CALCULUS III Surface Integrals. Paul Dawkins

Transcription

1 CALCULU III uface Integals Paul awkins

2 Table of Contents Peface... ii uface Integals... 3 Intoduction... 3 Paametic ufaces... 4 uface Integals... uface Integals of Vecto Fields... 9 tokes Theoem... 9 ivegence Theoem Paul awkins i

3 Peface Hee ae my online notes fo my Calculus III couse that I teach hee at Lama Univesity. espite the fact that these ae my class notes, they should be accessible to anyone wanting to lean Calculus III o needing a efeshe in some of the topics fom the class. These notes do assume that the eade has a good woking knowledge of Calculus I topics including limits, deivatives and integation. It also assumes that the eade has a good knowledge of seveal Calculus II topics including some integation techniques, paametic equations, vectos, and knowledge of thee dimensional space. Hee ae a couple of wanings to my students who may be hee to get a copy of what happened on a day that you missed.. Because I wanted to make this a faily complete set of notes fo anyone wanting to lean calculus I have included some mateial that I do not usually have time to cove in class and because this changes fom semeste to semeste it is not noted hee. You will need to find one of you fellow class mates to see if thee is something in these notes that wasn t coveed in class.. In geneal I ty to wok poblems in class that ae diffeent fom my notes. Howeve, with Calculus III many of the poblems ae difficult to make up on the spu of the moment and so in this class my class wok will follow these notes faily close as fa as woked poblems go. With that being said I will, on occasion, wok poblems off the top of my head when I can to povide moe examples than just those in my notes. Also, I often don t have time in class to wok all of the poblems in the notes and so you will find that some sections contain poblems that ween t woked in class due to time estictions. 3. ometimes questions in class will lead down paths that ae not coveed hee. I ty to anticipate as many of the questions as possible in witing these up, but the eality is that I can t anticipate all the questions. ometimes a vey good question gets asked in class that leads to insights that I ve not included hee. You should always talk to someone who was in class on the day you missed and compae these notes to thei notes and see what the diffeences ae. 4. This is somewhat elated to the pevious thee items, but is impotant enough to meit its own item. THEE NOTE ARE NOT A UBTITUTE FOR ATTENING CLA!! Using these notes as a substitute fo class is liable to get you in touble. As aleady noted not eveything in these notes is coveed in class and often mateial o insights not in these notes is coveed in class. 7 Paul awkins ii

4 uface Integals Intoduction In the pevious chapte we looked at evaluating integals of functions o vecto fields whee the points came fom a cuve in two- o thee-dimensional space. We now want to extend this idea and integate functions and vecto fields whee the points come fom a suface in theedimensional space. These integals ae called suface integals. Hee is a list of the topics coveed in this chapte. Paametic ufaces In this section we will take a look at the basics of epesenting a suface with paametic equations. We will also take a look at a couple of applications. uface Integals Hee we will intoduce the topic of suface integals. We will be woking with suface integals of functions in this section. uface Integals of Vecto Fields We will look at suface integals of vecto fields in this section. tokes Theoem We will look at tokes Theoem in this section. ivegence Theoem Hee we will take a look at the ivegence Theoem. 7 Paul awkins 3

5 Paametic ufaces Befoe we get into suface integals we fist need to talk about how to paameteize a suface. ab, and plugged them When we paameteized a cuve we took values of t fom some inteval [ ] into t x t i y t j z t k = + + and the esulting set of vectos will be the position vectos fo the points on the cuve. With sufaces we ll do something simila. We will take points, (, ) uv, out of some twodimensional space and plug them into uv, = x uv, i + y uv, j + z uv, k and the esulting set of vectos will be the position vectos fo the points on the suface that we ae tying to paameteize. This is often called the paametic epesentation of the paametic suface. We will sometimes need to wite the paametic equations fo a suface. Thee ae eally nothing moe than the components of the paametic epesentation explicitly witten down. x= x uv, y = y uv, z = z uv, Example etemine the suface given by the paametic epesentation uv, = ui + ucosvj + usin vk olution Let s fist wite down the paametic equations. x= u y = ucosv z = usin v Now if we squae y and z and then add them togethe we get, y + z = u cos v+ u sin v= u cos v+ sin v = u = x o, we wee able to eliminate the paametes and the equation in x, y, and z is given by, x = y + z Fom the Quadic ufaces section notes we can see that this is a cone that opens along the x-axis. We ae much moe likely to need to be able to wite down the paametic equations of a suface than identify the suface fom the paametic epesentation so let s take a look at some examples of this. Example Give paametic epesentations fo each of the following sufaces. (a) The elliptic paaboloid x= 5y + z -. [olution] (b) The elliptic paaboloid x= 5y + z - that is in font of the yz-plane. [olution] (c) The sphee x + y + z = 3. [olution] (d) The cylinde y + z = 5. [olution] 7 Paul awkins 4

