt F 0 ie F is curl free then you can nd such a g and when V gtlt F 31 0 then you cant See the very important Theorem 87 on p551 You can check that V gtlt F 0 for this problem

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