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## Multivariable Calculus

by: Lauriane Brown

29

0

1

# Multivariable Calculus MATH 222

Marketplace > Vassar College > Mathematics (M) > MATH 222 > Multivariable Calculus
Lauriane Brown

GPA 3.64

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
1
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 1 page Class Notes was uploaded by Lauriane Brown on Wednesday October 28, 2015. The Class Notes belongs to MATH 222 at Vassar College taught by Staff in Fall. Since its upload, it has received 29 views. For similar materials see /class/230533/math-222-vassar-college in Mathematics (M) at Vassar College.

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Date Created: 10/28/15
Final Review Sheet Hints7Math 222 Prof Frank Spring 2005 1 Let 0 6 7r 27139 and let 1 E 07r Then ltIgtt9 gt 4 3cos0sin gt5 3sin0sin gt 75 3cos gt The easy way to get the unit normal is to write it as 7 4 y 7 5 z 5 but you 71 1 could also compute TTe x T 3T x T9 2 You can get the derivative everywhere except where u 1 There are two cases If u gt u then Tu x Tu 1 71 1 If u lt u then Tu x Tu 1171 You can graph the gure by thinking about what happens on the line it v and then thinking about what happens to each boundary segment of D When you re done go to the surfaces applet and plot it You ll need to use that lu 7 v 1U 7 12 3 Since HT x TvH xg you don t need to break it into two regions The area is 4 In both cases we need a parametrization I D 7 R3 for which ltIgtD S Once we have that we can compute the normal vector Tu x Ty When we compute 1 15 as in 75 we are dealing with a realvalued function f R3 7 R and the de ntion is 1 d5 fltIgtuvHTu x Tull dudv Orientation is not a concern for this type of integral sincelit is only the length and not the direction of the normal vector which plays a role When we compute F 18 as in 76 we are integrating a vector eld F R3 7 R3 and the de nition is F dS FltIgtu 11 Tu x Tu dudv Orientation is an issue here switching the orientation will intFoduce a negative sign to the integral If we consider 1 R3 7 R to be a mass density function then computing 1 15 will give us the total mass of the surface S recall the big wobbly ball example If instead we consider F R3 7 R3 to be the velocity vector eld of a uid then the integral F 18 gives us the ux across the surface S S 5 The trick here is to notice that n and reduce to an integral from 75 Tu x Ty llTu X Tull C 7 WWW 7 In the rst case let Pmy 0 and Qmy m so that z dy P dz Q dy By 3D D LP 8y 6 Similarly notice that Tt and reduce to a path integral from 71 Green s theorem that equals 87639 D 8m the second part yourself dA 1 1A which is the area of D Try D 8 8P If you write Fm y Pmy Qzy then the scalar curl of F is 8762 7 87 which is a 39 scalar quantity which depends on The curl of F is a vector eld in R3 which gives an indication of the strength and direction of the rotation of the vector eld So we must think of F as embedded in R3 by writing Fzyz Pmy Qzy0 Go ahead and A 00 V compute V x F you ll see that you get Q 7 7 8m 8y Since S is closed it has no boundary so the integral over the boundary of S is zero Now apply Stokes theorem A to V

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