Math 32B Review Sheet

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1 Review heet Tau Beta Pi - Boelter 6266 Contents ouble Integrals 2. Changing order of integration Integrating over more general domains Triple Integrals 5 2. Change of Variables Line Integrals. Vetor Line Integrals Conservative Vetor Fields urfae Integrals 9 4. Vetor urfae Integrals Fundamental Theorems of Multivariable Calulus 2 5. Green s Theorem tokes Theorem ivergene Theorem

2 ouble Integrals The basi premise of alulating a double integral is alulating the volume underneath a surfae, just as the basi premise of alulating a single integral is finding the area under the line of a funtion. An example below shows a parabaloid where the funtion is z = x 2 + y 2. Figure : The graph of the surfae z = x 2 + y 2 is displayed. (Graph produed via Matlab We an see that many of the rules of alulating areas under urves also translate to alulate volumes under surfaes. For example, if we all the domain (, (, as (essentially x an take values from - to and y an take values from - to as shown in Figure, then the area under the paraboliod shown in Figure is x 2 + y 2 da In general, alulating the (possibly signed volume under the funtion z = f(x, y is f(x, y da As you might imagine, to alulate the volume over the paraboloid instead, for this image, the volume between x 2 + y 2 and 2 is 2 (x 2 + y 2 da ouble integrals seem to be fairly easy to visualize, so the hardest problem is usually omputation, whih has more nuanes than in the single variable ase. It is somewhat hard to show graphially, but the idea Page 2

3 behind omputing the double integral over a retangle is to treat da as the produt dxdy (or vie versa; more on that later. Then, the integral beomes x 2 + y 2 da = Holding y onstant while integrating along x, we obtain x 2 + y 2 da = x 2 + y 2 dxdy = x 2 + y 2 dxdy x + xy 2 ] dy = 2 + 2y 2 dy This last integral an be evaluated as a normal single integral from single variable alulus. x 2 + y 2 da = 2 + 2y 2 dy = 2 y + 2 y ] = 8 This might seem like an abnormally large number, but for referene, the box with the same base domain and height has volume 2 2 = 4 = 2 so this number is a bit easier to swallow. In general, we an ompute all volumes over retangles in this fashion, alulating d b a f(x, y dxdy where {x, y} (a, b (, d is the domain of integration. This idea is the same as starting out with the mixed derivative and finding f. Going forward, we have 2 f x y f f x y 2 x 2 f x y In the seond step, we differentiate with respet to x while holding y onstant, so to invert that step, we need to integrate x while holding y onstant. imilarly, in the first step, we differentiate y with x onstant, so to get bak the value of f, we want to integrate y while holding x onstant. ine we have equality of mixed derivatives, integrating the other way makes no differene in either ase. Another way to think about it is to onsider the inner integral as the area of a ross setion and to integrate that area over the range of ross setions. For example, when integrating d b a f(x, ydxdy we an first let (y = b f(x, ydx. is only a funtion of y beause we are integrating out x. For a onstant y, this orresponds to the area of a ross setion under the surfae. Then, we an integrate with respet to y, whih gives d b a f(x, ydxdy = d (ydy whih allows us to alulate the double integral as the familiar single integral we all know. Integrating the area of eah slie over y now gives us the volume underneath the surfae. Page