6 olution (a) The elliptic paaboloid x y z = This one is pobably the easiest one of the fou to see how to do. ince the suface is in the fom x= f yz, we can quickly wite down a set of paametic equations as follows, x= 5y + z - y = y z = z The last two equations ae just thee to acknowledge that we can choose y and z to be anything we want them to be. The paametic epesentation is then, yz, = 5y + z - i + yj + zk [Retun to Poblems] (b) The elliptic paaboloid x y z = that is in font of the yz-plane. This is eally a estiction on the pevious paametic epesentation. The paametic epesentation stays the same. yz, = 5y + z - i + yj + zk Howeve, since we only want the suface that lies in font of the yz-plane we also need to equie that x. This is equivalent to equiing, 5y + z - o 5y + z [Retun to Poblems] (c) The sphee x + y + z = 3. This one can be a little ticky until you see how to do it. In spheical coodinates we know that the equation of a sphee of adius a is given by, = a and so the equation of this sphee (in spheical coodinates) is = 3. Now, we also have the following convesion fomulas fo conveting Catesian coodinates into spheical coodinates. x= sinjcosq y = sinjsinq z = cosj Howeve, we know what is fo ou sphee and so if we plug this into these convesion fomulas we will aive at a paametic epesentation fo the sphee. Theefoe, the paametic epesentation is, qj, = 3sinjcosqi + 3sinjsinq j + 3cosjk All we need to do now is come up with some estiction on the vaiables. Fist we know that we have the following estiction. j p This is enfoced upon us by choosing to use spheical coodinates. Also, to make sue that we only tace out the sphee once we will also have the following estiction. q p [Retun to Poblems] 7 Paul awkins 5

7 (d) The cylinde y + z = 5. As with the last one this can be ticky until you see how to do it. In this case it makes some sense to use cylindical coodinates since they can be easily used to wite down the equation of a cylinde. In cylindical coodinates the equation of a cylinde of adius a is given by = a and so the equation of the cylinde in this poblem is = 5. Next, we have the following convesion fomulas. x= x y = sinq z = cosq Notice that they ae slightly diffeent fom those that we ae used to seeing. We needed to change them up hee since the cylinde was centeed upon the x-axis. Finally, we know what is so we can easily wite down a paametic epesentation fo this cylinde. x, q = xi + 5sinq j + 5cosq k We will also need the estiction q p to make sue that we don t etace any potion of the cylinde. ince we haven t put any estictions on the height of the cylinde thee won t be any estiction on x. [Retun to Poblems] In the fist pat of this example we used the fact that the function was in the fom x f ( yz, ) = to quickly wite down a paametic epesentation. This can always be done fo functions that ae in this basic fom. z = f xy fi xy = xi + yj + f xy k x= f yz fi xy = f yz i + yj + zk y = f xz fi xy = xi + f xz j + zk (, ) (, ) (, ) (, ) (, ) (, ) (, ) (, ) (, ) Okay, now that we have pactice witing down some paametic epesentations fo some sufaces let s take a quick look at a couple of applications. Let s take a look at finding the tangent plane to the paametic suface given by, uv, = x uv, i + y uv, j + z uv, k Fist, define x y z u( uv, ) = ( uv, ) i + ( uv, ) j + ( uv, ) k u u u x y z v( uv, ) = ( uv, ) i + ( uv, ) j + ( uv, ) k v v v 7 Paul awkins 6

8 Now, povided u v it can be shown that the vecto u v will be othogonal to the suface. This means that it can be used fo the nomal vecto that we need in ode to wite down the equation of a tangent plane. This is an impotant idea that will be used many times thoughout the next couple of sections. Let s take a look at an example. Example 3 Find the equation of the tangent plane to the suface given by uv, = ui + v j + u + v k at the point (,,3 ). olution Let s fist compute u v. Hee ae the two individual vectos. uv, = i + uk uv, = 4vj + k u Now the coss poduct (which will give us the nomal vecto n ) is, i j k n = = u =-8uvi - j + 4vk u v 4v v Now, this is all fine, but in ode to use it we will need to detemine the value of u and v that will give us the point in question. We can easily do this by setting the individual components of the paametic epesentation equal to the coodinates of the point in question. oing this gives, = u fi u = = v fi v=± 3 = u + v Now, as shown, we have the value of u, but thee ae two possible values of v. To detemine the coect value of v let s plug u into the thid equation and solve fo v. This should tell us what the coect value is. 3= 4+ v fi v=- Okay so we now know that we ll be at the point in question when u = and v =-. At this point the nomal vecto is, n = 6i - j -4k The tangent plane is then, 6 x- - y- -4 z- 3 = 6x- y- 4z = 8 You do emembe how to wite down the equation of a plane, ight? The second application that we want to take a quick look at is the suface aea of the paametic suface given by, 7 Paul awkins 7

9 ( uv, ) = x( uv, ) i + y( uv, ) j + z( uv, ) k and as we will see it again comes down to needing the vecto u. v o, povided is taced out exactly once as ( uv, ) anges ove the points in the suface aea of is given by, Let s take a look at an example. A= da u Example 4 Find the suface aea of the potion of the sphee of adius 4 that lies inside the cylinde x + y = and above the xy-plane. olution Okay we ve got a couple of things to do hee. Fist we need the paameteization of the sphee. We paameteized a sphee ealie in this section so thee isn t too much to do at this point. Hee is the paameteization. qj, = 4sinjcosqi + 4sinjsinq j + 4cosjk Next we need to detemine. ince we ae not esticting how fa aound the z-axis we ae otating with the sphee we can take the following ange fo q. q p Now, we need to detemine a ange fo j. This will take a little wok, although it s not too bad. Fist, let s stat with the equation of the sphee. x + y + z = 6 Now, if we substitute the equation fo the cylinde into this equation we can find the value of z whee the sphee and the cylinde intesect. x + y + z = 6 + z = 6 z v = 4 fi z =± Now, since we also specified that we only want the potion of the sphee that lies above the xyplane we know that we need z =. We also know that = 4. Plugging this into the following convesion fomula we get, z = cosj = 4cosj p cosj = fi j = 3 o, it looks like the ange of j will be, p j 3 7 Paul awkins 8

