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# METH APPL MATH I MATH 601

Texas A&M

GPA 3.92

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This 18 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 601 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 29 views. For similar materials see /class/226038/math-601-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15

Stokes39 Theorem Examples Stokes39 Theorem relates surface integrals and line integrals STOKES39 THEOREM Let F be a vector eld Let S be an oriented surface and let C be the boundary curve of S oriented ScurlFdA foFds using the righthand rule Then EXAMPLE 1 Let C be the curve defined by the parametric equations I0 y22cost 0 t 27r z22sint Use Stokes39 Theorem to evaluate 2551 dm I cos y dy 3y dz C SOLUTION The parametric equations above describe a circle of radius 2 on the yzplane Z Let D be the disc whose boundary is the given circle By Stokes Theorem 2551icosyj3ykds curlI2651iIcosyj3ykdA C D We compute the curl i j k curlI2551iIcosyj3yk 88I 88y 882 3i5I255 7 jcosyk 2551 I cos y 3y So we have to evaluate 3i 52551j cos yk dA D This integral can be evaluated geometrically The vector dA for the disc points in the positive I direction Stokes theorem uses the righthand rule if you curl the ngers of your right hand in the direction of C then your thumb points in the direction of dA So 3i5mze5zj cosyk dA 3dA Therefore 3i52551jcosykdA 3dA 3xtheareaofthedisc 127T I D D EXAMPLE 2 Let S be the surface 2 Iltl 7 Iy17 y for 0 S I S 1 and 0 S y S 1 Evaluate the integral Ik dA where dA is the upwardpointing normal vector 5 SOLUTION If we wish to use Stokes theorem we must express Ik as the curl of some vector field F The formula for the curl is i j k curlF 982 98y 982 FI Fy Fl 8FZ 78F 9FI 8FZ 8F 9FI 9y 927 92 927 92 9y So what we need is 8F 9Fy 7 9FI 8F 9Fy 9FI 9y 7570 92 92 92 9y 7 It is not hard to guess that F l 2 with FI FZ 0 Indeed j k cune j 882 aay 882 Ik 0 0 By Stokes Theorem we conclude that S Ik dA fc mz dy where C is the boundary curve of the surface 5 So what is the boundary of 5 Well the equation 2 Iltl 7 Iyl 7 y specifies a surface whose zcoordinate varies with horizontal position Z 4l1llllquot39 i y IIIIIIIIIIII I 1 v all h v 38 lt g k The allowed values of r and y are determined by the inequalities 0 g r g 1 and 0 g y g 1 which describe a square on the ryplane Ag 1 If x r l The sides of this square are along the lines at 0 y 0 r 1 and y 1 Looking at the equation we see that z 0 for these values of r and y so the boundary of the surface is just the boundary of the square on the ryplane Ag 1 A V A V53 V 1 We can evaluate f 5x2 dy geometrlcally C 1 2 f 1 2 f 1 2 1 2 1 2 r dy r dy r dy r dy r dy ii 2 top 2 bottom 2 left 2 right 2 The top and bottom sides are horizontal so dy 0 Furthermore 25 0 along the left edge and r 1 along the right edge so 12 l l Idy000 dy I 2 right2 2 EXAMPLE 3 Let S be the upper hemisphere of the unit sphere I2 y2 22 1 Use Stokes39 theorem to evaluate Iseyi 7 3l 2 5y dA where dA is the upwardpointing normal vector 5 3 SOLUTION If we wish to use Stokes39 theorem we must express I 51 i 7 3I25yj as the curl of some vector field F The formula for the curl is i j k 011le 992 99y 992 FI Fy FZ i 9in 9F179FZ 79FI 9y 927 92 9a7 92 9y So what we need is 9F 9F 3 92 I 6y 8F 8F 8y 9FI 9FZ i 3 2 y 92 92 7 I 6 92 9y 0 If we guess that Fy and FI are zero it is not too hard to gure out that FZ 3532 Indeed i j k curlI35y k aam aay 982 I361 i 7 3I25yj 0 0 361 By Stokes39 Theorem we conclude that 35 i 7 3l 26yj dA f I361 d2 