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}{PSTYLE "Normal" -1 259 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 5 "C3M17" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 22 "The Divergence Theorem" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "THE DIVERGENCE THEOREM: Let " }{TEXT 256 1 "Q" } {TEXT -1 51 " be a simple solid region whose boundary surface " } {TEXT 257 1 "S" }{TEXT -1 34 " is oriented by the unit normal " } {TEXT 258 1 "n" }{TEXT -1 24 " directed outward from " }{TEXT 259 1 " Q" }{TEXT -1 12 " , and let " }{TEXT 260 1 "F" }{TEXT -1 86 " be a v ector field whose component functions have continuous partial derivati ves on " }{TEXT 261 1 "Q" }{TEXT -1 8 " . Then" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 34 "the double integral over S of " }{TEXT 262 2 "F " }{TEXT -1 2 "\267 " }{TEXT 264 1 "n" }{TEXT -1 1 " " }{TEXT 263 1 "d" }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 41 " equals \+ the triple integral over " }{TEXT 267 1 "Q" }{TEXT -1 11 " of div (" }{TEXT 265 1 "F" }{TEXT -1 2 ") " }{TEXT 266 2 "dV" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "This states that the flux through a clos ed surface equals the total divergence throughout the solid." }}} {EXCHG {PARA 0 "" 0 "" {TEXT 268 9 "EXAMPLE 1" }{TEXT -1 31 " Conside r the solid cylinder " }{TEXT 269 1 "Q" }{TEXT -1 5 " = \{(" }{TEXT 270 5 "x,y,z" }{TEXT -1 4 "): " }{XPPEDIT 18 0 "x^2+y^2<=1" "6#1,&*$% \"xG\"\"#\"\"\"*$%\"yGF'F(F(" }{TEXT -1 8 " , 0 <= " }{TEXT 271 2 "z \+ " }{TEXT -1 29 "<= 1 \} and the vector field " }}{PARA 0 "" 0 "" {TEXT 272 1 "F" }{TEXT -1 1 "(" }{TEXT 273 5 "x,y,z" }{TEXT -1 6 ") = \+ < " }{XPPEDIT 18 0 "x*z,y,z" "6%*&%\"xG\"\"\"%\"zGF%%\"yGF&" }{TEXT -1 42 " > . The following Maple commands graph " }{TEXT 274 1 "Q" } {TEXT -1 2 " ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "with(plot s): side:=cylinderplot([1,th,z],th=0..2*Pi,z=0..1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "top:=cylinderplot([r,th,1],r=0..1,th=0..2 *Pi):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "bottom:=cylinderpl ot([r,th,0],r=0..1,th=0..2*Pi):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "display(side,top,bottom);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "The divergence of " }{TEXT 275 1 "F" }{TEXT -1 11 " is div(" } {TEXT 276 1 "F" }{TEXT -1 4 ") = " }{TEXT 278 1 "z" }{TEXT -1 9 " +1 + 1 = " }{TEXT 277 2 "z " }{TEXT -1 5 "+ 2 ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "So the triple integral over Q of div(" }{TEXT 279 1 "F " }{TEXT -1 2 ") " }{TEXT 280 4 "dV " }{TEXT -1 8 "equals " } {XPPEDIT 18 0 "Int(Int(Int((z+2)*r,z=0..1),r=0..1),theta=0..2*Pi)" "6# -%$IntG6$-F$6$-F$6$*&,&%\"zG\"\"\"\"\"#F-F-%\"rGF-/F,;\"\"!F-/F/;F2F-/ %&thetaG;F2*&F.F-%#PiGF-" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "The innermost integral gives us " }{XPPEDIT 18 0 "(z^2/2 +2z)*r" "6#*&,&*&%\"zG\"\"#F'!