10 Finally, we need to detemine q j. Hee ae the two individual vectos. q ( qj, ) =- 4sinjsinqi + 4sinjcosq j qj, = 4cosjcosqi + 4cosjsinq j -4sinjk j Now let s take the coss poduct. i j k q j =-4sinjsinq 4sinjcosq 4cosjcosq 4cosjsinq -4sinj =-6sin jcosqi -6sinjcosjsin q k -6sin jsinq j -6sinjcosjcos q k =-6sin jcosqi -6sin jsinq j - 6sinjcosj( sin q + cos q) k =-6sin jcosqi -6sin jsinq j -6sinjcosjk We now need the magnitude of this, 4 4 q j = 56sin jcos q + 56sin jsin q + 56sin jcos j 4 = 56sin j cos q + sin q + 56sin jcos j = 56sin j sin j+ cos j = 6 sin j = 6 sinj = 6sinj We can dop the absolute value bas in the sine because sine is positive in the ange of j that we ae woking with. We can finally get the suface aea. A= = p p 3 p p 3 6cosj = - = p = 6p 6sinj da Ú Ú Ú Ú 8dq 6sinjdjdq dq 7 Paul awkins 9

11 uface Integals It is now time to think about integating functions ove some suface,, in thee-dimensional space. Let s stat off with a sketch of the suface since the notation can get a little confusing once we get into it. Hee is a sketch of some suface. The egion will lie above (in this case) some egion that lies in the xy-plane. We used a ectangle hee, but it doesn t have to be of couse. Also note that we could just as easily looked at a suface that was in font of some egion in the yz-plane o the xz-plane. o not get so locked into the xy-plane that you can t do poblems that have egions in the othe two planes. Now, how we evaluate the suface integal will depend upon how the suface is given to us. Thee ae essentially two sepaate methods hee, although as we will see they ae eally the same. Fist, let s look at the suface integal in which the suface is given by z g( xy, ) case the suface integal is, =. In this ÛÛ Ê gˆ Ê gˆ f ( xyz,, ) d = ÙÙ f ( xyg,, ( xy, )) Á + Á + da ÙÙ ıı Ë x Ë y Now, we need to be caeful hee as both of these look like standad double integals. In fact the integal on the ight is a standad double integal. The integal on the left howeve is a suface integal. The way to tell them apat is by looking at the diffeentials. The suface integal will have a d while the standad double integal will have a da. In ode to evaluate a suface integal we will substitute the equation of the suface in fo z in the integand and then add on the often messy squae oot. Afte that the integal is a standad double integal and by this point we should be able to deal with that. 7 Paul awkins

12 = (with in the xz- Note as well that thee ae simila fomulas fo sufaces given by y g( xz, ) plane) and x g( yz, ) = (with in the yz-plane). We will see one of these fomulas in the examples and we ll leave the othe to you to wite down. The second method fo evaluating a suface integal is fo those sufaces that ae given by the paameteization, uv, = x uv, i + y uv, j + z uv, k In these cases the suface integal is, f ( xyz,, ) d = f ( uv, ) u v da whee is the ange of the paametes that tace out the suface. Befoe we wok some examples let s notice that since we can paameteize a suface given by z = g( xy, ) as, xy, = xi + yj + g xy, k we can always use this fom fo these kinds of sufaces as well. In fact it can be shown that, Ê g ˆ Ê g ˆ x y = Á + Á + Ë x Ë y fo these kinds of sufaces. You might want to veify this fo the pactice of computing these coss poducts. Let s wok some examples. Example Evaluate 6xyd font of the yz-plane. whee is the potion of the plane x y z + + = that lies in olution Okay, since we ae looking fo the potion of the plane that lies in font of the yz-plane we ae x= g yz,. This is easy enough going to need to wite the equation of the suface in the fom to do. x= - y- z Next we need to detemine just what is. Hee is a sketch of the suface. 7 Paul awkins

13 Hee is a sketch of the egion. Notice that the axes ae labeled diffeently than we ae used to seeing in the sketch of. This was to keep the sketch consistent with the sketch of the suface. We aived at the equation of the hypotenuse by setting x equal to zeo in the equation of the plane and solving fo z. Hee ae the anges fo y and z. y z - y Now, because the suface is not in the fom z g( xy, ) = we can t use the fomula above. Howeve, as noted above we can modify this fomula to get one that will wok fo us. Hee it is, ÛÛ Ê g ˆ Ê g ˆ f ( xyz,, ) d = ÙÙ f ( g( yz, ), yz, ) + Á + Á da ÙÙ ıı Ë y Ë z The changes made to the fomula should be the somewhat obvious changes. o, let s do the integal. 6xyd = 6- y- z y da Notice that we plugged in the equation of the plane fo the x in the integand. At this point we ve got a faily simple double integal to do. Hee is that wok. 7 Paul awkins

