Math 2263 Solutions for Spring 2003 Final Exam

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1 Math 6 Solutions fo Sping Final Exam ) A staightfowad appoach to finding the tangent plane to a suface at a point ( x, y, z ) would be to expess the cuve as an explicit function z = f ( x, y ), calculate the slope of the cuves at that point that ae the esult of coss-sections with planes paallel to the yz- and xz-planes, and then find the equation of the plane containing vectos with those slopes at that point. This may not be all that easy in pactice (o even possible), since we have seen in alculus II that cuves in the plane ae not always given by explicit functions of x. Fotunately, thee is an equivalent and moe diect way of finding the tangent plane to a point, as the gadient vecto points in a diection nomal to the tangent plane; we know how to find the equation fo the plane fom thee, using the components of a nomal vecto. So saying, we wite the given expession fo the suface as a function F ( x, y ) = x y y z + xz, fo which we find the gadient function as " F = #F # x, #F # y, #F #z = x y + z, x 4 yz, y + 6xz. The nomal vecto to the suface at the point (,, ) is thus " F (,, #) = # (#), + 4 (#), # + 6 (#) = 7, 8, #4. The equation of the plane pependicula to this vecto (that is, the plane having " F as a nomal vecto) is given by " F # x x, y y, z z = 7 ( x ) + 8 ( y ) + (4 ) (z [ ]) = " 7x # 4 + 8y # 8 # 4 z # 4 = " 7x + 8y # 4 z = 6. (D) ) Fo a space cuve descibed by the vecto function ( t ) = < x ( t ), y ( t ), z ( t ) >, the unit tangent vecto function is found fom T(t ) = ' (t ). Fo the cuve ' (t ) (t ) = e t + e "t, e t " e "t, t, the unit tangent vecto is T(t ) = = d dt e t + e "t, e t " e "t, t = ( dx dt ) + ( dy dt ) + ( dz dt ) e t " e "t, e t + e "t, (e t " e "t ) + (e t + e "t ) + () e t " e "t, e t + e "t, (e t " + e "t ) + (e t + + e "t ) + 4 e t e "t = et " e"t, et + e"t, (continued)

2 = e t " e "t, e t + e "t, # e t + + e "t = e t " e "t, e t + e "t, # (e t + e "t ) = e t " e "t, e t + e "t, (e t + e "t ). (In othe textbooks and in othe fields, such as physics, this is often notated with a caet, as ˆ T, to make it clea that a unit vecto is intended.) At the point coesponding to t =, the unit tangent vecto to ( t ) is thus T() = e " e ", e + e ", (e + e " ) o e " e, e + e, (e + e ). (D) ) Multivaiate functions follow a vesion of the hain Rule that involves patial deivatives and takes into account the possibility of each vaiable being, in tun, a function dependent on moe than one vaiable. In this Poblem, we have u as a function of x and y, each of which is a function of two vaiables, s and t ; we would need to expess this as u = f ( x ( s, t ), y ( s, t ) ). Fo a function of one vaiable, which is also a function of one vaiable, say, z = g ( v ( w ) ), we have seen that the hain Rule has the dz fom dw = dz dv " dv. The analogous fom fo the function fo diffeentiating the dw du function fo u with espect to t is = "u dt " x # " x + "u "t " y # " y "t, since we must now conside that t appeas as a vaiable in two sepaate vaiable functions. We have u = x + y, with x = e st and y = e s t, fo which we will need the "u " y = y, " x = se "t st, and " y = # e "t s#t. To evaluate patial deivatives "u " x = x, the total deivative du at s =, t =, we assemble all of these pieces to calculate dt du dt s =, t = = x " se st + y " (# e s#t ) { } s =, t = = { (e st ) " se st + (e s#t ) " (# e s#t )} s =, t = = (e " ) " " e " + (e # ) " (# e # ) = 6 (e ) " e + (e) " (# e) = 6 e 6 # e. (B)