\"\"\"\"\"*&F'F)F&F)F)F)%\"rGF)" }{TEXT -1 10 " from " }{XPPEDIT 18 0 "z=0" "6#/%\"zG\"\"!" }{TEXT -1 6 " \+ to " }{XPPEDIT 18 0 "z=1" "6#/%\"zG\"\"\"" }{TEXT -1 11 " so we get " }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "Int(Int( 5/2r,r=0..1),theta=0..2*Pi) " "6#-%$IntG6$-F$6$*(\"\"&\"\"\"\"\"#!\"\" %\"rGF*/F-;\"\"!F*/%&thetaG;F0*&F+F*%#PiGF*" }{TEXT -1 1 " " }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Now the inner integral gives us \+ " }{XPPEDIT 18 0 "5/4*r^2 " "6#*(\"\"&\"\"\"\"\"%!\"\"%\"rG\"\"#" } {TEXT -1 8 " from " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" }{TEXT -1 5 " to " }{XPPEDIT 18 0 "r=1" "6#/%\"rG\"\"\"" }{TEXT -1 10 " so we g et" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 3 " " }{XPPEDIT 18 0 "Int(5 /4,theta=0..2*Pi)=5*Pi/2" "6#/-%$IntG6$*&\"\"&\"\"\"\"\"%!\"\"/%&theta G;\"\"!*&\"\"#F)%#PiGF)*(F(F)F2F)F1F+" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Now we will compare our answer with the \+ sum of three surface integrals, each with the normal vector " }{TEXT 281 1 "n" }{TEXT -1 19 " pointing outward." }}}{EXCHG {PARA 0 "" 0 " " {TEXT 289 9 " bottom " }{TEXT -1 3 " " }{XPPEDIT 18 0 "z=0" "6#/% \"zG\"\"!" }{TEXT -1 4 " , " }{TEXT 282 1 "n" }{TEXT -1 4 " = -" } {TEXT 283 3 " k " }{TEXT -1 3 ", " }{TEXT 290 1 "F" }{TEXT -1 1 "(" } {TEXT 291 4 "x,y," }{TEXT 371 1 "0" }{TEXT -1 6 ") = < " }{XPPEDIT 18 0 "0,y,0" "6%\"\"!%\"yGF#" }{TEXT -1 5 " >, " }{TEXT 284 2 "F " } {TEXT -1 2 "\267 " }{TEXT 285 1 "n" }{TEXT -1 49 " = 0 so the double integral over the bottom of " }{TEXT 286 4 " F " }{TEXT -1 2 "\267 \+ " }{TEXT 288 1 "n" }{TEXT -1 1 " " }{TEXT 287 1 "d" }{XPPEDIT 18 0 "si gma" "6#%&sigmaG" }{TEXT -1 11 " is 0 . " }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 1 " " }{TEXT 293 4 " top" }{TEXT -1 3 " " }{XPPEDIT 18 0 "z=1" "6#/%\"zG\"\"\"" }{TEXT -1 3 ", " }{TEXT 292 1 "n" }{TEXT -1 3 " = " }{TEXT 294 1 "k" }{TEXT -1 3 " , " }{TEXT 295 1 "F" }{TEXT -1 1 "(" }{TEXT 296 3 "x,y" }{TEXT -1 8 ",1) = < " }{XPPEDIT 18 0 "x,y,1" " 6%%\"xG%\"yG\"\"\"" }{TEXT -1 6 " > , " }{TEXT 297 2 "F " }{TEXT -1 2 "\267 " }{TEXT 298 1 "n" }{TEXT -1 45 " = 1 so the double integral \+ over the top of " }{TEXT 299 4 " F " }{TEXT -1 2 "\267 " }{TEXT 301 1 "n" }{TEXT -1 1 " " }{TEXT 300 1 "d" }{XPPEDIT 18 0 "sigma" "6#%&sig maG" }{TEXT -1 16 " is its area, " }{XPPEDIT 18 0 "pi" "6#%#piG" } {TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 302 4 "side" }{TEXT -1 61 " As we saw in an early example, the side is para meterized by" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 " " }{XPPEDIT 18 0 "g(theta,z)=[cos(theta),sin(theta),z]" "6#/-%\"gG6$%&thetaG%\"zG7 %-%$cosG6#F'-%$sinG6#F'F(" }{TEXT -1 8 " 0 <= " }{XPPEDIT 18 0 "thet a" "6#%&thetaG" }{TEXT -1 4 " <= " }{XPPEDIT 18 0 "2Pi" "6#*&\"\"#\"\" \"%#PiGF%" }{TEXT -1 8 " , 0 <= " }{XPPEDIT 18 0 "z" "6#%\"zG" }{TEXT -1 12 " <= 1, so " }{XPPEDIT 18 0 "diff(g,theta)=[-sin(theta,cos(the ta),0]" "6#/-%%diffG6$%\"gG%&thetaG7#,$-%$sinG6%F(-%$cosG6#F(\"\"!!