14 Example Evaluate 6xyd = 3 6 y- y -zy da -y = 6 3Û - - ı Ú y y zydzdy -y Ê ˆ yz zy z y dy Û = 6 3Ù Á - - ı Ë = 6 3 Û Ù - + ı 3 y y y dy Ê 3 4ˆ 6 3Á y y y Ë = - + = zd whee is the uppe half of a sphee of adius. olution We gave the paameteization of a sphee in the pevious section. Hee is the paameteization fo this sphee. qj, = sinjcosqi + sinjsinq j + cosjk ince we ae woking on the uppe half of the sphee hee ae the limits on the paametes. p q p j Next, we need to detemine q j. Hee ae the two individual vectos. q ( qj, ) =- sinjsinqi + sinjcosq j qj, = cosjcosqi + cosjsinq j -sinjk j Now let s take the coss poduct. i j k q j =-sinjsinq sinjcosq cosjcosq cosjsinq -sinj =-4sin jcosqi -4sinjcosjsin q k -4sin jsinq j -4sinjcosjcos q k =-4sin jcosqi -4sin jsinq j - 4sinjcosj( sin q + cos q) k =-4sin jcosqi -4sin jsinq j -4sinjcosjk Finally, we need the magnitude of this, 7 Paul awkins 3

15 4 4 q j = 6sin jcos q + 6sin jsin q + 6sin jcos j 4 = 6sin j cos q + sin q + 6sin jcos j = 6sin j sin j+ cos j = 4 sin j = 4sinj = 4sinj We can dop the absolute value bas in the sine because sine is positive in the ange of j that we ae woking with. The suface integal is then, zd = cosj 4sinj da on t foget that we need to plug in fo x, y and/o z in these as well, although in this case we just needed to plug in z. Hee is the evaluation fo the double integal. p p zd = Ú 4sin( j) d d Ú p = - = Ú Ú p = 8p ( cos( j) ) 4dq p j q dq Example 3 Evaluate yd whee is the potion of the cylinde between z = and z = 6. x + y = 3 that lies olution We paameteized up a cylinde in the pevious section. Hee is the paameteization of this cylinde. z, q = 3cosqi + 3sinq j + zk The anges of the paametes ae, z 6 q p Now we need z q. Hee ae the two vectos. z( z, q) = k z, q =- 3sinqi + 3cosq j Hee is the coss poduct. q 7 Paul awkins 4

16 = z q i j k - 3sinq 3cosq =- 3cosqi - 3sinq j The magnitude of this vecto is, q = 3cos q + 3sin q = 3 z The suface integal is then, yd = p 6 p ( 8cosq) 3sinq 3 = 3 sinq dzdq = 3 6sinq dq = - = Ú Ú Ú p da Example 4 Evaluate y+ zd whee is the suface whose side is the cylinde x whose bottom is the disk x + y = 3, + y 3 in the xy-plane and whose top is the plane z = 4 - y. olution Thee is a lot of infomation that we need to keep tack of hee. Fist, we ae using petty much the same suface (the integand is diffeent howeve) as the pevious example. Howeve, unlike the pevious example we ae putting a top and bottom on the suface this time. Let s fist stat out with a sketch of the suface. 7 Paul awkins 5

17 Actually we need to be caeful hee. Thee is moe to this sketch than the actual suface itself. We e going to let be the potion of the cylinde that goes fom the xy-plane to the plane. In othe wods, the top of the cylinde will be at an angle. We ll call the potion of the plane that lies inside (i.e. the cap on the cylinde). Finally, the bottom of the cylinde (not shown hee) is the disk of adius 3 in the xy-plane and is denoted by 3. In ode to do this integal we ll need to note that just like the standad double integal, if the suface is split up into pieces we can also split up the suface integal. o, fo ou example we will have, y+ zd = y+ zd + y+ zd+ y+ zd 3 We e going to need to do thee integals hee. Howeve, we ve done most of the wok fo the fist one in the pevious example so let s stat with that. : The Cylinde The paameteization of the cylinde and z q is, ( z, q) = 3cosqi + 3sinq j + zk z q = 3 The diffeence between this poblem and the pevious one is the limits on the paametes. Hee they ae. q p z 4- y = 4-3sinq 7 Paul awkins 6

18 The uppe limit fo the z s is the plane so we can just plug that in. Howeve, since we ae on the cylinde we know what y is fom the paameteization so we will also need to plug that in. Hee is the integal fo the cylinde. ( 3sinq )( 3) y+ zd = + z da p 4-3sinq = 3 3sinq + zdzdq p = 3 3sinq 4-3sinq + 4-3sinq dq = p 3 3Ú 8- sin q dq = p 3 3Ú 8- ( -cos( q) ) dq 4 Ê9 3 = 3Á q + sin Ë 4 8 = Ú Ú Ú 9 3 : Plane on Top of the Cylinde p ( q) In this case we don t need to do any paameteization since it is set up to use the fomula that we gave at the stat of this section. Remembe that the plane is given by z = 4 - y. Also note that, fo this suface, is the disk of adius Hee is the integal fo the plane. ˆ p 3 centeed at the oigin. y+ zd = y+ 4- y da = 4dA on t foget that we need to plug in fo z! Now at this point we can poceed in one of two ways. Eithe we can poceed with the integal o we can ecall that da is nothing moe than the aea of and we know that is the disk of adius Hee is the emainde of the wok fo this poblem. 3 and so thee is no eason to do the integal. 7 Paul awkins 7

19 y+ zd = 4 da ( p ) = 4 3 = p 3 : Bottom of the Cylinde Again, this is set up to use the initial fomula we gave in this section once we ealize that the g xy, = and is the disk of adius 3 centeed at the equation fo the bottom is given by oigin. Also, don t foget to plug in fo z. Hee is the wok fo this integal. 3 () p 3 p 3 Û Ê 3 ˆ = Ù Á sinq ı Ë3 p = p y+ zd = y+ + + da = = Û ı = Ú =- yda Ú sinq ddq 3sinq dq 3cosq We can now get the value of the integal that we ae afte. dq y+ zd = y+ zd + y+ zd+ y+ zd 3 9 3p = + p + p = ( ) 7 Paul awkins 8