3 4) Fo this Poblem, we classify citical points of a two-vaiable function by using an extension of the Second Deivative Test fo functions of one vaiable. We calculate an index, which uses second patial deivatives, D = f xx f yy ( f xy ). When we used the Second Deivative Test fo a function of one vaiable, we could detemine the diection of concavity of the function to tell us, if the second deivative was not zeo, whethe a citical point is a local maximum o minimum. With a function of moe than one vaiable, the situation becomes moe complicated, since the function may have diffeent diections of concavity at a point along diffeent dimensions. Fo ou function, which has two vaiables, the Test becomes D > and f xx > : citical point is a local minimum ; D > and f xx < : citical point is a local maximum ; D < : citical point is a saddle point (concave upwad in one dimension, concave downwad in the othe dimension). Fo ou function, f ( x, y ) = x + y " x y + 9x " y, which is a continuous function, its citical points will be located whee the patial deivatives f x ae simultaneously zeo: f y and f x = x " y + 9 = f y = y " x " = # y = x + # x " (x + ) + 9 = " x # 9x + 6 = ( x # x + ) = ( x # ) ( x # ) = " x = " y = # + = 4 x = " y = # + = 7. So we have found citical points fo this function at (, 4 ) and (, 7 ). We will next find the vaious second patial deivatives of f ( x, y ) in ode to evaluate D at the two citical points: f xx = 6x, f yy =, f xy = f yx = " D = ( 6x ) ( ) = 6x 9 ; D (,4 ) = 6 " # 9 < (, 4 ) is a saddle point ; D (,7 ) = 6 " # 9 >, f xx (,7 ) = 6 " > " (, 7 ) is a local minimum. ()

4 5) The double integal x + " " f ( x, y ) dy dx instucts us to integate the function of two x vaiables f ( x, y ) ove a egion extending fom x = to x = and then above the cuve y = x and below the cuve y = occu fo x + ; since the intesections between these two cuves x = x + " x ( x + ) = 6 " x + x # 6 = ( x + ) ( x # ) =, we find that the limit x = is aived at automatically (the othe intesection at x = occus on the othe banch of the hypebola and so is not elevant hee). The egion of integation is shown in the gaph below, with the y-axis maked in ed and the two cuves, in blue. egion of integation is to the ight of the y-axis (in ed) and to the left of both cuves in blue If we choose to evese the ode of integation, then we will need to wok out the boundaies of the egion in tems of the vaiable y fist, and then expess the limits fo the vaiable x. We see that the cuves coss the y-axis at (, ) and (, ), so it would seem easonable that the limit of integation fo y should simply un fom y = to y =. Howeve, if we conside how the cuves behave in the positive x-diection, thee is the issue of thei intesection at (, ). Fom y = to y =, the uppe bounday in the positive x-diection is the line y = x, but it changes ove to the hypebola y = fom y = to y = ; the lowe bounday is the y-axis x + thoughout. (continued)

5 Since we will be integating with espect to y fist now, we need to descibe these cuves as functions of y, which we can accomplish by solving the given functions fo x : y = x " x = y, y = x + " x + = y " x = y #. The switch-ove of bounday functions means that we will have to wite two sepaate double integals now, both with the same lowe y-limit, x =, but each with a diffeent uppe limit cuve. The new desciption of the integation is theefoe y y # " " f ( x, y ) dx dy + " " f ( x, y ) dx dy. (A) 6) We ae asked to pefom a line integation of the given expession in the counteclockwise diection aound the semi-cicle, of adius and centeed on the oigin, which lies in the half-plane to the ight of the y-axis. The diect way to make this calculation is to sepaate the integation into two pats: one following staight down the y-axis fom (, ) to (, ), which keeps x = and dx =, the othe following along the semicicle, which will call fo the use of pola coodinates. So we will wite # (x + 8y ) dx + (x " 7y ) dy A = # (x + 8y ) dx + (x " 7y ) dy + # (x + 8y ) dx + (x " 7y ) dy B # x =, dx = = ( " + 8y ) " + ( " # 7y ) dy B A + ( cos " + 8 sin " ) (#sin " d" ) + ( cos " # 7sin " ) (cos " d" ) # on semi-cicle of adius : x = cos #, dx = sin # d# ; y = sin #, dy = cos # d#