\" \"" }{TEXT -1 7 " and " }{XPPEDIT 18 0 "diff(g,z)=[0,0,1]" "6#/-%%di ffG6$%\"gG%\"zG7%\"\"!F*\"\"\"" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "diff(g,theta)" "6#-%%diffG6$%\"gG %&thetaG" }{TEXT -1 4 " \327 " }{XPPEDIT 18 0 "diff(g,z)" "6#-%%diffG 6$%\"gG%\"zG" }{TEXT -1 89 " is given by a 3 \327 3 determinant with the previous vectors as the first two rows and [" }{TEXT 303 5 "i,j, k" }{TEXT -1 28 "] for the last row, yielding" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 304 1 "n" }{TEXT -1 5 " = < " }{XPPEDIT 18 0 "cos(theta),sin(theta),0" "6%-%$cosG6#%&thetaG-%$sinG6#F&\"\"!" } {TEXT -1 26 " > Is this a surprise ?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 305 1 "F" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "g(the ta,z)" "6#-%\"gG6$%&thetaG%\"zG" }{TEXT -1 6 ") = < " }{XPPEDIT 18 0 " z*cos(theta),sin(theta),z" "6%*&%\"zG\"\"\"-%$cosG6#%&thetaGF%-%$sinG6 #F)F$" }{TEXT -1 2 " >" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 306 4 " F " } {TEXT -1 2 "\267 " }{TEXT 307 1 "n" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " z*cos(theta)^2+sin(theta)^2" "6#,&*&%\"zG\"\"\"*$-%$cosG6#%&thetaG\"\" #F&F&*$-%$sinG6#F+F,F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "So our \+ surface integral is " }{XPPEDIT 18 0 "Int(Int(z*cos(theta)^2+sin(the ta)^2,z=0..1),theta=0..2*Pi)" "6#-%$IntG6$-F$6$,&*&%\"zG\"\"\"*$-%$cos G6#%&thetaG\"\"#F+F+*$-%$sinG6#F0F1F+/F*;\"\"!F+/F0;F8*&F1F+%#PiGF+" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 " The inner integral gives us " }{XPPEDIT 18 0 "z^2/2*cos(theta)^2+z*s in(theta)^2" "6#,&*(%\"zG\"\"#F&!\"\"-%$cosG6#%&thetaGF&\"\"\"*&F%F,*$ -%$sinG6#F+F&F,F," }{TEXT -1 8 " from " }{XPPEDIT 18 0 "z=0" "6#/%\" zG\"\"!" }{TEXT -1 7 " to " }{XPPEDIT 18 0 "z=1" "6#/%\"zG\"\"\"" } {TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "This gives us \+ " }{XPPEDIT 18 0 "Int(cos(theta)^2/2+sin(theta)^2,theta=0..2 *Pi)=Int((1+cos(2*theta))/4+(1-cos(2*theta))/2,theta=0..2*Pi)" "6#/-%$ IntG6$,&*&-%$cosG6#%&thetaG\"\"#F-!\"\"\"\"\"*$-%$sinG6#F,F-F//F,;\"\" !*&F-F/%#PiGF/-F%6$,&*&,&F/F/-F*6#*&F-F/F,F/F/F/\"\"%F.F/*&,&F/F/-F*6# *&F-F/F,F/F.F/F-F.F//F,;F6*&F-F/F8F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "which simplifies to " }{XPPEDIT 18 0 "Int(3/4-cos(2*theta)/4, theta=0..2*Pi)" "6#-%$IntG6$,&*&\"\"$\"\"\"\"\"%!\"\"F)*&-%$cosG6#*&\" \"#F)%&thetaGF)F)F*F+F+/F2;\"\"!*&F1F)%#PiGF)" }{TEXT -1 28 " . The \+ antiderivative is " }{XPPEDIT 18 0 "3/4*theta-sin(2*theta)/8" "6#,&*( \"\"$\"\"\"\"\"%!\"\"%&thetaGF&F&*&-%$sinG6#*&\"\"#F&F)F&F&\"\")F(F(" }{TEXT -1 45 " and plugging in the limits of integration " } {XPPEDIT 18 0 "theta=0" "6#/%&thetaG\"\"!" }{TEXT -1 7 " and " } {XPPEDIT 18 0 "theta=2Pi" "6#/%&thetaG*&\"\"#\"\"\"%#PiGF'" }{TEXT -1 26 " gives us an answer of " }{XPPEDIT 18 0 "3*Pi/2" "6#*(\"\"$\"\" \"%#PiGF%\"\"#!