20 uface Integals of Vecto Fields Just as we did with line integals we now need to move on to suface integals of vecto fields. Recall that in line integals the oientation of the cuve we wee integating along could change the answe. The same thing will hold tue with suface integals. o, befoe we eally get into doing suface integals of vecto fields we fist need to intoduce the idea of an oiented suface. Let s stat off with a suface that has two sides (while this may seem stange, ecall that the Mobius tip is a suface that only has one side!) that has a tangent plane at evey point (except possibly along the bounday). Making this assumption means that evey point will have two unit nomal vectos, n and n =-n. This means that evey suface will have two sets of nomal vectos. The set that we choose will give the suface an oientation. Thee is one convention that we will make in egads to cetain kinds of oiented sufaces. Fist we need to define a closed suface. A suface is closed if it is the bounday of some solid egion E. A good example of a closed suface is the suface of a sphee. We say that the closed suface has a positive oientation if we choose the set of unit nomal vectos that point outwad fom the egion E while the negative oientation will be the set of unit nomal vectos that point in towads the egion E. Note that this convention is only used fo closed sufaces. In ode to wok with suface integals of vecto fields we will need to be able to wite down a fomula fo the unit nomal vecto coesponding to the oientation that we ve chosen to wok with. We have two ways of doing this depending on how the suface has been given to us. Fist, let s suppose that the function is given by z g( xy, ) function, =. In this case we fist define a new f ( xyz,, ) = z- g( xy, ) In tems of ou new function the suface is then given by the equation f ( xyz,, ) =. Now, ecall that f will be othogonal (o nomal) to the suface given by f ( xyz,, ) =. This means that we have a nomal vecto to the suface. The only potential poblem is that it might not be a unit nomal vecto. That isn t a poblem since we also know that we can tun any vecto into a unit vecto by dividing the vecto by its length. In o case this is, f n = f In this case it will be convenient to actually compute the gadient vecto and plug this into the fomula fo the nomal vecto. oing this gives, f -gxi - gy j + k n = = f g + g + Now, fom a notational standpoint this might not have been so convenient, but it does allow us to make a couple of additional comments. x y 7 Paul awkins 9

21 Fist, notice that the component of the nomal vecto in the z-diection (identified by the k in the nomal vecto) is always positive and so this nomal vecto will geneally point upwads. It may not point diectly up, but it will have an upwads component to it. This will be impotant when we ae woking with a closed suface and we want the positive oientation. If we know that we can then look at the nomal vecto and detemine if the positive oientation should point upwads o downwads. Remembe that the positive oientation must point out of the egion and this may mean downwads in places. Of couse if it tuns out that we need the downwad oientation we can always take the negative of this unit vecto and we ll get the one that we need. Again, emembe that we always have that option when choosing the unit nomal vecto. Befoe we move onto the second method of giving the suface we should point out that we only z = g xy,. We could just as easily done the above wok fo did this fo sufaces in the fom sufaces in the fom y = g( xz, ) (so f ( xyz,, ) y g( xz, ) x= g( yz, ) (so f ( xyz,, ) = x- g( yz, ) ). = - ) o fo sufaces in the fom Now, we need to discuss how to find the unit nomal vecto if the suface is given paametically as, ( uv, ) = x( uv, ) i + y( uv, ) j + z( uv, ) k In this case ecall that the vecto u v will be nomal to the tangent plane at a paticula point. But if the vecto is nomal to the tangent plane at a point then it will also be nomal to the suface at that point. o, this is a nomal vecto. In ode to guaantee that it is a unit nomal vecto we will also need to divide it by its magnitude. o, in the case of paametic sufaces one of the unit nomal vectos will be, u v n = As with the fist case we will need to look at this once it s computed and detemine if it points in the coect diection o not. If it doesn t then we can always take the negative of this vecto and that will point in the coect diection. Finally, emembe that we can always paameteize any suface given by z g( xy, ) y = g( xz, ) o x g( yz, ) u 7 Paul awkins v = (o = ) easily enough and so if we want to we can always use the paameteization fomula to find the unit nomal vecto. Okay, now that we ve looked at oiented sufaces and thei associated unit nomal vectos we can actually give a fomula fo evaluating suface integals of vecto fields. Given a vecto field F with unit nomal vecto n then the suface integal of F ove the suface is given by, Fd g = Fnd g