6 " = # " 7y dy + ( cos " # 8 sin " # sin " cos " ) d". # With these simplifications made, we can commence evaluating each integal. The fist one " " yields # " 7y dy = & " 7 y ' ) ( = & " 7 ' ( ) " & " 7 ' ) = ( ; we can also see that this integal will be zeo, since we ae integating an odd function ove an inteval symmetic about y =. The second integal can be dealt with by applying the tigonometic double-angle identities to obtain & ( " [ + cos # ] 8 " [ cos # ] 5 sin # ) d# = (" 5 + cos # " 5 sin # ) d# = (" 5 # + 4 sin # + 5 cos # ) " " = (" 5 # + 4 # + 5 # [ "]) " (" 5 #[ " ] + 4 # + 5 # [ "]) = " 5. This then is the value of the oiginal integal as well. (B) This is cetainly an effot making good use of ou studies in alculus II. But we now have a much moe efficient tool fo calculating line integals on closed cuves: Geen s Theoem. Fo a line integal of the fom " P( x, y ) dx + Q( x, y ) dy, with the integation following counte-clockwise aound a simple closed cuve, which is piecewise-continuous, and the functions P and Q have continuous patial deivatives in the open egion enclosed by, we find " P( x, y ) dx + Q( x, y ) dy = "". D #Q # x #P ( ' & # y * da ) Fo ou line integal, these deivatives ae given by P( x, y ) = x + 8y " #P # y = 8, Q( x, y ) = x 7y " #Q # x =, making the integal in question equal to the double integal ## ( " 8) da = (" 5) da D ##. Since the integand simply becomes a constant, this can be evaluated as ( 5 ) times the aea enclosed by the semi-cicle of adius ; hence, D # (x + 8y ) dx + (x " 7y ) dy = ## (" 5) da = ( " 5) (+) ( ) = " 5 D W aea of cicle.

7 7) Hee we ae also pefoming a line integal, but it is along an open, piecewise continuous cuve: we follow the paabola y = x fom A (, 9 ) to B (, 4 ) and then the line y = x fom B (, 4 ) to the oigin (, ) [the path is shown on the gaph below]. The integal to be calculated is " x y dx + ( x + ) dy, which will have to be split into two integals, since the cuve is assembled piecewise. We essentially teat each integal as if it wee following a paametic cuve (as we did in alculus II), with x being the paamete: B " x y dx + ( x + ) dy + " x y dx + ( x + ) dy A B = # x " x dx + ( x + ) (x dx ) + # x " x dx + ( x + ) ( dx ) y = x " dy = x dx y = x " dy = dx = " (4x + 4 x ) dx + " (6x + 4 ) dx = ( x 4 + x ) + (x + 4 x ) = [ ( 4 + " ) # ( 4 + " ) ] + [ ( " + 4 " ) # ( " + 4 " ) ] = " 8 " " 6 " 8 = " 99. (D)