\"\"" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Thus, adding these three parts, the total surface integral ove r " }{TEXT 311 2 " S" }{TEXT -1 7 " of " }{TEXT 308 2 "F " }{TEXT -1 2 "\267 " }{TEXT 310 1 "n" }{TEXT -1 1 " " }{TEXT 309 1 "d" } {XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 7 " is " }{XPPEDIT 18 0 "0+Pi+3Pi/2=5Pi/2" "6#/,(\"\"!\"\"\"%#PiGF&*(\"\"$F&F'F&\"\"#!\"\"F& *(\"\"&F&F'F&F*F+" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "which agrees with the volume integral, as predicted bythe Divergen ce theorem." }}}{EXCHG {PARA 258 "" 0 "" {TEXT 312 1 "F" }{TEXT -1 1 " (" }{TEXT 313 5 "x,y,z" }{TEXT -1 7 ") = < " }{XPPEDIT 18 0 "1,1,z" " 6%\"\"\"F#%\"zG" }{TEXT -1 19 " > for the solid " }{TEXT 314 1 "Q" } {TEXT -1 34 " that lies above the paraboloid " }{XPPEDIT 18 0 "z=x^2 +y^2" "6#/%\"zG,&*$%\"xG\"\"#\"\"\"*$%\"yGF(F)" }{TEXT -1 23 " and be low the plane " }{XPPEDIT 18 0 "z=4" "6#/%\"zG\"\"%" }{TEXT -1 2 " . " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "The Maple command below plots the solid." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "cylinderplot (\{[r,t,r^2],[r,t,4]\},r=0..2,t=0..2*Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 315 17 "Volume integral " }{TEXT -1 11 "Since div(" }{TEXT 316 1 "F" }{TEXT -1 34 ") = 1 , the triple integral over " }{TEXT 318 1 "Q" }{TEXT -1 11 " of div(" }{TEXT 317 1 "F" }{TEXT -1 2 ") \+ " }{TEXT 319 2 "dV" }{TEXT -1 45 " is the triple integral of 1 and \+ it equals" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "Int(Int(Int(1,z=(z=x^2+y^2)..(z=4)),x=(x=-sqrt(4-y^2))..(x=sqrt(4-y^2 ))),y=-2..2)" "6#-%$IntG6$-F$6$-F$6$\"\"\"/%\"zG;/F,,&*$%\"xG\"\"#F**$ %\"yGF2F*/F,\"\"%/F1;/F1,$-%%sqrtG6#,&F6F**$F4F2!\"\"F@/F1-F<6#,&F6F** $F4F2F@/F4;,$F2F@F2" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 6 " = " }{XPPEDIT 18 0 "Int(In t(Int(r,z=(z=r^2)..(z=4)),r=(r=0)..(r=2)),theta=0..2*Pi)" "6#-%$IntG6$ -F$6$-F$6$%\"rG/%\"zG;/F,*$F*\"\"#/F,\"\"%/F*;/F*\"\"!/F*F0/%&thetaG;F 6*&F0\"\"\"%#PiGF<" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 257 "" 0 "" {TEXT -1 5 " = " }{XPPEDIT 18 0 "Int(Int(4 r-r^3,r=(r=0)..(r=2)),theta=0..2*Pi)" "6#-%$IntG6$-F$6$,&*&\"\"%\"\"\" %\"rGF+F+*$F,\"\"$!\"\"/F,;/F,\"\"!/F,\"\"#/%&thetaG;F3*&F5F+%#PiGF+" }{TEXT -1 1 " " }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 6 " = " } {XPPEDIT 18 0 "Int(4,theta=0..2*Pi)" "6#-%$IntG6$\"\"%/%&thetaG;\"\"!