where C is the boundary curve of the surface S S C Since 5 is the upper hemisphere of the unit sphere C is just the unit circle on the Iyplane By the righthand rule C is oriented counterclockwise It would be clockwise if we had started with the downwardpointing dA Then I361 d2 is zero since 2 is not changing over the course of the circle C We conclude that Isey i 7 3l 26yj dA 0 I s EXAMPLE 4 Let S be the surface de ned by 2 I2 y2 for 2 S 4 Use Stokes39 Theorem to evaluate 3l 22 i 23 k dA where dA is the upwardpointing normal vector 5 SOLUTION If we wish to use Stokes39 theorem we must express 3l 22 i 23 j as the curl of some vector field F The formula for the curl is i j k 011le 992 99y 992 FI Fy Fl i 9in 9F179FZ 79FI 7 9y 927 92 9a7 92 9y So what we need is 9FZ 9Fy 2 9FI 9FZ 7 0 9Fy 7 9FI 7 3 9y 92 92 92 7 92 9y 72 If we guess that FI and F1 are zero it is not hard to gure out that Fy 23 Indeed i j k curlm23j aam 3311 332 73m22i 23k 0 I23 0 By Stokes39 Theorem we conclude that 3l 22ld 23 k dA fr dy where C is the s C boundary curve of the surface S So what is the boundary of 5 Well the equation for the surface S can be expressed as 2 T2 where 2 S 4 This appears as a parabola on the T2plane 27 4 Therefore the surface S is a bowlshaped paraboloid Its boundary curve C is a counterclockwise circle on the plane 2 4 with radius 2 with parameterization I 2 cost y2sint 0 t 27r 2 4 The circle is counterclockwise by the righthand rule Thus 27v f I23 dy 2 cos t432 cos t dt C o 27v 256 cosztdt o 27v 1 256 1 cos 2t dt doubleangle formula 0 1 27v 128 t sin2t 2 o 256W Inner Products and Fourier Series The Inner Product If f and g are continuous functions on the interval 11 b the standard inner product of f and g is the following quantity 7 mmfmmmM The standard inner product of two functions is analogous to the dot product of two vectors ltfggt islike vw EXAMPLE 1 Find the inner product of the functions f I and g I2 1 on the interval 07 2 SOLUTION We have ltf79gt 02fIgIdI 02II21dm 6 I PROPERTIES OF THE STANDARD INNER PRODUCT l ltffgt Z 0 and ltffgt 0 only iff is the constant zero function 2 ltf79gt lt97fgt 3UampMUm gt 4 f cggt cltf ggt for any scalar c When discussing the inner product of functions we use much of the same terminology as we do for the dot product of vectors For example I 17 The norm ofafunctionf is the quantity ltffgt fm2 dm We refer to afunction f as a unit vectorif 1 ie f fgt 1 Two functions f and g are orthogonal if lt f ggt 0 Functions f1 fquot are orthonormal if each fl is a unit vector and each pair f1 fj are orthogonal Hf1H1 and ltf1fjgt0 foriy j EXAMPLE 2 Let S SpanI 2 on the interval 0 3 Find an orthonormal basis for S SOLUTION We use the GramSchmidt process First the norm of the function I is M 993 thde 3 1 Since I has length 3 the function I is a unit vector This reduces the bas1s to s s 1 2 an I I P 3 7 1 Next we nd the inner product of I2 With gl 1 3 1 I2 I I2 I dI a 3 O 3 4 2 9 1 7 I is orthogonal to I This reduces the bas1s 27 1 Therefore the function I2 7 I I 4 3 4 3 to 1 9 S Span I I2 7 9 Finally we must make I2 7 Zl into a unit vector We have 9 9 3 9 2 9 3 ltI27ZII27ZIgt O I27ZI dI 9 9 3 4 5 9 5 4 Since I2 7 ZI has length A g the function 7 E EI2 7 I is a unit vector Therefore is an orthonormal basis for S An orthonormal basis of functions works just like an orthonormal basis of vectors In particular recall that we can decompose any vector v using an orthonormal basis 111 un v vu1u1vu2uzvunun An analogous result holds for functions FOURIER DECOMPOSITION Let V be a vector space of functions on