4 . Changing order of integration Although the order of integration does not make a differene in the final value of the double integral, it an make a big differene in how easy it is to evaluate the integral analytially. For example 2 xe xy dxdy Tehnially, we an integrate with respet to x first holding y onstant using integration by parts. However, it seems muh easier to evaluate the integral with respet to y first instead. 2 xe xy dydx = 2 xe xy ] dx = 2 e x dx = e 2 ometimes, the integral might be impossible to evaluate analytially one way. Take 2 2 y e x y dydx This integral is impossible to evaluate the way it is written. Calulating it the other way however, 2 2 y e x y dxdy = Integrating over more general domains y 2 (e 2 y e y dy = (e y 2 e 2 y ] 2 = e 2 2 e + 2 e2 While integrating over retangles is quite useful, we have a muh greater variety of domains that we did in single variable alulus. All domains of integration in single variable alulus had to be intervals of the form [a,b], but in 2-dimensions, we have retangles, irles, ellipses, and so on whih give muh greater variety of domains to integrate over. Figure 2 from Rogawski sums up how to deal with the situation fairly well. If we make the retangles very small so that they approximate the domain very well, we an see that the top value of the retangle will be the top funtion g(x and the bottom value the retangles will reah will be the bottom funtion h(x. Therefore, the sum of the integrals of some funtion f(x, y over this domain beomes f(x, yda = b g(x a h(x f(x, ydydx Here, a and b are the smallest and largest values x an take respetively and still remain inside the domain. The funtion g(x and h(x take are of the values that y an take inside the domain. Notie that this integral eventually evaluates to a number, so swapping the integral in this ase to somehow put g(x and h(x on the outside and integrate with respet to x first does not really make sense. Instead, we have to find the upper and lower bounds on x as a funtion of y and then use those as the bounds (more on this later. It would be easiest to see these integrals evaluated with an example: x dydx This domain is desribed by the piture in Figure. The double integral gives the area of the triangle, sine it gives the volume under the surfae of height, whih gives the same numerial value as the area. oing the integral, we get x dydx = y] x dx = xdx = 2 We know the triangle represents the bounds on the integral beause y goes from to x, whih is the boundary y = x and x goes from to. If we wanted to swith the bounds, we need the bounds on x as a funtion of y first and then the overall bounds on y. Looking at Figure, the largest x an beome is. However, the smallest x an beome is y. The values y an take also range between and. Thus, we have Page 4

5 Figure 2: The graph of an arbitrary domain approximated using retangles. (Graph adapted from Rogawski, rd edition, pg. 849 x 2 Triple Integrals dydx = y dxdy = x] ydy = ydy = 2 After learning double integrals, triple integrals largely use the same tools, but simply add an extra step of integration. If double integrals are defined over an area domain, triple integrals are defined over a volume domain whih means it is not quite possible to visually see the value being alulated, but it is still useful to see the domain graphially. We an evaluated triple integrals over boxes and more general domains just as before: x 2 + y 2 + z 2 dxdydz = x + xy 2 + xz 2 ] dydz = Page 5 + y2 + z 2 dydz

6 Figure : The domain of integration {(x, y [, ] [, ] : y x}. (Graph produed via Matlab At this point, we an solve the triple integral just as a double integral. + y2 + z 2 2 dydz = + z2 dz = A sometimes useful trik in both the double and triple integral ase is integrating over a box where the integrand is a produt of single variable funtions. In that ase, we have f d b e a ( b f(xg(yh(zdxdydz = a f f(xdx e d ( b g(yh(zdydz = a ( d ( f f(xdx g(ydy h(zdz e This allows us to integrate a produt of single variable funtions as just a produt of the single variable integrals. An example is shown below. 2 2 ( 2 xy z 2 dxdydz = ( ( z 2 dz y dy xdx = 2 ( 7 ( 77 4 As an example of more ompliated domains, onsider the volume given in Figure 4. If we want to alulate the integral xdv W ( = Page 6

7 Figure 4: The domain of integration, W, taken from Rogawski, rd edition, page 865.The domain is the volume found in between the surfaes whih is over the first quadrant in the x-y plane. then hoosing a good order of integration would do wonders for easiness of the alulation. ine the volume is bounded by the two surfaes whih give the bounds on z as funtions of x and y, it makes sense to make z the inner bounds. W xdv = 4 x 2 y 2 x 2 +y 2 xdzda For the outer bounds, we need to see the largest values that x and y an take while still remaining inside the domain. This is given by the region where the two surfaes interset, where x 2 + y 2 = 4 x 2 y 2 x 2 + 2y 2 = 2 This domain gives the ellipse shown in Figure 5. The domain is fairly symmetri, so we an hoose to either find the bounds for x in terms of y or vie versa. We hoose x in terms of y beause the bounds of y are simpler and the this will yield even polynomial orders of y after the x integration, whih will be nie for anelling out the square root. The bounds on x are then = and the bounds on y are [, ]. Thus, the integral is W 2 2y 2 xdv = 2 2y 2 4 x 2 y 2 ( 4x y 2 2x dxdy = x 2 2y 2 x 2 +y 2 xdzdxdy = 2 2y 2 2x 2( y 2 ] 2 2y2 2 x4 dy = ( x 4 4y 2 2x 2 dxdy ( 2 2 2y 2( y 2 ( 2 2y 2 2 dy 2 It is the not the point of this lass to solve ompliated single integrals, but in ase you are interested, this simply involves quite a bit of expansion and polynomial integration. Page 7