* &\"\"#\"\"\"%#PiGF-" }{TEXT -1 1 " " }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 6 " = " }{XPPEDIT 18 0 "8Pi" "6#*&\"\")\"\"\"%#PiGF%" } {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 320 17 "Surface integrals" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "(a) Top " }{XPPEDIT 18 0 "z=4" "6#/%\"zG\"\"%" }{TEXT -1 5 " , " }{TEXT 321 1 "n" }{TEXT -1 3 " = " }{TEXT 322 1 "k" }{TEXT -1 14 " = <0,0,1> , " }{TEXT 323 1 "F" }{TEXT -1 1 "(" }{TEXT 324 3 "x,y" } {TEXT -1 16 ",4) = <1,1,4>, " }{TEXT 325 2 "F " }{TEXT -1 2 "\267 " } {TEXT 326 2 "n " }{TEXT -1 42 "= 4 , the double integral over S of \+ 4 " }{TEXT 327 1 "d" }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 42 " is 4 times the area of the top, or 4(" }{XPPEDIT 18 0 "4Pi" " 6#*&\"\"%\"\"\"%#PiGF%" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "16Pi" "6#*& \"#;\"\"\"%#PiGF%" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "(b) Bottom " }{XPPEDIT 18 0 "z=x^2+y^2" "6#/%\"zG,&*$%\"xG\"\" #\"\"\"*$%\"yGF(F)" }{TEXT -1 50 " , but we will parameterize in pola r coordinates." }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "g(r,theta)=[r*cos(theta),r*sint(theta),r^2]" "6#/-%\"gG6$%\"rG%& thetaG7%*&F'\"\"\"-%$cosG6#F(F+*&F'F+-%%sintG6#F(F+*$F'\"\"#" }{TEXT -1 8 " 0 <= " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 16 " <= 2 and \+ 0 <= " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 5 " <= " } {XPPEDIT 18 0 "2Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 2 " ." }}} {EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "diff(g,r" "6# -%%diffG6$%\"gG%\"rG" }{TEXT -1 4 " \327 " }{XPPEDIT 18 0 "diff(g,the ta)" "6#-%%diffG6$%\"gG%&thetaG" }{TEXT -1 7 " = < " }{XPPEDIT 18 0 "-2r^2*cos(theta),-2r^2*sin(theta),r" "6%,$*(\"\"#\"\"\"*$%\"rGF%F&-%$ cosG6#%&thetaGF&!\"\",$*(F%F&*$F(F%F&-%$sinG6#F,F&F-F(" }{TEXT -1 3 " \+ >" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "It is important to note tha t our normal vector points inward, so we must choose its negative! Th us" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{TEXT 328 1 "n" } {TEXT -1 7 " = < " }{XPPEDIT 18 0 "2r^2*cos(theta),2r^2*sin(theta),- r" "6%*(\"\"#\"\"\"*$%\"rGF$F%-%$cosG6#%&thetaGF%*(F$F%*$F'F$F%-%$sinG 6#F+F%,$F'!\"\"" }{TEXT -1 3 " >" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "We have " }{TEXT 329 1 "F" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "g(r,the ta)" "6#-%\"gG6$%\"rG%&thetaG" }{TEXT -1 10 ") = < 1,1," }{XPPEDIT 18 0 "r^2" "6#*$%\"rG\"\"#" }{TEXT -1 5 " >, " }{TEXT 330 2 "F " }{TEXT -1 2 "\267 " }{TEXT 331 1 "n" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2r^2*c os(theta)+2r^2*sin(theta)-r^3" "6#,(*(\"\"#\"\"\"*$%\"rGF%F&-%$cosG6#% &thetaGF&F&*(F%F&*$F(F%F&-%$sinG6#F,F&F&*$F(\"\"$!\"\"" }{TEXT -1 13 " , so we have" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "Int(Int(2r^2*cos(theta)+2r^2*sin(theta)-r^3,r=0..2),theta=0..2*P i)" "6#-%$IntG6$-F$6$,(*(\"\"#\"\"\"*$%\"rGF*F+-%$cosG6#%&thetaGF+F+*( F*F+*$F-F*F+-%$sinG6#F1F+F+*$F-\"\"$!\"\"/F-;\"\"!F*/F1;F<*&F*F+%#PiGF +" }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "The inner i ntegral gives us " }{XPPEDIT 18 0 "2r^3/3*cos(theta)+2r^3/3*sin(theta )-r^4/4" "6#,(**\"\"#\"\"\"*$%\"rG\"\"$F&F)!