the interval 11 and suppose that f1 fquot is an orthonormal basis for V Then for any function f E V at ltf7f1gtf1I ltf7f2gtf2I ltf7fngtfnI EXAMPLE3 Let f1I 1 f2I ex 71 and 16333 6 I2 7 I on the interval 07 1 Given that f17 f2 f3 is an orthonormal basis for P3 express the function f I2 as a linear combination off1f2f3 SOLUTION Wehave ltf7f1gtltm21gtOl dx lt12 f2gt ltm27 lt2m71gtgt Almwgmm ltmsgt ltm26 3xtx gt AZZMWHW 673 Therefore Fourier Series Fourier series are a special kind of Fourier decomposition on the interval 77T 7T using the following inner product ltmgt ijfmgwx THEOREM ORTHONORMAL TRIG FUNCTIONS With respect to the above inner product the following functions are all orthonormal on the interval 77T 7T 1 The constant function 2 2 The functions cos I cos 2m cos 3m and 3 The functions sinm sinZI sin3m This theorem is really just saying something about the values of certain integrals For example sin 3m and sin 2m are orthogonal because 1 739 lts1n3m sm2mgt s1n3s1n2md 0 7T 7 Similarly sin 3m is a unit vector because l 7 ltsm3m s1n3mgt s1n23mdm 1 7T 7w NOTE The sorts of integrals that come up here can be evaluated using the following obscure trig identities l sinzsiny COSI T y coslt1 y 2 coszsiny W 3 coszcosy W For example 7r 1 7r ltsin31sin2zgt sin315in2zdz 0051700551 dz 0 4 r and 7r 1 7r HsinBzH ltsir1317 sin31gt sin315in31dz 17 cos6zdz 4 4r EXAMPLE 4 Given that cos 6 Span1 cos I cos 2m cos 3m cos 4m express cos4m as a linear combination of 1 cos I cos 2m cos 3m and cos 4m SOLUTION Using a calculator or computer algebra system we have 1 1 cos4xdr 3 W W 4V5 1 7r cos4m cos Igt cos5m dm 0 7T 7 4 1 4 1 cos I cos 2m cos ICOS 2m dm 5 7T aw lt lt gt Z ltcos4m cos 3gt j cos4m cos 3m dm 0 lt gt 1 cos cos 4m cos4m cos 4m dm aw Therefore 3 1 1 1 3 1 1 cos cos2m cos4x g Ecos2m gcoszlm I EXAMPLE 5 On the interval inn it is possible to express I3 as an infinite linear combination of the functions sin 7m x I3 E ansinna quot1 Assuming this fact find the values of the coef cients an SOLUTION Wehave 1 quot 1 3 2 2 7 2 39 2 2 7 6 ltl 37 Sinnatgt IsSlIl l dl n I SmnI 7 I01 I cosnI 7T 7 7T 714 n3 77 Now sin 7m 0 for any integer value of 71 while cos 7m 71quot Therefore this integral evaluates to 7 1001279 i 6 1quot 7 00127T2 671quot 71quot27127T2 6 7T 3 Therefore In the last example it s not at all obvious that I3 can be expressed as a Fourier sine series But assuming that it can then we know that we ve found the coef cients It turns out that any continuous function on the interval 771 71 can be expressed as a Fourier series FOURIER SERIES Any continuous function f on the interval 771 71 can be expressed as a Fourier series x a ancosnm chsinnm quot1 n1 This series converges for all values of I in 77171 The above theorem also holds for functions with nitely many jump discontinuities except that the series may not converge to the correct value at the jump point EXAMPLE 6 Let f be the following step function 0 lf77T S a lt 0 fI1 if0lt 7T Find the Fourier series for f on the interval 771 71 SOLUTION Wehave 1 1 quot fa 1 7T 1 d ltf gt lmEm WxE 1 739 l 739 l I 7 ltfcosnmgt fcosnmdm cosnmdm Smi 0 7T 7 7T 0 7T 71 O 1 icosnmy 1 7r 7 l ltfsinngt fsinnmdm sinnde 7T 7 7T 0 7T 71 o 2 The last express1on is equal to 0 when n is even and 1s when n is odd Therefore 777T 1 2 2 2 sm s1n3m s1n5m 2 7T 371 571 There are other advantages to having an orthonormal basis For example recall that a1u1 anun blul bnun 111171 1an for any orthonormal basis 111 uquot Similarly Halul anunH These rules work for functions as well EXAMPLE 7 Let xl2cosm3cos2m5cos3m