22 whee the ight hand integal is a standad suface integal. This is sometimes called the flux of F acoss. Befoe we wok any examples let s notice that we can substitute in fo the unit nomal vecto to get a somewhat easie fomula to use. We will need to be caeful with each of the following fomulas howeve as each will assume a cetain oientation and we may have to change the nomal vecto to match the given oientation. =. In this case let s also assume that the vecto field is given by F = Pi + Qj + Rk and that the oientation that we ae Let s fist stat by assuming that the suface is given by z g( xy, ) afte is the upwads oientation. Unde all of these assumptions the suface integal of F ove is, Fd g = ÛÛ Ê ˆ Á -gxi - gy j + k = ÙÙ ( Pi + Qj + Rk) g ( g ) x + gy + da ÙÙ ÙÙ Á ( gx) + ( gy) + ıı Ë = Pi + Qj + Rk -g i - g j + k da Fnd g ( g x y ) = -Pg - Qg + RdA x y Now, emembe that this assumed the upwad oientation. If we d needed the downwad oientation then we would need to change the signs on the nomal vecto. This would in tun change the signs on the integand as well. o, we eally need to be caeful hee when using this fomula. In geneal it is best to edeive this fomula as you need it. When we ve been given a suface that is not in paametic fom thee ae in fact 6 possible z g xy, y g xz, x= g yz,. integals hee. Two fo each fom of the suface =, = and Given each fom of the suface thee will be two possible unit nomal vectos and we ll need to choose the coect one to match the given oientation of the suface. Howeve, the deivation of each fomula is simila to that given hee and so shouldn t be too bad to do as you need to. Notice as well that because we ae using the unit nomal vecto the messy squae oot will always dop out. This means that when we do need to deive the fomula we won t eally need to put this in. All we ll need to wok with is the numeato of the unit vecto. We will see at least one moe of these deived in the examples below. It should also be noted that the squae oot is nothing moe than, g + g + = f x so in the following wok we will pobably just use this notation in place of the squae oot when we can to make things a little simple. y 7 Paul awkins

23 Let s now take a quick look at the fomula fo the suface integal when the suface is given paametically by ( uv, ). In this case the suface integal is, Fd g = Fnd g ÛÛ Ê ˆ = ÙÙ g ıı Ë = Fg da u v F Á u v da u v u Again note that we may have to change the sign on u v to match the oientation of the suface and so thee is once again eally two fomulas hee. Also note that again the magnitude cancels in this case and so we won t need to woy that in these poblems eithe. Note as well that thee ae even times when we will used the definition, diectly. We will see an example of this below. Let s now wok a couple of examples. Example Evaluate Fd g whee F = yj -zk paaboloid y = x + z, y and the disk x oientation. v Fd g = Fnd g, and is the suface given by the + z at y =. Assume that has positive olution Okay, fist let s notice that the disk is eally nothing moe than the cap on the paaboloid. This means that we have a closed suface. This is impotant because we ve been told that the suface has a positive oientation and by convention this means that all the unit nomal vectos will need to point outwads fom the egion enclosed by. Let s fist get a sketch of so we can get a feel fo what is going on and in which diection we will need to unit nomal vectos to point. 7 Paul awkins

24 As noted in the sketch we will denote the paaboloid by and the disk by. Also note that in ode fo unit nomal vectos on the paaboloid to point away fom the egion they will all need to point geneally in the negative y diection. On the othe hand, unit nomal vectos on the disk will need to point in the positive y diection in ode to point away fom the egion. ince is composed of the two sufaces we ll need to do the suface integal on each and then add the esults to get the oveall suface integal. Let s stat with the paaboloid. In this case we y = g xz, so we will need to deive the coect fomula since the have the suface in the fom one given initially wasn t fo this kind of function. This is easy enough to do howeve. Fist define, f xyz,, = y- g xz, = y-x - z We will next need the gadient vecto of this function. f = - x,, - z Now, the y component of the gadient is positive and so this vecto will geneally point in the positive y diection. Howeve, as noted above we need the nomal vecto point in the negative y diection to make sue that it will be pointing away fom the enclosed egion. This means that we will need to use - f x, -,z n = = - f f Let s note a couple of things hee befoe we poceed. We don t eally need to divide this by the magnitude of the gadient since this will just cancel out once we actually do the integal. o, because of this we didn t bothe computing it. Also, the dopping of the minus sign is not a typo. When we compute the magnitude we ae going to squae each of the components and so the minus sign will dop out. : The Paaboloid Okay, hee is the suface integal in this case. 7 Paul awkins 3

25 Ê x, -,z ˆ Fd g = ( yj - zk) g f da Á f Ë = -y- z da x z z da = =- x + 3z da on t foget that we need to plug in the equation of the suface fo y befoe we actually compute the integal. In this case is the disk of adius in the xz-plane and so it makes sense to use pola coodinates to complete this integal. Hee ae pola coodinates fo this egion. x= cosq z = sinq q p Note that we kept the x convesion fomula the same as the one we ae used to using fo x and let z be the fomula that used the sine. We could have done it any ode, howeve in this way we ae at least woking with one of them as we ae used to woking with. Hee is the evaluation of this integal. Fd g =- x + 3z da p p ( cos 3 sin ) =- Û q + q ddq ı Ú ( cos 3sin ) 3 =- Û q + q ddq ı Ú p Ê 3 ˆÊ 4ˆ q q dq Û =- Ù Á ( + cos( )) + ( -cos( )) Á ı Ë Ë4 p =- 4 cos 8Ú - =- 4 - sin 8 =-p : The Cap of the Paaboloid ( q) ( q ( q) ) p dq We can now do the suface integal on the disk (cap on the paaboloid). This one is actually faily easy to do and in fact we can use the definition of the suface integal diectly. Fist let s notice that the disk is eally just the potion of the plane y = that is in font of the disk of adius in the xz-plane. Now we want the unit nomal vecto to point away fom the enclosed egion and since it must 7 Paul awkins 4