\"\"-%$cosG6#%&thetaGF&F&* *F%F&*$F(F)F&F)F*-%$sinG6#F.F&F&*&F(\"\"%F5F*F*" }{TEXT -1 8 " from \+ " }{XPPEDIT 18 0 "r=0" "6#/%\"rG\"\"!" }{TEXT -1 6 " to " }{XPPEDIT 18 0 "r=2" "6#/%\"rG\"\"#" }{TEXT -1 11 " so we get" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 2 " " }{XPPEDIT 18 0 "Int(16/3*cos(theta)+16/3*s in(theta)-4,theta=0..2*Pi)=-8*Pi" "6#/-%$IntG6$,(*(\"#;\"\"\"\"\"$!\" \"-%$cosG6#%&thetaGF*F**(F)F*F+F,-%$sinG6#F0F*F*\"\"%F,/F0;\"\"!*&\"\" #F*%#PiGF*,$*&\"\")F*F;F*F," }{TEXT -1 2 " " }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 62 "And as was predicted, the double integral over all of \+ S of " }{TEXT 332 2 "F " }{TEXT -1 2 "\267 " }{TEXT 334 1 "n" } {TEXT -1 1 " " }{TEXT 333 1 "d" }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" } {TEXT -1 7 " is " }{XPPEDIT 18 0 "16Pi-8Pi=8Pi" "6#/,&*&\"#;\"\"\"% #PiGF'F'*&\"\")F'F(F'!\"\"*&F*F'F(F'" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 335 16 "Example 3, Maple" }{TEXT -1 158 ": We w ill illustrate the equivalence of the two integrals in the Divergence \+ Theorem with a challenging problem. The solid is bounded by the obliq ue plane 2 " }{TEXT 337 1 "x" }{TEXT -1 3 " + " }{TEXT 336 1 "z" } {TEXT -1 30 " = 6, the vertical cylinder " }{XPPEDIT 18 0 "x^2+y^2=4 " "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"\"%" }{TEXT -1 12 " , and the \+ " }{TEXT 338 2 "xy" }{TEXT -1 32 "-plane. The vector function is " } {TEXT 339 1 "F" }{TEXT -1 1 "(" }{TEXT 340 5 "x,y,z" }{TEXT -1 8 ") = \+ < " }{XPPEDIT 18 0 "x^2+z,x*y+2z,3y" "6%,&*$%\"xG\"\"#\"\"\"%\"zGF', &*&F%F'%\"yGF'F'*&F&F'F(F'F'*&\"\"$F'F+F'" }{TEXT -1 233 " > . We w ill begin by sketching the solid, then compute the divergence over the solid, and end with the three surface integrals. The easiest approac h is to use cylindrical coordinates. The following commands need to b e executed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "restart: wit h(student): with(plots): with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "S:=cylinderplot([2,theta,z],theta=0..2*Pi,z=0..6-2*2* cos(theta),color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "T :=cylinderplot([r,theta,6-2*r*cos(theta)],r=0..2,theta=0..2*Pi,color=b lue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "xaxis:=spacecurve( [t,0,0],t=0..3,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "yaxis:=spacecurve([0,t,0],t=0..3,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "display(S,T,xaxis,yaxis);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "F:=(x,y,z)->[x^2+z,x*y+2*z,3*y];" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "divF:=diverge(F(x,y,z),[x, y,z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "grand:=subs(x=r*c os(theta),y=r*sin(theta),z=z,divF);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "divint:=Tripleint(grand*r,z=0..6-2*r*cos(theta),r=0.. 