and g27cosm3cos2cos3m Find the inner product lt f ggt on the interval 77T 7T SOLUTION In terms of our orthonormal basis H A H V l 1 i 2cosm 3cos2m 5cos3m and g 7 cosm 3cos2m cos3m ltf79gt 2N4 33 50 16 EXAMPLE 8 Compute sinZI 2 sin4x 3 sin 6m2 dm 7 SOLUTION This integral is related to the norm of sin 2x 2 sin 4m 3 sin 6m sin2m2sin4x3sin62dm 7THsi112I2sin4m3sin6IH2 7T12 22 32 147T Divergence Theorem Examples Gauss39 divergence theorem relates triple integrals and surface integrals GAUSS39 DIVERGENCE THEOREM Let F be a vector eld Let S be a closed surface and let R be the region inside of S Then SFdA RdivFdV EXAMPLE 1 Evaluate 3l l Zyj dA where S is the sphere I2 12 22 9 s SOLUTION We could parameterize the surface and evaluate the surface integral but it is much faster to use the divergence theorem Since 8 8 8 d39339239 3 2 05 1vltm m 8Ilt xgt8ylt ygt82ltgt the divergence theorem gives 3Ii2yj dA 5dV 5 x thevolumeofthesphere 1807T I s R EXAMPLE 2 Evaluate yzzi ysj m2 k dA where S is the boundary of the cube de ned by 71 I 171 y lad0 2 2 SOLUTION Since divy22i ysj 2 k 83 122 3y2 a I 9y 92 the divergence theorem gives Sy22iy3jmzkdA R3y2xdv Azll3y2xdxdydz 1 26y2dy8 I 71 EXAMPLE 3 Let R be the region in R3 bounded by the paraboloid z I2 y2 and the plane 2 1 and let S be the boundary of the region R Evaluate yi Ij 22k dA s SOLUTION Here is a sketch of the region in question Z z 1 11 Since divyi 21 22k 22 122kdA122dv It is easiest to set up the triple integral in cylindrical coordinates 27v 1 1 22dV 22szdrd0 R 0 0 72 1 1 7 2 7 ZWj Z T172 air the divergence theorem gives In general you should probably use the divergence theorem whenever you wish to evaluate a vector surface integral over a closed surface The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals This depends on finding a vector field whose divergence is equal to the given function EXAMPLE 4 Find a vector eld F whose divergence is the given function f a fI 1 b fI 211 0 fI v 2 22 SOLUTION The formula for the divergence is 8FI 9Fy 8FZ 92 9y 92 divF V F We get to choose F1 Fy and F1 so there are several possible vector elds with a given divergence This is similar to the freedom enjoyed when nding a vector eld with a given rotation a F Ii works as does F yj F 2k and so forth l l b Three poss1ble solutions are F Emsyi F imzyzj and F IZyZ k c It is dif cult to integrate v I2 22 with respect to I or 2 but we can integrate with respect to y togetFyI222j I EXAMPLE 5 Let R be the region de ned by I2 y2 22 S 1 Use the divergence theorem to evaluate 22 dV R SOLUTION Let S be the unit sphere I2 12 22 1 By the divergence theorem mamw 1 where F is any vector eld whose divergence is 22 One poss1ble choice is F gzsk 22dvzskdA 1 All that remains is to compute the surface integral Ezsk dA s We have parameterized the sphere many times by now Z 250 ICOSUCOSt TCOSU so ycosusint T zs1nu zs1nu ml 7T ogtgw and iggu u 1 27f 7r2 E sinu3 cosusinududt 0 77r2 27T 7r2 4 s1n ucosudu 3 77r2 2 1 5 y s1n u 3 5 77r2 47F I 15 Of course in the last example it would have been faster to simply compute the triple integral In reality the divergence theorem is only used to compute triple integrals that would otherwise be difficult to set up EXAMPLE 6 Let S be the surface obtained by rotating the curve around the zaXis r z 005 sin 2n 7T 7 Su 2 ml Use the divergence theorem to find the volume of the region inside of S u so 27v 7r2 zkdA sin2ucosusinududt s 0 42 7r2 27T sin2ucosusinudu 7r2 7r2 l 2w 7 sin 410 according to my calculator 42 to

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