8 8) Because both the cone and the cylinde have symmety about the z-axis in evey diection (what is efeed to as axial symmety ), it will be convenient to wok with the sufaces in this Poblem in cylindical coodinates. So the cylinde x + y = 4 can be descibed by = 4, o just = (the absence of the z-coodinate tells us that this suface is the esult of the cicle = in the xy-plane being caied upwad and downwad indefinitely along the z-axis). The cone x + y = z becomes = z in cylindical coodinates, by a simila agument. It is then clea that the two sufaces intesect at z =, that is, at z = and z =. We wish to wite an integal expession fo the volume that is within the cylinde, but outside the cone. The sufaces both possess symmety about the xy-plane, so we could conside the volume fo z " only, and then double the esult. This allows us to egad the integation of the volume as the integation of a height function ove the cicle = in the xy-plane, V = z d d" ##, that function being z =. The axial symmety of the sufaces means that the function z has no dependence upon #, so we may simply integate the angle diffeential ove # # <. The integation ove adius is caied fom the oigin, =, to the adius at which the sufaces meet, =. Thus, the integal fo the complete volume between the cone and the cylinde is V = " d d# o d d#. 9) The potential function associated with a vecto field is a function which poduces a numeical value (what physicists call a scala potential ), so choices (A) and () can immediately be discaded. We seek a function f ( x, y, z ) fo which the gadient, which is a vecto function, is " f = F = 6x # 5y # 6xz, #x y + z, # 6x z + 6yz. Since each component of the gadient function is a patial deivative of f with espect to each of the thee vaiables, we could integate the components ove each coesponding vaiable, in ode to uncove the potential function: " F x dx = " 6x # 5y # 6xz d x = x # 5y x # x z + D( y, z ), " F y dx = " (#x y + z ) d y = # 5 x y + z y + E( x, z ), " F z dx = " (# 6x z + 6yz ) d z = # x z + yz + F( x, y ). (Hee, F x, F y, and F z epesent the components of F, athe than being patial deivatives, and D, E, and F ae abitay constant functions, in the sense that they have no dependence upon the vaiable being integated.) We can now pull togethe the unique tems that ae poduced in the thee integations; we note that tems which did not appea in any one integal ae those which ae functions only of the othe two vaiables (fo example, the tem yz, which appeas in the second and thid lines, does not depend on x, and so must be the abitay constant function D ( y, z ) ). Assembling the unique tems tells us that the associated potential function fo F is f ( x, y, z ) = x 5xy x z + yz. (E) (A)

9 ) To find the centoid of a egion of the xy-plane having unifom (constant) density, we need to find the aea of the egion and the moments of the egion about each coodinate axis. The coodinates of the centoid ae then found fom x = M y " A, y = M x " A. The bounday cuves fo the egion, y = x and y = 4x x intesect at the points given by x = 4x x " x x = " x ( x ) = " x = o ; so the intesections lie at (, ) and (, ). In the inteval < x <, the paabola is above the line, so the aea integal is A = # (4x " x ) " x dx = # (x " x ) dx = ( x " x ) = ( " # " ) # ( " # " ) = 7 # 9 # + = 9. The moment integals fo this egion ae M y = " x # (x x ) dx = " (x x ) dx = " ( x 4 x 4 ) = " [ ( # 4 4 ) # ( # 4 4 ) ] = " (7 # 8 4 # + ) = 7 4 ", M x = " #[ (4x x ) x ] dx = " ( 5 x 4 x + x 4 ) dx = " ( 5 x # x 4 + x 5 ) = " [ ( 5 # 4 + # 5 ) ( 5 # 4 + # 5 ) ] = " ( 5 # # + # ) = 54 5 ". At last, we find the coodinates of the centoid to be x = 7 4 " " # & 9 ' ) ( = 7 4 # 9 =, y = 54 5 " " # & 9 ' ) ( = 54 5 # 9 = 5. (D) A gaph of the egion, with the centoid maked, is pesented below.