2,theta=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "divans wer:=value(divint);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 173 "It should be obvious that we have just finished the easy half of this problem. \+ We will work from top to bottom on the surfaces. We begin with the o blique plane. Note how " }{TEXT 341 3 " u " }{TEXT -1 20 " plays the \+ role of " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 6 " and " }{TEXT 342 3 " v " }{TEXT -1 11 " that of " }{XPPEDIT 18 0 "theta" "6#%&the taG" }{TEXT -1 3 " ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "g1 :=[u*cos(v),u*sin(v),6-2*u*cos(v)];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g1u:=map(diff,g1,u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g1v:=map(diff,g1,v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "fcp1:=crossprod(g1u,g1v);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "fcp1:=simplify(fcp1,symbolic);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 19 "Fatg1uv:=F(op(g1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "grand1:=innerprod(Fatg1uv,fcp1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "ans1:=Doubleint(grand1,u=0..2,v=0.. 2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "flux1:=value(ans1 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now we will work on the cur ved side." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "g2:=vector([2* cos(u),2*sin(u),v]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g2u :=map(diff,g2,u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "g2v:=m ap(diff,g2,v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "fcp2:=cro ssprod(g2u,g2v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Fatg2uv :=F(op(g2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "grand2:=inn erprod(Fatg2uv,fcp2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "an s2:=Doubleint(grand2,v=0..6-2*2*cos(u),u=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "flux2:=value(ans2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Let\222s think about the bottom surface for a moment . The " }{TEXT 344 7 "outward" }{TEXT -1 26 " unit normal vector is \+ -" }{TEXT 343 1 "k" }{TEXT -1 12 " = < 0 ,0 , " }{XPPEDIT 18 0 "-1" " 6#,$\"\"\"!\"\"" }{TEXT -1 4 " > ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Fonbot:=F(x,y,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "n:=vector([0,0,-1 ]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "grand3:=innerprod(Fonbot,n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "grand3:=subs(x=r*cos(theta),y=r*sin(theta),grand 3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "ans3:=Doubleint(gran d3*r,r=0..2,theta=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "flux3:=value(ans3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 " Fluxtotal:=flux1+flux2+flux3;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 239 "Once again, the divergence integral is seen to be the same as the flu x integral. At this point, the reader should begin to develop an appr eciation for how the divergence