26 also be othogonal to the plane y = then it must point in a diection that is paallel to the y-axis, but we aleady have a unit vecto that does this. Namely, n = j the standad unit basis vecto. It also points in the coect diection fo us to use. Because we have the vecto field and the nomal vecto we can plug diectly into the definition of the suface integal to get, Fd g = yj - zk g j d= yd At this point we need to plug in fo y (since is a potion of the plane y = we do know what it is) and we ll also need the squae oot this time when we convet the suface integal ove to a double integal. In this case since we ae using the definition diectly we won t get the canceling of the squae oot that we saw with the fist potion. To get the squae oot well need to acknowledge that y = = g xz, and so the squae oot is, The suface integal is then, Fd g = ( g ) + + ( g ) x = + + da= yd At this point we can acknowledge that is a disk of adius and this double integal is nothing moe than the double integal that will give the aea of the egion so thee is no eason to compute the integal. Hee is the value of the suface integal. Fd g = p Finally, to finish this off we just need to add the two pats up. Hee is the suface integal that we wee actually asked to compute. Fd g = Fd g + Fd g =- p + p = Example Evaluate x + y + z = 9 and the disk x oientation. Fd g whee 4 F = xi + yj + z k z da and is the uppe half the sphee + y 9 in the plane z =. Assume that has the positive olution o, as with the pevious poblem we have a closed suface and since we ae also told that the suface has a positive oientation all the unit nomal vectos must point away fom the enclosed egion. To help us visualize this hee is a sketch of the suface. 7 Paul awkins 5

27 We will call the hemisphee and will be the bottom of the hemisphee (which isn t shown on the sketch). Now, in ode fo the unit nomal vectos on the sphee to point away fom enclosed egion they will all need to have a positive z component. Remembe that the vecto must be nomal to the suface and if thee is a positive z component and the vecto is nomal it will have to be pointing away fom the enclosed egion. On the othe hand, the unit nomal on the bottom of the disk must point in the negative z diection in ode to point away fom the enclosed egion. : The phee Let s do the suface integal on fist. In this case since the suface is a sphee we will need to use the paametic epesentation of the suface. This is, qj, = 3sinjcosqi + 3sinjsinq j + 3cosjk ince we ae woking on the hemisphee hee ae the limits on the paametes that we ll need to use. p q p j Next, we need to detemine q j. Hee ae the two individual vectos and the coss poduct. q ( qj, ) =- 3sinjsinqi + 3sinjcosq j qj, = 3cosjcosqi + 3cosjsinq j -3sinjk j i j k q j =-3sinjsinq 3sinjcosq 3cosjcosq 3cosjsinq -3sinj =-9sin jcosqi -9sinjcosjsin q k -9sin jsinq j -9sinjcosjcos q k =-9sin jcosqi -9sin jsinq j - 9sinjcosj( sin q + cos q) k =-9sin jcosqi -9sin jsinq j -9sinjcosjk 7 Paul awkins 6

28 Note that we won t need the magnitude of the coss poduct since that will cancel out once we stat doing the integal. Notice that fo the ange of j that we ve got both sine and cosine ae positive and so this vecto will have a negative z component and as we noted above in ode fo this to point away fom the enclosed aea we will need the z component to be positive. Theefoe we will need to use the following vecto fo the unit nomal vecto. q j 9sin jcosqi + 9sin jsinq j + 9sinjcosjk n =- = q j q j Again, we will dop the magnitude once we get to actually doing the integal since it will just cancel in the integal. Okay, next we ll need 4 F( ( qj, )) = 3sinjcosqi + 3sinjsinq j + 8cos jk qj, Remembe that in this evaluation we ae just plugging in the x component of vecto field etc. into the We also may as well get the dot poduct out of the way that we know we ae going to need. F qj, g = 7sin jcos q + 7sin jsin q + 79sinjcos j ( ) ( q j) 3 5 = 7sin j+ 79sinjcos j Now we can do the integal. ÛÛ Ê u ˆ v Fd g = ÙÙ Fg q j da ÙÙ Á ıı Ë q j p p p p p 3 5 = Û Ù 7sin j+ 79sinjcos jdjdq ı Ú p = Û Ù 7sin - cos + 79sin cos ı Ú 5 j j j jdjdq p j 3 ˆ 79 6 ˆ j j dq Û Ê Ê =- Á7Ácos - cos + cos Ù Ë Ë 3 6 ı = Û Ù ı = 79p 79 : The Bottom of the Hemi-phee dq Now, we need to do the integal ove the bottom of the hemisphee. In this case we ae looking at the disk x + y 9 that lies in the plane z = and so the equation of this suface is actually z =. The disk is eally the egion that tells us how much of the suface we ae going to use. 7 Paul awkins 7

29 This also means that we can use the definition of the suface integal hee with n =-k We need the negative since it must point away fom the enclosed egion. The suface integal in this case is, 4 Fd g = xi + yj + z k g -k d = - 4 z d Remembe, howeve, that we ae in the plane given by z = and so the suface integal becomes, 4 Fd g = - z d = d = The last step is to then add the two pieces up. Hee is suface integal that we wee asked to look at. Fd g = Fd g + Fd g = 79p + = 79p We will leave this section with a quick intepetation of a suface integal ove a vecto field. If v is the velocity field of a fluid then the suface integal vd g epesents the volume of fluid flowing though pe time unit (i.e. pe second, pe minute, o whateve time unit you ae using). 7 Paul awkins 8