10 ) This Poblem is of a sot simila to Poblem 8, fo which we need to find a way to descibe the geometic egion involved, in ode to be able to wite the volume integal. We ae told that the egion is in the fist octant of R, which equies that x ", y ", and z " (this makes the additional emak in the poblem statement about bounday planes x = and y = edundant). Thee is anothe bounding plane at z = a, which seves to put a ceiling on the egion. One bounday suface is the cylinde x + y = a, which intesects the planes at z = and z = a in quate-cicles based on the same equation. Finally, the second bounday suface is the sphee x + y + z = a, which acts as a floo fo the egion, since it is above the z = plane. Because a sphee is involved, it may seem easonable to use spheical coodinates fo ou desciption. But the cylinde only has axial symmety, making it inconvenient to descibe in such a system. So we will use cylindical coodinates instead, as the sphee is somewhat less toublesome to descibe using that system. Analogously to what we did in Poblem 8, we will think about the quate of the cicle x + y = a in the xy-plane using pola coodinates and integate the volume of the egion using a height function above that quate-cicle. To aange that, we will expess the suface of the sphee in the fist octant in cylindical coodinates as x + y + z = a " z = a # ( x + y ) z = a #. positive squae oot only The cylinde, which povides the bounding wall of the egion in the fist octant, is simply x + y + z = a = a " = a. As fo the volume integal itself, woking in cylindical coodinates leads us to V = " dv = """ dz d d#, using the volume diffeential in that system (we ae using an invisible function f (, #, z ) =, since we only wish to integate the volume itself and not a function ove that egion). So the ange of heights, ove which we will integate z, uns fom the spheical suface z = a " upwad to the ceiling at z = a. The integation fo adius extends fom the oigin = (popely speaking, fom

11 the z-axis) out to the cylinde at = a. Finally, as we ae woking in the fist octant (o fist quadant of the xy-plane), the integation in azimuth angle uns fom # = to " = #. Hence, the volume of ou egion is given by V = / a a # # # dz d d. () a " ) Fo a poblem of this kind, we have to look ove each statement choice and conside whethe it is tue (only one is coect): (A) A gadient field is just the vecto field poduced by applying the gadient opeation to a (scala) function, so thee is no special popety of this field. Taking the divegence of this field will be possible as long as the patial deivatives of the component functions fo the vectos in the field ae defined. So the divegence of the vecto field need not be zeo. (The divegence of the gadient of a vecto field is epesented by what is called the Laplace opeato o, often, just the Laplacian on the field, " # (" F ) = " F, which, in geneal, is not zeo.) (B) If the second-ode patial deivatives of the component functions of a vecto " # (" F ) is zeo field F ae continuous, then we ae guaanteed that div cul F o (Theoem, Section 6.5, Stewat, 7 th ed.). Thee is no dependence on the domain fo the tuth of this statement. () The divegence of a vecto field is a scala function. Since the gadient is a vecto opeato, it cannot be applied to the divegence of a field. The statement fo this choice thus descibes a meaningless combination of opeations. (D) By the natue of the cul opeato, its application to the gadient of a scala function f automatically gives a value of zeo [ " # (" f ) = ], povided that the second-ode patial deivatives of that function ae continuous (Theoem, Section 6.5, Stewat, 7 th ed.). ) To locate the maximum o minimum of a multivaiate function on a suface in thee dimensions, we need to use the method of Lagange multiplies : essentially, we seek citical points on the suface whee the nomal vecto is paallel to the gadient vecto of the function. (In alculus I, with functions of only one vaiable, we find extema by what amounts to a geatly-educed vesion of this method.) So saying, we will compute the gadient of the function to be extemized, &f, and the gadient of the function fo the suface, &g, and solve the equations poduced fo the set of components, &f = ' &g, with ' being an unspecified constant; an additional equation, often povided by the equation of the suface itself, will also be needed (this is efeed to as the constaint equation ). Fo this Poblem, we find f = x + 7y " z # f =, 7, ", g = x + 7y + z " # g = 4 x, 4 y, 6z ; (continued)

