Curves: We always parameterize a curve with a single variable, for example r(t) =

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1 Final Exam Topics hapters 16 and 17 In a very broad sense, the two major topics of this exam will be line and surface integrals. Both of these have versions for scalar functions and vector fields, and both have various versions of the fundamental theorem of calculus that apply to them. Parameterizations: etting up a line or surface integral requires us to parameterize the curve or surface we are integrating over. Here are a few parameterization techniques it would be good to be familiar with. urves: We always parameterize a curve with a single variable, for example r(t) = x(t), y(t). If our curve is the graph of a function y = f(x) then we can set x(t) = t and y(t) = f(t). In this case, whatever range of values x covered is also our range for t. Ex: If the curve is the graph of y = x 4 with 2 x 4, we can parameterize this as r(t) = t, t 4 with 2 t 4. Lines are also the graph of a function, but sometimes it is easier to parameterize them by finding a direction vector v and a point P on the line. In this case the parameterization is r(t) = vt + P. Ex: To parameterize the line between P = (1, 2, 5) and Q = (3, 5, 9) we find a vector between these points v = 3 1, 5 2, 9 5 = 2, 3, 4. If we just want the segment of the line between these points, the parameterization r(t) = 2, 3, 4 t + 1, 2, 5 = 2t + 1, 3t + 2, 4t + 5 with t 1 will start at P and end at Q. ircles can be parameterized with trig. functions. The segment of the circle of radius R between the angles θ 1 and θ 2 is parameterized by r(t) = R cos(t), R sin(t) with θ 1 t θ 2. If we are in a 3 setting, just add a z-coordinate to this parameterization. Ex: The circle of radius 3 at height z = 4, with x and y could be parameterized by r(t) = 3 cos(t), 3 sin(t), 4 with t π 2. urfaces: We parameterize surfaces with two variables, for example G(u, v) = x(u, v), y(u, v), z(u, v). We also need to specify what domain (2 region) u and v come from. If the surface is the graph of a function of two variables z = f(x, y), we can parameterize by setting x = x, y = y and z(x, y) = f(x, y). Whatever domain in the xy-plane the surface sits over is the domain of parameterization. Ex: We can parameterize the part of the ellipsoid 2x 2 + 3y 2 + z 2 = 1 sitting over the xy plane by solving for z = f(x, y) = 1 2x 2 3y 2 and letting our parameterization be G(x, y) = u, v, 1 2x 2 3y 2. In the xy plane z =, so 2x 2 + 3y 2 = 1. We want the region inside this ellipse, so a domain for x and y would be 2x 2 + 3y 2 1. Parameterizing a cylinder is similar to parameterizing a circle. If the height of the cylinder is given by a z b and the radius is R we can parameterize it with G(θ, z) = R cos(θ), R sin(θ), z with a z b and θ 2π. If we only want part of the cylinder, we can restrict θ to a smaller range. To parameterize a sphere of radius R we can use our spherical coordinates functions. Let θ and φ be our parameters, so our parameterization is G(θ, φ) = R cos(θ) sin(φ), R sin(θ) sin(φ), R cos(φ) with θ 2π and φ π. epending on the part of our sphere we want, we may restrict the range on θ and φ.

2 Line Integrals: A line integral is the integral of a scalar function or a vector field over a 1 curve. uppose we have parameterized our curve as r(t) with a t b. calar Line Integrals: If f(x, y, z) is a scalar function, then Vector Line Integrals: If F (x, y, z) is a vector field, then f(x, y, z) ds = F (x, y, z) dr = Ex: If is r(t) = 3 cos(t), 3 sin(t), 4 with t π 2 (the circle from before) then r (t) = 3 sin(t), 3 cos(t), and c (t) = 3. The integral of f(x, y, z) = x 2 + z 2 over is π 2 π 2 9 sin 2 (t) + 12 cos(t) dt. (9 cos 2 (t) + 16)(3) dt. The integral of F (x, y, z) = y, z, over is urface Integrals: A surface integral is the integral of a scalar function or a vector field over a 2 surface. uppose we have parameterized our surface as G(u, v) with (u, v) in domain. b a b a f(r(t)) r (t) dt. F (r(t)) r (t) dt. Tangent vectors to this surface are given by T u = G u product, we get a normal vector n(u, v) = T u T v. and T v = G v. If we take their cross calar urface Integrals: If f(x, y, z) is a scalar function, then f(x, y, z) d = f(g(u, v)) T u T v da Vector urface Integrals: If F (x, y, z) is a vector field, then F (x, y, z) N d = F (G(u, v)) (T u T v ) da Ex: If is the cylinder G(u, v) = 3 cos(u), 3 sin(u), v then T u = 3 sin(u), 3 cos(u), and T v =,, 1. The normal vector is T u T v = 3 cos(u), 3 sin(u), and T u T v = 3. The integral of f(x, y, z) = x 2 + z 2 over is (9 cos 2 (u) + v 2 )(3) da The integral of F (x, y, z) = y, z, over is 9 cos(u) sin(u) + 3v sin(u) da For both of these, you would need to set up bounds for the domain. Look back to hapter 15 for help on setting up double integral bounds.

3 The Fundamental Theorem of alculus in multiple dimensions: onservative fields, Green s Theorem, tokes Theorem and the ivergence Theorem. Besides just setting up line and surface integrals with parameterizations, we also have a number of theorems for transforming these integrals. In truth these are all versions of the fundamental theorem of calculus. All of these theorems deal with vector integrals. Terminology and notation: If is a closed curve we put a circle on the integral symbol to show this, and we sometimes refer to F dr as the circulation of F around. For any surface and vector field F we can refer to our usual vector surface integral F N d as the flux of F across. Remember: The curl of a vector field, F is itself a vector field, so we can also talk about the flux of the curl of F which would be ( F ) N d. onservative Fields A vector field F is conservative if there exists a scalar function f, called the potential of F, such that F = (f). For a 3 vector field F (x, y, z) = P, Q, R this would mean that P = f x, Q = f f y and R = z. Ex: The function f(x, y, z) = x 2 + y 2 + z 2 is a potential for the vector field F (x, y, z) = 2x, 2y, 2z. To check if a vector field is conservative, take its curl. If F =,, then F is conservative. If the curl is not zero, the field is not conservative. For 2 vector fields, all we have to check is that Q y =. Ex: The vector field F (x, y) = x+y, x y is conservative, because Q P x = 1 and y = 1. ometimes we can find a potential function just by looking at F. If you are able to do this for a conservative field on the exam, at least verify that it has the correct partial derivatives. Ex: The vector field F (x, y) = x + y, x y has a potential because F is conservative. A potential for F is f(x, y) = 1 2 x2 + xy 1 2 y2 because f f x = x + y and y = x y. If you cannot find a potential function by observation then you can try the following process for finding potential f. These instructions are for a 3 vector field F (x, y, z) = P, Q, R. The procedure for 2 vector fields is similar, but requires less steps. (i) Integrate P dx, Q dy and R dz. (ii) Build a potential function f(x, y, z) which is the sum of all unique terms in these three integrals. In other words, put each term into the potential, but only once. (iii) ee the example on page 97 of your book for another slightly different technique. Ex: The vector field F (x, y, z) = y 2 + yz 2, 2xy + xz 2 + e y, 2zyx + 1 is conservative (you can check that F = ). To find a potential, first integrate y 2 + yz 2 dx = xy 2 + xyz 2 2xy + xz 2 + e y dy = xy 2 + xyz 2 + e y 2zyx + 1 dz = xyz 2 + z

4 Our final version of the potential is f(x, y, z) = xy 2 + xyz 2 + e y + z The main reason we want a potential function is to evaluate line integrals of our vector field. The reason for this is the following version of the fundamental theorem of calculus. If F = (f) (so f is a potential for F ) and is a curve going from P to Q, then F dr = f(p ) f(q) This means that if you recognize a field is conservative, and find a potential function for it, you can skip the parameterization process and evaluate line integrals just by plugging into the potential. Ex: Let be the curve r(t) = 2t + 1, t, 3t with t 4 and F (x, y, z) = y 2 + yz 2, 2xy + xz 2 + e y, 2zyx + 1 (from the previous example). The potential f(x, y, z) for this field was found in the previous example. This curve starts at r() = 1,, and ends at r(4) = 9, 4, 12. By the fundamental theorem for conservative fields given above, F dr = f(9, 4, 12) f(1,, ) = (9(4 2 ) + (9)(4)(12 2 ) + e ) ( + + e + ) = e 4 If is a closed curve, the initial and end points are the same. Without even calculating the potential, we can say that the circulation F dr = for a conservative field.

5 Green s Theorem Green s Theorem applies to the circulation around a closed curve in 2. all the region inside the curve. We sometimes use the notation = for the boundary of. Assume that has a parameterization which is continuous with continuous derivatives. Also assume that it does not intersect itself. Green s Theorem says that if F is continuous on and is oriented with exterior boundaries counterclockwise and interior boundaries clockwise then Q F dr = y da It may be the case that I would tell you I want you to use Green s theorem on a particular problem, but it may also be the case that I don t tell you to use Green s even though you should. This applies to all of the integral theorems. For this reason, you should be familiar with the types of setups and situations where these theorems apply. With Green s theorem, here is what to look for. If you are asked about the circulation of F around a closed curve, and F is not conservative, be suspicious that I want you to use Green s theorem. In particular, if F is a nasty vector field but Q y is simple, or if is complicated but setting up a double integral over would be easy, it s time for Green s. Ex: uppose we want to know the circulation of F (x, y) = x arctan(x) + y, e y2 + x 2 around the triangle with corners at (, ), (, 2) and (1, ). This is a terrible vector field, and the curve has to be parameterized in three pieces. On the other hand, Q y = 2x + 1. To set up bounds for the inside of the triangle, notice that the hypotenuse is the line between (, 2) and (1, ). This is y = 2 2x. o by Green s theorem we could compute the circulation as 1 2 2x F dr = 2x + 1 dy dx ince Green s (and several other topics) require you to set up a multiple integral, this may be something good to look back to hapter 15 and review as well.

6 The last two integral theorems apply to vector fields in 3. tokes Theorem: tokes theorem applies to vector surface integrals (i.e. flux) on surfaces with boundary. Generally we say is the surface and is the boundary curve of. Assume both and have parameterizations which are continuous with continuous partial derivatives. Also assume that they do not intersect themselves. To apply tokes we need to be oriented in a particular way with respect to the normal vectors on. For the correct orientation, we want the surface to be on the left as we move around the boundary facing in the direction of the normal vector. You can imagine walking your fingers around the boundary. Make a finger-person with the index and middle fingers on your right hand. If your finger-person stands lined up with the normal vectors and walks around the boundary, the surface should be on the same side as your thumb. tokes theorem says that if F is continuous and has continuous partial derivatives on, and is oriented as described above, then ( F ) N d = F dr In words you can say that the flux of the curl of F over is equal to the circulation of F around the boundary. When do you want to use this? Any time I ask you for ( F ) N d, and is easy to parameterize, you should try tokes theorem. If you use this theorem, you don t even need to compute F because it is just F that goes into the line integral. Ex: Let be the cylinder x 2 +y 2 = 1 with 1 z 3. uppose this cylinder has a cap at the upper end but not at the lower end, and that its normal vectors point out of the cylinder. Let F = y 2, y 2, z. We can use tokes theorem to compute ( F ) N d as a line integral. That way we don t have to parameterize both the sides and top of the cylinder, and we don t have to compute the curl. The boundary of is the circle at the bottom of the cylinder. We want this circle to be oriented counterclockwise to use tokes theorem. Imagine walking your right hand around the base of the cylinder. Your thumb should point up toward the surface of the cylinder, so your fingers walk counterclockwise. This circle has radius 1 and sits at height z = 1, so we parameterize it as r(t) = 1 cos(t), 1 sin(t), 1 with t 2π. This matches the correct orientation. From here we need to set up the line integral. o we plug r(t) into F and compute r (t) and obtain ( F ) N d = F dr = F (r(t)) = 1 sin 2 (t), 1 sin 2 (t), 1 r (t) = 1 sin(t), 1 cos(t), 2π 1 1 sin 2 (t)(cos(t) sin(t)) dt

7 ivergence Theorem The ivergence theorem applies to closed surfaces. These could be whole spheres or boxes, cylinders with both caps, or the surface surrounding any volume. The only assumption we need about orientation is that the normal vectors on are oriented out of the volume inside. The ivergence theorem says that if is a closed surface (with nice parameterization as usual) and F is continuous on the volume V inside then F N d = F dv Generally, any time you have a surface integral over a closed surface it is going to be easier to use the ivergence theorem. This is because setting up the triple integral is usually easier than messing around with parameterizations. If we think of the divergence F as the instantaneous flux, this theorem takes on a nice interpretation. It says the flux across the surface is equal to the net instantaneous flux inside. Ex: Let be the surface of the region above the graph z = 2 + x 2 + y 2 and under the plane z = 6. Let F (x, y, z) = xy, yz, xy 2. We can compute the flux across, which is F N d by setting up a triple integral over this region. The region is bounded between 2 + x 2 + y 2 z 6. Where these surfaces intersect, 6 = 2 + x 2 + y 2 so this volume sits over top of the circle x 2 + y 2 = 4 in the xy-plane. The divergence is F (x, y, z) = y + z. While it is not totally necessary, this integral is easier to set up in cylindrical coordinates as 2π 2 6 V 2+r 2 r sin(θ) + z dz dr dθ Last Notes: Here are a few more miscellaneous tips. The integral of 1 over any geometric object (like a surface, curve or volume) gives us the measure of that object (so for curves it gives us the length, for surfaces the surface area, etc.). o if I ask for the length of a curve I want 1 ds Or if I ask for the surface are of I want 1 d These are both scalar integrals, so in both cases you should have the magnitude of something inside the integral. This also works the other direction. If you wind up with the integral of a constant, pull the constant out of the integral and now it represents the measure of something. For example 5 da = 5 1 da = 5Area() Use this fact to make evaluating integrals easier!