30 tokes Theoem In this section we ae going to take a look at a theoem that is a highe dimensional vesion of Geen s Theoem. In Geen s Theoem we elated a line integal to a double integal ove some egion. In this section we ae going to elate a line integal to a suface integal. Howeve, befoe we give the theoem we fist need to define the cuve that we e going to use in the line integal. Let s stat off with the following suface with the indicated oientation. Aound the edge of this suface we have a cuve C. This cuve is called the bounday cuve. The oientation of the suface will induce the positive oientation of C. To get the positive oientation of C think of youself as walking along the cuve. While you ae walking along the cuve if you head is pointing in the same diection as the unit nomal vectos while the suface is on the left then you ae walking in the positive diection on C. Now that we have this cuve definition out of the way we can give tokes Theoem. tokes Theoem Let be an oiented smooth suface that is bounded by a simple, closed, smooth bounday cuve C with positive oientation. Also let F be a vecto field then, Fgd = cul Fg d Ú C In this theoem note that the suface can actually be any suface so long as its bounday cuve is given by C. This is something that can be used to ou advantage to simplify the suface integal on occasion. Let s take a look at a couple of examples. 7 Paul awkins 9

31 Example Use tokes Theoem to evaluate and is the pat of Calculus III = above the plane z x y olution Let s stat this off with a sketch of the suface. cul F g d whee 3 3 F = z i - 3xy j + xy k z =. Assume that is oiented upwads. In this case the bounday cuve C will be whee the suface intesects the plane z = and so will be the cuve = 5-x - y x y z + = 4 at = o, the bounday cuve will be the cicle of adius that is in the plane z =. The paameteization of this cuve is, t = ti + tj + k t p cos sin, The fist two components give the cicle and the thid component makes sue that it is in the plane z =. Using tokes Theoem we can wite the suface integal as the following line integal. p cul Fgd = Fgd = F t g t dt Ú Ú C ( ()) () o, it looks like we need a couple of quantities befoe we do this integal. Let s fist get the vecto field evaluated on the cuve. Remembe that this is simply plugging the components of the paameteization into the vecto field. F( ( t) ) = i - 3cos ( t)( sint) j + ( cost) 3 ( sin t) 3 k 3 3 = i - costsintj + 64cos tsin tk 7 Paul awkins 3

32 Next, we need the deivative of the paameteization and the dot poduct of this and the vecto field. ( t) =- sinti + costj F t g t =-sint-4sintcos t ( ()) () We can now do the integal. cul g sin 4sin cos p F d = - t- t tdt Ú 3 ( cost 8cos t) = + = Example Use tokes Theoem to evaluate C p Ú F g d whee the tiangle with vetices (,, ), (,, ) and = + + and C is F z i y j xk,, with counte-clockwise otation. olution We ae going to need the cul of the vecto field eventually so let s get that out of the way fist. i j k culf = = zj - j = ( z-) j x y z z y x Now, all we have is the bounday cuve fo the suface that we ll need to use in the suface integal. Howeve, as noted above all we need is any suface that has this as its bounday cuve. o, let s use the following plane with upwads oientation fo the suface. 7 Paul awkins 3

33 ince the plane is oiented upwads this induces the positive diection on C as shown. The equation of this plane is, x+ y+ z = fi z = g xy, = -x- y Now, let s use tokes Theoem and get the suface integal set up. Fgd = cul Fgd Ú C ( z ) = - jd g f = ( z- ) j g f da f Okay, we now need to find a couple of quantities. Fist let s get the gadient. Recall that this comes fom the function of the suface. f ( xyz,, ) = z- g( xy, ) = z- + x+ y f = i + j + k Note as well that this also points upwads and so we have the coect diection. Now, is the egion in the xy-plane shown below, We get the equation of the line by plugging in z = into the equation of the plane. o based on this the anges that define ae, x y - x+ The integal is then, Ú C Fgd = z- j i + j + k da ( ) g - x+ = -x- y -dydx on t foget to plug in fo z since we ae doing the suface integal on the plane. Finishing this out gives, 7 Paul awkins 3

34 Ú C - x+ Fgd = -x-ydydx Ú ( ) = Ú x - Ê = Á x - x Ë3 = x+ = y- xy- y dx In both of these examples we wee able to take an integal that would have been somewhat unpleasant to deal with and by the use of tokes Theoem we wee able to convet it into an integal that wasn t too bad. xdx ˆ 7 Paul awkins 33

35 ivegence Theoem In this section we ae going to elate suface integals to tiple integals. We will do this with the ivegence Theoem. ivegence Theoem Let E be a simple solid egion and is the bounday suface of E with positive oientation. Let F be a vecto field whose components have continuous fist ode patial deivatives. Then, Fd g = div FdV Let s see an example of how to use this theoem. Ú Example Use the divegence theoem to evaluate and the suface consists of the thee sufaces, x E Fd g whee F = xyi - y j + zk = , z 4 z x y + y =, z on the sides and z = on the bottom. olution Let s stat this off with a sketch of the suface. on the top, The egion E fo the tiple integal is then the egion enclosed by these sufaces. Note that cylindical coodinates would be a pefect coodinate system fo this egion. If we do that hee ae the limits fo the anges. z 4-3 q p We ll also need the divegence of the vecto field so let s get that. divf = y- y+ = 7 Paul awkins 34

36 The integal is then, Fd g = Ú E Û = Ù ı p p dd p Ê 3 4ˆ d p div FdV = Û 4-3 ı Ú Û = Ù Á - ı Ë 4 = Û Ù ı 5 = p Û Ù ı Ú 5 dq 4 dzddq q q 7 Paul awkins 35