12 fom these esults, we constuct thee component equations fom &f = ' &g : = ' 4x, 7 = ' 4y, = ' 6z " # = x = y = z The ole of ' vaies among diffeent sots of poblems: sometimes, we want to detemine its value, wheeas hee, it seves to show the elationship among the coodinates at the extemal points on the suface. We see fom this multi-pat equation that we must have x = y = z. Hee is whee we bing in the equation fo the suface; inseting this appopiately into that equation, we obtain x + 7y + z = 6 " x + 7x + (# x ) = 6 x = 6. " x = y = ±, z = m. Fom this, we have located two extemal points on the suface, one at (,, " ), the othe at (", ", ). To esolve the issue of whee the maximum lies, we will use both sets of coodinate values in the function f ( x, y, z ) : f (,, " ) = # & # ( + 7 ' & # ( " " & # ' ( = ' & ( = 6 " maximum, ' f (", ", ) = # " & # ( + 7 " & # ' ( " ' & # ( = " ' & ( = " 6 " minimum. ' Inteestingly, the maximal and minimal values of f on the suface have exactly opposite-signed values and fall at points in exactly opposite diections fom the oigin. This should not be too supising, as the level cuves of the function ae a set of paallel planes and the ellipsoid is a suface which is symmetic about the oigin. 4) a) Fo the paametic suface defined by = < u cos v, u sin v, v >, we find the nomal vecto at a point by locating the tangent plane thee fist. Fo this, we need the tangent vectos u = " "u = cos v, sin v,, v = " = # u sin v, u cos v,. "v The nomal vecto to the suface is then pependicula to both of these vectos, fo which we can find a desciption by computing the coss poduct u " v " = " i j k cos v sin v # u sin v u cos v the deteminant is not a definition, but simply a notational convenience = sin v, # cos v, u cos v + u sin v = sin v, " cos v, u. (continued)

13 We want to make a unit nomal vecto of this, so we detemine its length and divide though by that esult: ˆ n = u " v u " v = sin v, #cos v, u = sin v + (#cos v ) + u sin v, #cos v, u + u. b) We now wish to compute the suface integal "" x + y ds ove the S suface S coveed by # u # and # v #. The suface aea diffeential is given by ds = u " v du dv = + u du dv, so the integal becomes # # x + y " + u du dv # # = (u cos v ) + (u cos v ) " + u du dv # # = " u + u du dv " # dv w dw = v " " w = + u, dw = u du ( w / ) = " # # ( / / ) = " ( ). 5) a) The solid unde consideation is descibed in atesian coodinates by the inequality z " ( x + y + z ) # x + y + z, which is not all that easy to visualize. We will use some of the coodinate tansfomation equations to convet this expession to spheical coodinates, namely, x + y + z = and z = cos (.* This pemits us to simplify the expession of the inequality consideably: z " ( x + y + z ) # x + y + z cos & " # " cos # & " ' + cos #. * Be awae that in othe textbooks and othe fields, # denotes the pola angle and ( the azimuthal angle fo spheical coodinates. (continued)

14 The equation = + cos ( descibes a cadioid oiented with the maximum adius = along the z-axis ( ( = ) in, say, the yz-plane (as shown in the gaph below); because this equation has no dependence on the azimuthal angle, #, it descibes a suface of evolution about the z-axis in thee-dimensional space. Since the inequality coves adii equal to o smalle than those on the cadioid, it epesents the inteio of the cuve with the cadioid as its bounday. Oveall, the solid geneally esembles, pehaps, an apple o tomato sitting stem end down with its ounded end facing upwad. coss-section of the solid in the yz-plane b) Since the adius function = + cos ( is always defined, the full inteval of pola angles is allowed, so # ( #. In that inteval, we will thus find that # # + cos ( (the ange of possible values coveed is # # ). The adius equation does not depend on azimuth, so the entie inteval fo that angle is applicable, hence # # <. c) In spheical coodinates, the diffeential volume is dv = d sin ( d( d#. So the volume integal fo this solid, coveing the full inteio, is & & + cos# V = " d" sin # d# d = & & ( " + cos# ) sin # d# d = ( + cos" ) sin " d" d# & d# u (' du ) u = + cos (, du = sin ( = " # ( ) " & 4 u 4 ( ' ) + = * " " (4 4 ) = " (6) = 8.

15 6) We ae asked hee to pefom a line integal of a vecto function aound a closed cuve in thee-dimensional space. This cuve is defined by the intesection of the plane x + y + z = with the thee coodinate planes; each of the intesections is a staight line, so the cuve is a tiangle tilted with espect to the xy-, yz-, and xz-planes. The integal is " F # d = " y x, x z, x y # d : to evaluate it, we would need to wok out a vecto epesenting each of the thee sides of the tiangle, fom a diffeential vesion of it, calculate its dot poduct with F, and compute the esulting integal along the espective leg of the tiangle. Fo example, along the leg in the xy-plane fom A (,, ) to B (,, ), we have the vecto <,, >, with which we could descibe this path paametically by < t, + t, >, poduce the diffeential d = < dt, dt, >, and at last calculate y " x, x " z, x " y # d = t " ( " t ), ( " t ) ", ( " t ) " t # " dt, dt, A B = # (" 6t + 4 ) + ( " t ) dt = # 6 " 8t dt = (6t " 4t ) =. ontinuing on fom B to (,, ), the path is given by <, t, t >, with the diffeential d = <, dt, dt >, making the integal y " x, x " z, x " y # d = ( " t ) ", " t, " ( " t ) #, " dt, dt B = # (t ) + (" + t ) dt = # 4t " dt = (t " t ) =. Finally, closing the cuve by etuning fom to A, the path followed is < t,, t >, which has the diffeential d = < dt,, dt > ; the emaining integal is y " x, x " z, x " y # d = " (t ), (t ) " ( " t ), (t ) " # dt,, " dt A = # (" 4t ) + (" 4t ) dt = # " 8t dt = (" 4t ) = " 4. Adding the esults fo the thee legs of the tiangle togethe yields F " d = + + (# 4 ) = #. (Notice that we do not use unit vectos fo the diections of the paths: the lengths contain infomation about the size of the coesponding paths, and so ae impotant in evaluating the line integal coectly.) (continued)

16 Thee is a athe less tedious way to obtain this value. It is possible to extend Geen s Theoem into highe-dimensional spaces, becoming what is known as Stokes Theoem: fo simila conditions to those equied fo Geen s Theoem, as discussed in Poblem 6 above, we have # F " d = ## ( F ) " d S, with d S being the unit S nomal vecto associated with a diffeential patch of aea da. We know fom alculus II that the nomal vecto to the plane containing is <,, >, so d S = <,, > da anywhee in the tiangle. The cul of the vecto field F is found fom " # F " = " i j k x y z y x x z x y = () (),, =,,. We can now apply Stokes Theoem in vaious ways. We can wite the integal as (" # F ) d S =, &,,, da =, &,,, dy dx, S S with S being the pojection of the tilted tiangle onto the xy-plane. The pojection has legs connected the points (,, ), (,, ), and (,, ), so one of the sides lies along the line x + y =, o y = x. This leads to the suface integal, ", #,, dy dx = (" ) dy dx S " x " = (" ) y # x dx = (" ) # " x dx = (" ) ( x " 4 x ) = (" ) # = ". Since the pojected tiangle is a ight tiangle with sides of lengths and, and the integand is just the constant ( ), we could also simply evaluate this suface integal as ## (" ) dy dx = (" ) A poj = (" ) ( ) = ". S (" # F ) ˆ Some texts also give this suface integal as n being S the unit nomal vecto to the egion bounded by the cuve; in this case, the aea to be integated is the full aea of the egion, athe than its pojection onto a coodinate plane. The unit nomal to the tilted tiangle is ˆ n = S,,, with ˆ + + = 6,,. Fo the aea of this tiangle, we may note that it is an isosceles tiangle with conguent sides of length 5 and a base (the side unning fom to A ) of length 8 =. If we constuct the altitude to this base, we poduce conguent ight tiangles with bases of

17 length and hypotenuses of length 5 ; by the Pythagoean Theoem, the altitudes have length. The aea of the tilted tiangle is theefoe A = = 6. The suface integal again has a constant integand and so will have a value which is just a multiple of the aea of the tilted tiangle, (" # F ) n ˆ da =, &, T T 6,, da # = " & ( )) da 6 ' = # " & 6 ' ( A = # " & 6 ( * 6 = ", as was found above. ' T G. Ruffa 6/