8 Figure 5: The outer domain,, whih desribes the bounds on x and y. (Produed via Matlab ( 2 2 2y 2( y 2 2 (2 2y 2 2 dy = = 2y 4 y y5 ] ( 4 8y 2 + 4y 4 ( 2 4y 2 + 2y 4 dy = = = = y 2 + 2y 4 dy We were fortunate enough to avoid trigonometri substitution in the last problem and as it happens, trigonometri substitution very rarely shows up in multiple integration due to our ability to make variable substitutions instead. 2. Change of Variables The proof of hange of variables is something done using linear algebra and so is left out of this lass, but an example explains the onept fairly well. Take the parallelogram displayed in Figure 6 as the domain. We would like to alulate da Computing this integral with respet to either x or y would be messy as either the upper bound or the lower bound would hange twie, leading to the sum of three double integrals, something we would all like to avoid doing. Consider the 2A way of alulating the area given by two parallelograms. It was alulated using the magnitude of the ross produt of the two vetors that made up the sides, in this ase, Page 8

9 Figure 6: The inside of the parallelogram represents the domain,, of integration. The area is not shaded to emphasize the boundary instead. (Produed via Exel ( [ 2 ] A = det 2 = Beause the two vetors are v = 2, and v 2 =, 2. Looking at it differently, let us define two new oordinates, u and v in terms of x and y given by the following formulas. x = 2u + v y = u + 2v This an also be written in terms of the matrix above as [ ] [ ] [ ] x 2 u = y 2 v Then, u =, v = gives the origin in (x, y; (u, v = (, gives (x, y = ( 2,, (u, v = (, gives (x, y = (, 2 and (u, v = (, gives (x, y = (, 5, whih are all the 4 verties of the parallelogram defined above. Therefore, the domain of integration in terms of (u,v is that in Figure 7, a square of sides. o, the integral beomes simply the area of the square,, multiplied by the determinant whih appropriately sales this value to get the value of the original integral. In general, a hange of oordinates in 2-dimensions is done through expression x and y as funtions of u and v and then finding the Jaobian matrix of the transformation, [ x = y] [ ] g(u, v = Ja = h(u, v Page 9 [ x u y u x v y v ]

10 Figure 7: The transformation above defines the following square of sides in (u,v oordinates whih is now the new domain,. (Produed via Matlab Then, the double integral f(x, ydxdy an be found using f(x, ydxdy = f(x(u, v, y(u, v Ja(u, v dudv where the Ja(u, v is the (absolute value of the determinant of the Jaobian. generalized to -dimensions using the -dimensional Jaobian instead. x x x x g(u, v, w u v w y = h(u, v, w = Ja = y y y u v w z k(u, v, w z u z v z w This idea an be Then, the integral beomes f(x, y, zdv (x, y, z = f(x(u, v, w, y(u, v, w, z(u, v, w Ja(u, v, w dv (u, v, w It is again easiest to see this with an example, suh as the area of an ellipse. Let us take a general ellipse given by the equation ( 2 x + a ( 2 y = b If we take the entire ellipse as the domain, we have to solve the following integral to get the area. da Page

11 Instead of solving this by expressing x in terms of y or vie versa, let us instead make the substitution u = x a and v = y a. This turns the (u,v domain into a irle of radius. In terms of x and y, x = au and y = bv. This transformation is shown in Figure 8. Figure 8: The defined transformation from an ellipse in (x,y oordinates to the disk in (u,v oordinates. The determinant of the Jaobian is a b = ab Then, the integral beomes da(x, y = abda(u, v = ab da(u, v = πab imilarly, try showing that the volume of an ellipsoid whih is surrounded by the surfae ( x 2 ( a + y 2 b + ( z 2 = is πab. It is very hard to know what kind of substitution will make your life easier, but there are a few ommon ones that are frequently useful. The one we just performed was a saling of the axes. Another one, in 2-dimensions, is a hange to polar oordinates, given by the transformation x = r os(θ and y = r sin(θ. This transformation has the Jaobian matrix determinant x x r θ y y = os(θ r sin(θ sin(θ r os(θ = r r θ In -dimensions, we an define this same polar oordinate transformation while keeping the z-oordinate and thus using ylindrial oordinates; x = r os(θ, y = r sin(θ and z = z. x x x u v w y y y u v w z z = os(θ r sin(θ sin(θ r os(θ = r u z v w Finally, another transformation that is ommon is using spherial oordinates, x = ρ os(θ sin(φ, y = ρ sin(θ sin(φ, and z = ρ sin(φ. The physial interpretation of the symbols ρ, θ, and φ in this ase an be seen in Figure 9. The Jaobian in this ase is x x x u v w y y y u v w z z = os(θ sin(φ ρ sin(θ sin(φ ρ os(θ os(φ sin(θ sin(φ ρ os(θ sin(φ ρ sin(θ os(φ os(φ ρ sin(φ = ρ2 sin(φ u z v w In this ase, the evaluation of the determinant is negative. However, this is simply a hoie of the ordering of the oordinates sine swapping two rows in a determinant flips the sign. Thus, we will take the absolute value of the determinant and get Ja = ρ 2 sin(φ. Page

12 Figure 9: The φ angle indiates deviane from the vertial axis, the θ angle indiates rotation in the x-y plane as usual and ρ indiates distane from the origin. As an example of spherial oordinates, onsider the funtion f(x, y, z = (x 2 + y 2 + z 2 =fra2 with the domain x 2 + y 2 + z 2 4 and z x 2 + y 2. Then, the integral f(x, y, zdv an be evaluated as ρ 2 ρ 2 sin(φdv (ρ, θ, φ = ρ sin(φdv (ρ, θ, φ We have to find the bounds in terms of ρ, θ, and φ. ine the original domain is a one, ρ bounds simply go from to 4 sine there is no other upper and lower bound. Full rotation in the x-y plane is allowed, so θ bounds are to 2π. Finally, for the φ bounds, we have that φ is bounded below by and bounded above by the value of φ when z = x 2 + y 2. ine z = ρ os(φ and x 2 + y 2 = ρ sin(φ, we end up with the equation tan(φ = whih an be solved to get φ = π 4 as an upper bound. Thus, the integral beomes ρ sin(φdv (ρ, θ, φ = 2π π 4 4 ρ sin(φdρdφdθ = 2π π 4 = 6π (2 2 Page 2 sin(φdφ ρ sin(φdρ = 2π 6 2

13 Line Integrals Line integrals are a very speial type of integral unlike the other ones we have seen before. Figure presents the essential idea of how we view an integral. Figure : The physial intuition of an integral. The integral finds the area, A, between an arbitrary funtion f(x and the x axis between the x values of a and b. We sum up over dx, the differential element that desribes how to hange from a to b. In the line integral ase, we instead have a surfae z = f(x, y defined over a urve in the two dimensional ase. The urve is in the x-y plane and the surfae is above the x-y plane and denotes a height for every point along the urve. To find the area of the urtain that now drops from the surfae to the urve, we sum up the areas of the retangles, whih have individual areas A = f(x, y s where s desribes the length of the retangle, whih is the ar-length of the segment of the urve and f(x,y is the value of the surfae whih is also the height of the urve. Then, to find the area, we would need take the limit as all the s beome small, whih is the same as omputing the integral over the urve,. f(x, yds This area is defined independent of a parametrization given to the urve sine it is simply an intrinsi property of the urve and the surfae over it. However, this integral is in pratie omputing by expressing both x and y in terms of a parameter and thus omputing the integral as suh. Let the parametrization be r(t = x(t, y(t and the beginning and ending points of the integral be the points on the urve given by r(t and r(t respetively. Then, f(x, yds = t t f(x(t, y(t ds dt dt ine s desribed the ar length of the urve, (dx 2 ( 2 ds dy dt = (t = + dt dt t (dx 2 ( 2 dy f(x, yds = f(x(t, y(t + dt dt dt t Again, integrating the funtion f(x, y = will give you the length of the urve. As an example, let us integrate the funtion f(x, y = x along the path defined by the parabola given by y = x 2 from the point Page

14 , to the point,. A parametrization of this urve an simply be given by Then, x(t, y(t = t, t 2 ds dt = + 4t 2 The funtion in terms of the parameter t is Thus, we have f(x, yds = f(x(t, y(t = t t + 4t 2 dt = 8 5 [ ] 5 2 udu = 8 u 2 = This idea extends niely to three dimensions, whih is beyond our ability to visualize, but still an be omputed. t f(x, y, zds = f(x(t, y(t, z(t ds t (dx 2 ( 2 ( 2 dy dz t dt dt = f(x(t, y(t, z(t + + dt t dt dt dt As an example, onsider the funtion f(x, y, z = x 2 z over the path r(t = e t, 2t, e t from t = to t =. Then, we have f(x, y, zds = e t e 2t e 2t dt = e t( e t + e t dt = e 2t + dt = ] 2 e2t + t = e2 + 2 The ds dt term in the line integral makes most integrals of this form impossible to ompute analytially, but there is a speial lass of integrands that do not have this problem that are ommon; vetor line integrals.. Vetor Line Integrals Before disussing a vetor line integral itself, we first need to have a onept of a vetor field. A vetor field is a funtion that takes possibly multiple inputs and returns outputs as a vetor. An example is F (x, y = x 2 + y 2, xy. In general, it is of the form F (x, y = f(x, y, g(x, y, or in three dimensions, F (x, y, z = f(x, y, z, g(x, y, z, h(x, y, z. In two dimensions, this is equivalent to taking every point inside the domain inside the x-y plane and attahing a 2-dimensional vetor to it. An example vetor field, F (x, y = x, y is shown in Figure. Another ommon example is the vetor field given by F (x, y = y, x, whih is Figure 2. Finally, a more omplex vetor field, F (x, y = y, sin(x, is shown in Figure. Notie how as x approahes, the arrows beome horizontal as sin( = and as y approahes, the arrows beomes vertial.the arrows also flip from left to right moving right on the x=axis just as sin(x flips from negative to positive. Finally, a three-dimensional example is rarely drawn, but it is instrutive to see at least one, shown in Figure 4 of F (x, y, z = x, y, z. A vetor line integral is that whih alulates ( F T ds Page 4

15 Figure : The vetor field F (x, y = x, y is displayed. magnitudes. The length of the vetors orrespond to the where T is the unit tangent vetor of the urve. This expression is still intrinsi to the urve and the vetor field and not to the hoie of parametrization, However, when we parametrize, we find the omputation of this urve muh simpler. ( t ( F T ds = F ds t T t dt dt = t ( F d r dt dt This expression gets rid of the inherent square root that usually appears in the line integral. To denote that this expression is independent of the parametrization hosen, it is also written as F d r As an example, let F = xy, 2, z defined on the helix given by r(t = os(t, sin(t, t from t = to t = π. Then, F d r = π F ( r(t = sin(t os(t, 2, t d r dt sin(t os(t, 2, t sin(t, os(t, = = sin(t, os(t, π = π4 4 ] π sin 2 (tos(t+2 os(t+t dt = sin (t +2 sin(t+ t4 4 As another example, onsider the vetor field F = x, y integrated over the unit irle, or the path r(t = os(t, sin(t from t = to t = 2π. We have Page 5

16 Figure 2: The vetor field F (x, y = y, x is displayed. F d r = 2π F ( r(t = os(t, sin(t d r dt = sin(t, os(t os(t, sin(t sin(t, os(t dt = 2π dt = This answer somewhat makes sense beause we have onstruted a path that starts at a point,, and ends at the same point. However, this is atually not always the ase, as we an see integrating the vetor field F = y, x. We instead get F ( r(t = sin(t, os(t 2π sin(t, os(t sin(t, os(t dt = 2π dt = 2π Therefore, the vetor field determines whether line integrals around losed paths are. losed path, the following notation is ommon. F d r To denote a.2 Conservative Vetor Fields When a vetor field integrates suh that its line integral is for all losed loops, we all it a onservative vetor field. This happens when the vetor field is the gradient of some funtion f(x,y,z beause, from 2A, df( r(t = f d r dt dt Page 6

17 Figure : The vetor field F (x, y = y, x is displayed. F d r = t t f d r t ] dt dt = t df( r(t = f( r(t = t t The last expression is beause the losed line integral implies that r(t = r(t. A useful way of testing whether a vetor field is the gradient of some funtion is heking the url of the funtion, whih heks all mixed derivatives. From 2A, we have y ( f = x x ( f y so, for 2-dimensions, if F = F (x, y, F 2 (x, y we must hek y x θ F F 2 = F 2 x F y = For -dimensions, F = F, F 2, F we must hek F î ĵ k ( = x y z F F 2 F = F y F ( 2 F î + z z F ( F2 ĵ + x x F î =,, y If this is the ase, then we an find the funtion f itself and not bother with alulating the line integral diretly. For example, let us alulate the line integral of F = y, x, z over the path r(t = t, t 2, t from t = to t =. In heking whether the field is onservative, we obtain î ĵ k x y z y x z = ( î + ( ĵ + ( î =,, To find the field itself, let f x, f y, f = y, x, z z Page 7

18 Figure 4: The vetor field F (x, y, z = x, y, z is displayed. Integrating with respet to x, we get f x = y f(x, y, z = xy + g(y, z ine differentiating with respet to x sends all funtions of only y and z to, we obtain a onstant of integration that is a funtion of y and z. We an use the other omponents of the vetor field to find what this funtion is. x + g y = f y = x g y = g(y, z = h(z f(x, y, z = xy + h(z Finally, we set the z derivative equal to the last omponent, z to obtain what the expression for f is. dh dz = f z = z h(z = z4 4 + C Page 8

19 f(x, y, z = xy + z4 4 + C The final C is truly a onstant independent of x, y, and z. It will anel out in the differene when doing the integral. Thus, for any path starting at,, and ending at,,, the line integral for this vetor field is 4 urfae Integrals F d r = f(,, f(,, = ( ( + 4 = We an also think about what it means to define an integral over a surfae just as we did for a line. We do not have a visual representation in this ase beause surfaes are already -dimensional objets, but we an imagine a funtion that gives a value for every (x,y,z oordinate on a surfae. Then, we would somehow want to ompute f(x, y, z for all small surfae area elements we divide the surfae into and take the limit as goes to to get f(x, y, zd The ds in the line integral ase denoted the differential ar length element and so d in this ase must represent the differential surfae area. From the example in Figure 6, we remember that the area of a parallelogram is given by the magnitude of the ross produt of the vetors that make up the sides. ine the surfae area an be approximated as parallelograms that beome inreasingly aurate as the surfae area is divided up more and more, we an find d as loal ross produt of the tangent vetors. uppose r(t, s = x(t, s, y(t, s, z(t, s desribes the surfae. An example of this would be how r(x, y = x, y, x 2 +y 2 desribes the paraboloid z = x 2 +y 2. Then, the surfae area element is given by the magnitude of the ross produt of the tangent vetors multiplied by area element in terms of (t,s oordinates, or d = r t r s da(t, s f(x, y, zd = f(x(t, s, y(t, s, z(t, s r t r s da(t, s As an example, let f(x, y, z = x + y + z. Let us integrate this funtion over the surfae of the one z 2 = x 2 + y 2 from z = to z = 4. The natural way to express the one is in ylindrial oordinates, sine we already have the equation z = r. Therefore, it seems natural to use x = r os(θ and y = r sin(θ as our two parameters. The bounds on theta are [, 2π] in that ase and for r, we have r = z 4, so the bounds on r are [, 4]. Therefore, we have our parametrization r(r, θ = r os(θ, r sin(θ, r and our r and θ bounds as speified above. Now we must alulate the ross produt. r r r θ = r = os(θ, sin(θ, r r = r sin(θ, r os(θ, θ î ĵ k os(θ sin(θ = r os(θ, r sin(θ, r r sin(θ r os(θ Page 9

20 r r r θ = 2r ( ( ( f(x(r, θ, y(r, θ, z(r, θ = r os(θ + r sin(θ + (r = r + sin(θ + os(θ ( f(x, y, zd = r + sin(θ + os(θ 2rdA(r, θ = 2π 4 2 r 2( + sin(θ + os(θ drdθ = ( 4 ( 2π 2 r 2 dr 4. Vetor urfae Integrals ( + sin(θ + os(θ dθ = π = 28π 2 Just as we previously defined a vetor line integrals as taking the integrand as a vetor field dotted with the unit tangent vetor, we aomplish a similar task with a surfae integral. It does not make sense to alulate the amount of a vetor field is parallel to a surfae, but it does make sense to ompute how muh of a vetor field is normal to a surfae, or F n. Then, a vetor surfae integral is ( F n d Again, this quantity is independent of parametrization, but the parametrization greatly simplifies its alulation sine n = r t r s is already perpendiular to the surfae as it is found through the ross produt of the tangent vetors. Then, ( ( F n d = F n n da(t, s = F nda(t, s As a shorthand, to delare that this value is independent of parametrization, it is often written as F d A physial intuition that an be given to this type of integral is thinking of the vetor field F as a field that desribes the veloity of water at every point and as a surfae in the water. The value of F d desribes how muh water passes through the surfae. As an example, let F = y, x, z and let the plane x + 2y + z = be the surfae inside the first otant with an upward pointing normal. A parametrization that an be given to this surfae is r = x, y, x 2y. Then, r =,, x r =,, 2 y r x r y = î ĵ k =, 2, 2 This is the normal vetor we would have gotten from just taking the oeffiients in front of the variables as usual from a plane, but that shortut only works if the plane equation is expressed in the form ax+by+z =. F d = F ( r(x, y = y, x, x 2y y, x, x 2y, 2, da(x, y = + x yda(x, y The domain in this ase is the area in the positive x-y plane where x 2y or where y x Thus, the x bounds are [, ] and the y bounds are [, x 2 ]. 2. F d = x 2 ( + x ydydx = F ( r(x, y = y, x, x 2y 2 ( x2 4 ( x2 dx = 4 Page 2 ( + x 2 dx = ] ( + x = 7

21 Figure 5: An area domain,, is surrounded by a urve,. 5 Fundamental Theorems of Multivariable Calulus imilar to the fundamental theorem of alulus, we also have fundamental theorems of multivariable alulus. As a reminder, if f(x = F '(x, then we had b a f(xdx = F (b F (a This gives us a way of turning a higher order (in this ase st integral into a lower (th order integral, whih is just a differene. Green s theorem has the same idea in mind by turning a double integral over an area into a line integral around the boundary. 5. Green s Theorem Green s theorem involves the situation present in Figure 5. When omputing the line integral of F = F, F 2, F d r, we have the hoie of doing it diretly or instead evaluating it as a double integral through the formula F d r = ( F2 x F da y whih works for all simply onneted (no holes in the domain or vetor field domains. As an example, onsider the example we had before with the urve given by r(t = os(t, sin(t from t = to t = 2π. This losed loop surrounds the unit irle of area. We had two vetor fields that we alulated this line integral Page 2

22 for, F = x, y and F 2 = y, x. We an see that F d r = F 2 d r = ( ( x x y y ( y x x y da = da = 2dA = 2 2dA = 2π whih is muh easier to ompute than the line integral ase. One important thing to remember is that the diretion that the urve is oriented is ounterlokwise. Otherwise, the opposite diretion must instead be taken by adding a minus in front of the line integral. Another important thing to note is that the domain must be simply onneted; neither the vetor field nor the domain an have holes inside of it. As an example, take the vetor field y F = x 2 + y 2, x x 2 + y 2 along the same unit irle path as before. We have F 2 x x F = In fat, this vetor field is the gradient of the funtion f(x, y = tan ( y x. However, F d r = 2π Therefore, Green s theorem fails in this ase. 5.2 tokes Theorem F ( r(t = sin(t, os(t ṛ = sin(t, os(t dt sin(t, os(t sin(t, os(t dt = 2π dt = 2π tokes theorem generalizes the ideas from Green s theorem to -dimensional line integrals. We annot expet a -dimensional losed loop to surround an area inside the x-y plane anymore, so it instead relates the line integral to a surfae integral of the url instead. F d r = ( F d In Green s theorem, we were only able to pik one area that was enlosed by the losed loop; however, here, we an atually pik ANY surfae whih has the losed loop as its boundary. Figure 6 presents some usual ases. For the first figure, we an easily apply tokes theorem as written. A few important points should be noted however. The diretion of the surfae integral an be oriented either way. However, the line integral will then be oriented in that diretion. For example, with the surfae on the left, if we want the normal vetor to point outward, this signals that we should want the diretion of the urve to be ounterlokwise whereever the boundary touhes the surfae. Exusing a poor drawing, onsider the same surfae now in Figure 7. Therefore, tokes theorem holds normally and we an write F d r = ( F d Page 22

23 Figure 6: The first surfae is bounded by only one urve, so we an diretly apply tokes theorem. The seond surfae is instead bounded by three urves, so we must use a modified version of tokes. If we instead wanted the normal vetor to point inward and still keep the orientation of the line integral as it is, we would instead have ( F d r = F d For the seond surfae, we use the same idea of orientation. uppose we want the outward pointing surfae integral again. Then, the drawing beomes that in Figure 8. Thus, in this ase, sine all the urves point in the diretion they should (even though it appears as if is pointing in the wrong diretion, we have ( F d = F d r + F d r + F d r 2 As an example of tokes theorem, onsider the portion of the plane x 2 + y + z = in the first otant. We would like to find the line integral around the boundary of this surfae of the vetor field F = yz,, x. The boundary is not very ontinous; omputing the line integral diretly would require doing integrals. However, the surfae integral is muh more manageable. Reall that the normal vetor for a surfae integral of a plane of the form ax + by + z = is simply the vetor a, b,. In our ase, that is 2.., To use this normal vetor, we must also use the parametrization r(x, y = x, y, ax by ; in our ase, r(x, y = x, y, x 2 y. Then F î ĵ k = x y z =, y, z =, y, + x yz x 2 + y The bounds for the domain in terms of x and y are where x, y, and x 2 + y. This gives the bounds for x as [, 2] and for y as [, 2 x]. F d r = F d = s 2 = 2 2 x ( 2 x 4 5. ivergene Theorem, y, + x 2 + y ( 2 x + ( 2 x 2 dx = 2,, dydx = x 2 x dx = 4 x2 x ] 2 ( 2 x+ 2 y 4 dydx Finally, we have ome to the last fundamental theorem of multivariable alulus, ivergene Theorem (also known as Gauss s theorem. ivergene theorem uses a similar idea to Green s theorem and tokes theorem Page 2 =

24 Figure 7: An outward pointing vetor prompts a ounterlokwise orientation for line integrals bounding the surfae; the line integral here is oriented orretly beause the arrows on the normal vetor and urve point along the same diretion where they touh. by relating a surfae integral to a volume integral. divergene of a vetor field F = F, F 2, F. Before that, however, we first need the definition of The divergene of a vetor field, F, is defined as F = F x + F 2 y + F z Then, by divergene theorem, we have, for a volume, W, and the surfae that bounds it, W, F d ( = F dv W There are surprisingly few nuanes with this definition and the only thing to really keep trak of is the fat that the surfae must be oriented outwards the way this definition is written. Also, the domain must be simply onneted and the vetor field must be defined everywhere on the inside just like Green s theorem. Let us see an example where we ompute it both ways. W Consider the vetor field F = x, y, z and the surfae given by x 2 + y 2 + z 2 = that enloses the unit ball entered around the origin. Let us ompute the surfae integral first. A parametrization of the surfae is given from spherial oordinates, Then, r(θ, φ = os(θ sin(φ, sin(θ sin(φ, os(φ r = sin(θ sin(φ, os(θ sin(φ, θ Page 24

25 Figure 8: An outward pointing vetor prompts a ounterlokwise orientation for line integrals bounding the surfae; all line integrals here are oriented orretly beause the arrows on the normal vetors and urves point along the same diretion where they touh. r θ r φ = r = os(θ os(φ, sin(θ os(φ, sin(φ φ î ĵ k sin(θ sin(φ os(θ sin(φ os(θ os(φ sin(θ os(φ sin(φ = sin2 (φ os(θ, sin 2 (φ sin(θ, sin(φ os(φ It is hard to tell if this normal vetor is pointing outward or not, so we an test it by heking a single point; θ, φ =, π 2. ( sin 2 π 2 ( os(, sin 2 π ( π ( π sin(, sin os =,, o the normal vetor is pointing inward, whih means we really want the negative of this vetor, so n = sin 2 (φ os(θ, sin 2 (φ sin(θ, sin(φ os(φ Notie we did not try θ, φ =, instead as would be anyone s first try. It gives a normal vetor of,, whih does not tell us anything about the diretion, F ( r(θ, φ = os(θ sin(φ, sin(θ sin(φ, os(φ Page 25

26 W F d = 2π π os(θ sin(φ, sin(θ sin(φ, os(φ sin 2 (φ os(θ, sin 2 (φ sin(θ, sin(φ os(φ dφdθ = 2π π sin(φdφdθ = 2π π sin(φdφ = 4π Now that we have done this the hard way, let us see the easy way to do it. W F d = W F = (x x + (y y + (z z = ( ( F 4 dv = dv = dv = π = 4π W This is not to say that you an always get away with a volume integral being easier than the surfae integral, but it usually is easier to just find the bounds on a volume than to go through finding a parametrization of a surfae. W Page 26