integral is usually easier to evaluate than the surface integral(s)." }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 14 "C3M17 problems" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "1. Use Map le to evaluate directly the flux integral over " }{TEXT 357 1 "S" } {TEXT -1 7 " of " }{TEXT 345 2 "F " }{TEXT -1 2 "\267 " }{TEXT 347 1 "n" }{TEXT -1 1 " " }{TEXT 346 1 "d" }{XPPEDIT 18 0 "sigma" "6#%&sig maG" }{TEXT -1 41 " which equals the double integral over " }{TEXT 356 1 "D" }{TEXT -1 6 " of " }}{PARA 0 "" 0 "" {TEXT -1 1 " " } {TEXT 348 1 "F" }{TEXT -1 2 "( " }{XPPEDIT 18 0 "g(u,v)" "6#-%\"gG6$% \"uG%\"vG" }{TEXT -1 4 " ) \267" }{TEXT 349 1 " " }{TEXT -1 2 " (" } {TEXT 351 1 " " }{XPPEDIT 18 0 "diff(g,u)" "6#-%%diffG6$%\"gG%\"uG" } {TEXT -1 5 " \327 " }{XPPEDIT 18 0 "diff(g,v)" "6#-%%diffG6$%\"gG%\" vG" }{TEXT -1 3 " ) " }{TEXT 350 11 "du dv . " }{TEXT 352 1 "F" } {TEXT -1 1 "(" }{TEXT 353 5 "x,y,z" }{TEXT -1 6 ") = < " }{XPPEDIT 18 0 "y,-x,1" "6%%\"yG,$%\"xG!\"\"\"\"\"" }{TEXT -1 6 " > , " }{TEXT 354 1 "n" }{TEXT -1 12 " outward, " }{TEXT 355 1 "S" }{TEXT -1 32 " \+ is the surface of the sphere " }{XPPEDIT 18 0 "x^2+y^2+z^2=1" "6#/,( *$%\"xG\"\"#\"\"\"*$%\"yGF'F(*$%\"zGF'F(F(" }{TEXT -1 2 " ." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Use Maple and the Divergence Theor em to evaluate the given flux integrals over S of " }{TEXT 358 2 "F \+ " }{TEXT -1 2 "\267 " }{TEXT 360 1 "n" }{TEXT -1 1 " " }{TEXT 359 1 "d " }{XPPEDIT 18 0 "sigma" "6#%&sigmaG" }{TEXT -1 46 " . (Do NOT evalu ate the integrals directly!)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "2. " }{TEXT 361 1 "F" }{TEXT -1 1 "(" }{TEXT 362 5 "x,y,z" }{TEXT -1 9 ") = < " }{XPPEDIT 18 0 "3x^2,x*y,z" "6%*&\"\"$\"\"\"*$%\"xG\"\"#F% *&F'F%%\"yGF%%\"zG" }{TEXT -1 7 " > , " }{TEXT 363 1 "S" }{TEXT -1 18 " bounds the solid " }{TEXT 364 1 "Q" }{TEXT -1 5 " = \{(" }{TEXT 365 5 "x,y,z" }{TEXT -1 4 "): " }{XPPEDIT 18 0 "x+y+z=1" "6#/,(%\"xG \"\"\"%\"yGF&%\"zGF&F&" }{TEXT -1 4 " , " }{XPPEDIT 18 0 "0<=x" "6#1 \"\"!%\"xG" }{TEXT 366 5 ",y,z " }{TEXT -1 1 "\}" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 4 "3. " }{TEXT 367 1 "F" }{TEXT -1 1 "(" }{TEXT 368 6 "x,y,z)" }{TEXT -1 6 " = < " }{XPPEDIT 18 0 "x^3,y^3,z^3" "6%*$%\"x G\"\"$*$%\"yGF%*$%\"zGF%" }{TEXT -1 6 " > , " }{TEXT 369 1 "Q" } {TEXT -1 33 " is the solid above the cone " }{XPPEDIT 18 0 "z^2=x^ 2+y^2" "6#/*$%\"zG\"\"#,&*$%\"xGF&\"\"\"*$%\"yGF&F*" }{TEXT -1 25 " a nd inside the sphere " }{XPPEDIT 18 0 "x^2+y^2+z^2=9" "6#/,(*$%\"xG\" \"#\"\"\"*$%\"yGF'F(*$%\"zGF'F(\"\"*" }{TEXT -1 36 " .\nHint: use sphe rical coordinates.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 370 15 "Acknowle dgement" }{TEXT -1 148 ": Some of the problems for these notes have b een borrowed from various texts. The authors include W.F.Trench, Earl Swokowski, and several others." }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }