P1 Calculus II. Partial Differentiation & Multiple Integration. Prof David Murray. dwm/courses/1pd

Transcription

1 P / 58 P1 Calculus II Partial Differentiation & Multiple Integration Prof David Murray david.murray@eng.ox.ac.uk dwm/courses/1pd 4 lectures, MT 2017

2 P / 58 4 Multiple Integration

3 P / 58 Lecture contents In this lecture we are concerned with Double Integrals as a template for Multiple Integration 4.2 Changing variables in Double Integration 4.3 Triple (and higher) integration 4.4 Standard transformations revisited 4.5 Physical definitions involving integration

4 P / Double Integration: a template for Multiple Integration

5 P / 58 The integral for a single variable In Calculus I we considered a function f (x) defined over some bounded region R of x. We divided up R into n subregions, where δx i denotes the width of the ith subregion. f(x i) Region R δx i x Letting f i be the associated function value, the integral is the limit f (x)dx = x i lim n δx i 0 n f i δx i. i=1

6 P / 58 Double Integration f (x, y) is defined in some bounded region R of the (x, y) plane. Let R be divided up into n subregions, where δa i denotes the area of the ith subregion. Let f i be the function value associated with the ith subregion. If the sum exists and is finite as n and δa i 0, then its limit is the double integral: f (x, y)da = lim n δa i 0 n f i δa i. i=1 f i y da i x Function surface Region R in x,y plane Assume it is irrelevant how the region is subdivided and how the x i, y i chosen. This is always true for a continuous function f (x, y).

7 P / 58 Example: Double Integration Q: Consider a thin plate, whose surface density (mass per unit area) is a function σ(x, y). What is the mass of region R of the plate? A: The little bit of mass in area δa at (x, y) must be or in the limit Hence the total mass is M = δm = σ(x, y)δa dm = σ(x, y)da R dm = R σ(x, y)da. The setting up of integrals in this way is important when solving applied problems.

8 P / 58 Tiling the region To progress we have to (i) define what da is. (ii) define the region of integration. You can think of placing the da s as generating a tiling across the region. y Tile dxdy in general position Region in the x,y plane x In Cartesian coordinates, the xy plane is divided into a cake rack of x-constant, y-constant lines. The tiles are rectangles, so that δa = δxδy, and in the limit da = dxdy. Eg Mass = σ(x, y)dxdy. You start by placing a tile in general position at an x, y value that is not special in any way. R

9 P / 58 Summing tiles into strips First sum the dxdy tiles into a strip parallel to the y-axis. For each strip this involves holding x constant and summing up the length along y from one boundary intersection to the other. y 2 (x) y 1 (x) In our example, the mass of the thin strip is [ ] y=y2 (x) dm strip = σ(x, y)dy dx y=y 1 (x) x Sum over tiles in a strip x+dx x Region in the x,y plane Note carefully the that limits are functions of x. They change as x changes.

10 P / 58 Summing strips into the region Then sum the strip values up from smallest to largest x value. Sum over strips in the region Region in the x,y plane x1 x 2 In our example, the total mass of the region is [ x2 ] y=y2 (x) Mass of region = σ(x, y)dy dx x=x 1 y=y 1 (x) Note that during integration wrt x, the length of the strip varies automatically as x changes because the inner limits are function of x.

11 P / 58 Evaluating the integral This is done in two stages, as a repeated integral. [ y=y2 (x) First, consider the integral y=y 1 (x) ]. σ(x, y)dy Because x is held constant, simply treat it no differently from any other constant. This yields some new function of (x, y) which is evaluated between the limits y = y 1 (x) and y = y 2 (x) in turn, giving some function of x alone, g(x) say. We then integrate the function g(x) over x and evaluate it between the limits x 1 and x 2. Thus the double integral is broken down into two single integrals.

12 P / 58 Example: Evaluating the integral Q: Derive the area of the shaded triangular region of the y xy plane by integration. 1 y (x) 2 A: Obviously the result should be 1... A = R da = 2 x=0 [ y=1 dy y=x/2 ] dx = = 2 x=0 [1 12 x ] dx y 1(x) x 1 2 [x 14 2 x 2 = = 1 0

13 P / 58 The order of integration does not matter You can just as well sum tiles into strips along the x-direction, then sum strips in the y-direction. You must take care to redefine the region. y 1 x 1(y) x 1 x (y) 2 2 A = R da = 1 y=0 [ x=2y x=0 ] dx dy = 1 y=0 2ydy = [ y = 1

14 P / 58 A standard recipe for success... 1 Sketch the region of integration. 2 Draw the tile in general position 3 Think about limits as you build the tile in a strip, then the strip into the region. 4 Think about the δ quantity associated with the tile. 5 Then write down the integral, and finally 6 Do it.

15 P / 58 Example: A standard recipe for success Q: The mass per unit area of the thin triangular plate varies as σ(x, y) = k xy, where k is constant. Find the plate s mass. Steps 1,2: Sketch & place tile Step 3: Limits Tile strip: y goes from x/2 1 Strip region: x from 0 2. Step 4: delta quantity dm = σ(x, y)da = k xy dydx. Step 5: Write down integral M = k x=2 x=0 x [ y=1 ydy y=x/2 ] dx. y 1 Step 6: Do it M = k = k x=2 y 2 (x) x=0 x=2 x=0 x=2 y 1(x) x 1 2 [ y=1 x ydy y=x/2 [1 x 2 x [x x 3 ] dx ] dx = k ] dx 2 x=0 4 = k [ x x = k 0 2

16 P / 58 Complicated regions (a) For a continuous function it does not matter how you subdivide the region. But there are several types of complicated regions where care must be taken. Suppose we wish to find the mass of a plate covering a region R comprised of two sub-regions with different density functions. σ 1 σ 2 R 1 R 2 (a) Perform two separate double integrations and add up the results: σda = σ 1 da + σ 1 da. R R 1 R 2 NB! σ 1 and σ 2 can be discontinuous at the boundary.

17 P / 58 Complicated regions (b) Regions must be split up if there are lines of constant x and/or constant y that cut the boundary more than twice. Integrals only have two limits! R 1 R 2 (b) Subdivide the region R into subregions not troubled by this shape problem, integrate over the subregions and then sum the results.

18 P / 58 Complicated regions (c) Sometimes the boundary is not defined by a single function. In the example shown, summing tiles into strips in the y direction would be straightforward, as the two boundaries provide the lower and upper limits. However, summing tiles along the x-direction would require splitting the region into two. One region (c) Needs two regions

19 P / Change of variables in Double Integration

20 P / 58 Change of variables in double integrals The definition of the double integral involved a subdivision of the region into arbitrarily shapes tiles da i. Then, for a function f (x, y), it was noted that the obvious tiling had da = dxdy. However, often the shape and symmetry of a problem suggests a different tiling or mesh, introduced by a transformation to a new set of variables. x = x(u, v) ; y = y(u, v). Let us keep the function in xy space and consider the new tiling.

21 P / 58 Change of variables in double integrals y y x The tiling is no longer one of straight lines of constant x separated by δx and constant y separated by δy but rather one of curves of constant u separated by δu and constant v separated by δv. Two key things to note are that. these tiles are NOT rectangular, and they do NOT have area dudv δv δu x

22 P / 58 Change of variables in double integrals To transform the integral we require three things: 1. The function value in (u, v) coordinates. 2. The tile area 3. The region defined in (u, v) space. We look at these in turn. 1. A function value for each (u, v). This is obtained by substituting for x and y using x = x(u, v) and y = y(u, v). Thus F (u, v) = f (x(u, v), y(u, v))

23 P / The area of the tile The tile is not a rectangle, but (in the limit) is a parallelogram. Consider the parallelogram ABCD. Its area is given by the modulus of the vector product: da = du dv y C δv D A δu B Using î, ĵ and ˆk as unit vectors in the x, y and z directions, we have x δu = (x B x A )î + (y B y A )ĵ x du = u du î + y u du ĵ The vector product is thus ( du dv = ˆk x u du y v δv = (x C x A )î + (y C y A )ĵ dv = x v dv î + ) x y dv du v u dv y v dv ĵ

24 P / 58 Change of variables in double integrals Reminder: The vector product is thus ( du dv = ˆk x u du y v ) x y dv du v u dv Taking the modulus (k is a unit vector!), and tidying up... ( x y da = u v x ) y v u dudv. But the... term is the modulus of the Jacobian! Important result: da = dxdy = mod ( ) (x, y) dudv. (u, v)

25 P / 58 Change of variables in double integrals 3. The region of integration We have to express the region R in uv space using the limits of integration of u and v to exactly cover the physical region R previously expressed in terms of x and y limits. v R inuv space u Need specific examples to explain further.

26 P / 58 Example: Change of variables Q: Use integration in Cartesians to find the area of a semicircle of radius a shown A: Steps 1,2: dxdy tile in general position x, y. Step 3: Integrate tiles into strips along y Tiles strips: a 2 x 2 y + a 2 x 2. Strips region: x ranges from 0 x a. Step 4: da = dxdy Step 5: So the integral is A = da = x=a y=+ a 2 x 2 x=0 Trig subst: eg x = a sin p and dx = cos p dp y= a 2 x 2 dydx = x=a x=0 y a 2 x 2 x 0 a 2 a 2 x 2 dx. a 2 x 2 x=a 2 π/2 a 2 x 2 dx = 2a 2 cos 2 pdp = a2 x=0 p=0 2 π/2 p=0 (1 + cos 2p)dp = a2 π 2.

27 P / 58 Example: now in plane polars Transformation is x = rcos φ, y = rsin φ. 1. Function is f (x, y) = 1, F (r, φ) = MOD Jacobian. x r = Cφ, mod x φ = rsφ, y r = Sφ, y φ = rcφ (x, y) (r, φ) = r cos2 φ ( r sin 2 φ) = r = r. (x, y) da = mod dr dφ = r dr dφ (r, φ) 3. Region. This is 0 r a and π/2 φ π/2 So the new integral is entirely separable ( nice) y r φ x 0 a A = +π/2 φ= π/2 a r=0 r dr dφ = +π/2 φ= π/2 a dφ r dr = 1 r=0 2 πa2.

28 P / 58 Example #2: Change of variables Q: The number of dopant atoms per unit area in a flat semicircular semiconducting wafer of radius a is α(x 2 + y 2 ) 3/2. Determine the average dopant level per unit area. A: We must work out the total number of atoms and divide by the area. Screaming out for plane polars! Function: f (x, y) = α(x 2 + y 2 ) 3/2 = αr 3. Mod Jacobian: as before, is r, da = r dr dφ Region: A rectangle 0 r a, π/2 φ π/2. π/2 a N = α r 3 r dr dφ = απ a5 φ= π/2 r=0 5. The average number per unit area is then N a = N πa 2 /2 = α2a3 5.

29 P / Triple and higher integration

30 P / 58 Triple integrals: just continue the ideas... Definition: The triple integral is defined as R f (x, y, z)dv = lim n n f i δv i i=1 Volume element: In Cartesians, we can write dv = dxdydz and evaluate as a repeated integral. Region of Integration: Defining the region of integration tends to be harder in 3D, but the plan is the same. You place a volume element in general position (x, y, z), integrate the element first into a rod, then integrate the rod into a lamina or plate, and finally integrate the plate into the complete volume. Changing variables: One can also change variables, and again the pattern is exactly the same...

31 P / 58 Triple (and higher) integrals Suppose we wish and we wish to change to variables u, v, w where x = x(u, v, w) y = y(u, v, w) z = z(u, v, w) Transforming the integral involves 1 Finding the new function by substitution F (u, v, w) = f (x(u, v, w), y(u, v, w), z(u, v, w)). 2 Finding the modulus of the Jacobian mod (x, y, z) (u, v, w) = 3 Defining the new region R mod x u x v x w y u y v y w z u z v z w 4 And doing the integral I = F (u, v, w) (x, y, z) R (u, v, w) dudvdw.

32 P / 58 Example #1 Q: Find the mass of that part of the solid sphere x 2 + y 2 + z 2 a which occupies the octant x 0, y 0, z 0, and which has volume density ρ = ρ(x, y, z) = xyz. A: This is an obvious candidate for a transformation, but let s set up the integral first in Cartesian coordinates. dm = ρ(x, y, z)dv M = Rxyz dxdydz. Place the volume element in general position (x, y, z), integrate into a rod over x, with y and z constant, then integrate over y into a plate, then finally over z. z Volume element dxdydz y max y x x max

33 P / 58 Example #1 Elemental cuboid into rod: Integrate into a rod along x. Variable x changes from x min = 0 to x max =? at which point we need to think hard! Volume element dxdydz y max x x max Note immediately that x max a. The extreme position for the volume element is on the surface of the sphere, so xmax 2 + y 2 + z 2 = a 2 z y x max = a 2 y 2 z 2

34 P / 58 Example #1 Rod into plate: When integrating the rod along y direction y changes from y = 0 to y max =? z Volume element dxdydz y max y x x max Note immediately that y max a. Rather (x = 0) 2 + ymax 2 + z 2 = a 2 y max = a 2 z 2

35 P / 58 Example #1 z Volume element dxdydz y max y x x max The plate is integrated over z going from 0 to a. Thus the mass is M = = 1 2 = 1 2 a a 2 z 2 z=0 a z z=0 a z=0 y=0 x=0 a 2 z 2 z z y=0 a 2 z 2 y=0 a 2 y 2 z 2 y x dx dy dz a 2 y 2 z 2 y [ x 2 dy dz 0 y(a 2 y 2 z 2 ) dy dz

36 P / 58 Example #1 M = 1 2 = 1 2 = 1 2 = 1 2 = 1 8 a z=0 a z=0 a z=0 a z=0 a z=0 z z a 2 z 2 y=0 a 2 z 2 y=0 y(a 2 y 2 z 2 ) dydz y(a 2 z 2 ) y 3 dydz [ y 2 z 2 (a2 z 2 ) y 4 a 2 z 2 4 dz y=0 [ 1 z 2 (a2 z 2 ) 2 1 ] 4 (a2 z 2 ) 2 dz z(a 2 z 2 ) 2 dz

37 P / 58 Example #1 M = 1 8 a z=0 a z(a 2 z 2 ) 2 dz = 1 za 4 2z 3 a 2 + z 5 dz 8 z=0 = 1 [ z2 a z4 a z6 = 1 48 a6 a z=0 We will return to perform this integral in spherical polar coordinates later.

38 P / 58 Notice this general rule The integral M = a z=0 a 2 z 2 y=0 a 2 y 2 z 2 xyz dx dy dz x=0 provided a specific example of a more generally true thing... M = h2 z=h 1 g2 (z) y=g 1 (z) x=f2 (y,z) x=f 1 (y,z) F (x, y, z) dx dy dz The limits of the first integration can be functions of the remaining two variables. The limits of the second integration can be functions of the remaining 1 variable. The limits of the last integration must be constants.

39 P / Standard transformations revisited

40 P / 58 Standard transformations revisited We specialize the general theory to look at the shapes of the area and volume elements generated by the standard transformations. 1. 2D Cartesian to plane polars. The Jacobian is r which is always positive so that its modulus is r. Thus the area element is da = r dr dφ. y da =rdrdφ dφ r φ dr x

41 P / D Cartesian to cylindrical polars. The Jacobian is r, and, as r > 0 so is the modulus of the Jacobian. Thus the volume element is dv = rdrdφdz. Note also that the volume element is dimensionally correct. r dφ x z φ dz r dr dφ y

42 P / D Cartesian to spherical polars The Jacobian is r 2 sin θ (see Lecture 3). Now θ ranges from 0 to π, so that the Jacobian is always positive. Thus the volume element is dv = r 2 sin θ dr dθ dφ. The volume element has sides: dr rdθ r sin θdφ, generated by swinging the arm of length r sin θ through the change in azimuthal angle dφ. x z φ θ rsinθ r dr dφ rsinθdφ dθ r dθ y

43 P / 58 Example, now in Spherical Polars A: With region R being the positive octant, M = R x y z dxdydz = ( ) (x, y, z) R (r sin θ cos φ)(r sin θ sin φ)(r cos θ) mod dr dθ dφ (r, θ, φ) = R r 3 (sin 2 θ cos θ)(sin φ cos φ)(r 2 sin θ)dr dθ dφ = R r 5 (sin 3 θ cos θ)(sin φ cos φ)dr dθ dφ = = a r=0 [ 1 6 r 6 = 1 48 a6 r 5 dr a r=0 π/2 θ=0 sin 3 θ cos θdθ [ 1 4 sin4 θ π/2 θ=0 π/2 φ=0 [ 1 2 sin2 φ sin φ cos φdφ π/2 φ=0

44 P / 58 Covering the region z This happens automatically (if the limits are correct!!) but it is useful to be able to imagine in your head... ORDER 1: First r, then θ, then φ θ z θ z θ z r r r x φ y x φ x φ x ORDER 2: First φ, then θ, then r Easier to imagine if you think about an entire sphere, then chop out the octant dv to a ring, ring to shell, then shell to sphere ORDER 3: First φ, then r, then θ... DIY!... etc

45 P / Physical definitions involving integration

46 P / 58 The definition of several integral properties It is important to be able to set up integrals from first principles. You should focus first on defining the the d(quantity) first the total quantity is then simply the integral over the region.

47 P / 58 In 2D... Consider a lamina of uniform thickness and mass per unit area σ(x, y) occupying the region R in the xy plane. Then (1) Area is R dxdy. (2) Mass is R σ(x, y)dxdy. (3) Moment of inertia about the y-axis is R σ(x, y) x 2 dxdy. (4) Polar moment of inertia about origin is R σ(x, y) (x 2 + y 2 )dxdy. (5) Centroid x-coordinate is xdxdy/ dxdy. (6) Centre of mass x-coord is σ(x, y) x dxdy/ σ(x, y) dxdy.

48 P / 58 In 3D... and in more detail... (1) Volume: dv = dxdydz V = R dxdydz (2) Mass: Given a volume density function ρ(x, y, z), dm = ρ(x, y, z) dxdydz M = ρ(x, y, z) dxdydz (3) Centroid: Average value of x, y or z. For example x = 1 x dxdydz V R

49 P / 58 (4) Centre of Mass: This is the point (x COM, y COM, z COM ) at which placing the entire mass of a body gives the same moment or torque Γ as integrating over the torque of the distributed mass. z x y dm Dealing with just x COM, the contribution of an element of mass dm at (x, y, z) (ie, in general posn) to the torque about x = 0 is (assuming an acceleration g) dγ = x g dm Γ = g xdm = gx COM M x COM = 1 M R R xρ(x, y, z) dxdydz

50 P / 58 (5) Moment of inertia about an axis: The value depends on the axis of rotation. z r y dm Supposing this to be the z-axis, the element of moment of inertia is dj z = r 2 dm = (x 2 + y 2 )dm J z = (x 2 + y 2 )ρ(x, y, z)dxdydz. About x, the value is J x = (y 2 + z 2 )ρ(x, y, z)dxdydz. R R

51 P / More examples

52 P / 58 Example #2 Triple integration Q: Compute the moment of inertia J x about z the x-axis of the solid cylinder x 2 + y 2 a 2 bounded by z = 0 and z = b. Assume uniform volume density ρ. A: The moment of inertia of an element dxdydz about the x-axis is a b dj x = ρ(y 2 + z 2 )dxdydz J x = ρ (y 2 + z 2 )dxdydz. R Geometry hints at cylindrical polars, x = rcos φ, y = rsin φ, z = z. Function: (y 2 + z 2 ) is (r 2 sin 2 φ + z 2 ). Mod Jacobian: From earlier, r. Region: 0 r a, 0 φ 2π and 0 z b. a J x = ρ = ρ r=0 a r=0 rdr rdr 2π φ=0 2π φ=0 b dφ (r 2 sin 2 φ + z 2 )dz z=0 (br 2 sin 2 φ + b 3 /3)dφ x

53 P / 58 Example /ctd Reminder... J x = ρ = ρ a r=0 a r=0 rdr rdr 2π φ=0 2π φ=0 b dφ (r 2 sin 2 φ + z 2 )dz z=0 (br 2 sin 2 φ + b 3 /3)dφ Now so that 2π 0 sin 2 φdφ = 1 2 2π 0 (1 cos 2φ)dφ = π a J x = ρ rdr(r 2 πb + b3 r=0 3 2π) [ a 4 πb = ρ + b3 2πa 2 ] 4 6 = ρ a2 bπ 12 (3a2 + 4b 2 ).

54 P / 58 Example #3 Triple integration Q: A spherical settling tank has radius a and, as shown in the diagram, is filled to a depth 3a/2 with a fluid suspension whose volume density is ρ 0 (3a/2 z) where z is measured upwards from the bottom of the tank. a Calculate the mass of fluid suspension in the tank. z In general posn at r,φ,z z=3a/2 r max a z a r max a A: The spherical tank pulls you towards spherical polars, but the dependence of the density on z alone should be enough to make you realize that cylindrical polars are easier! z=0

55 P / 58 Example #3 /ctd z In general posn at r,φ,z z=3a/2 r max a z a r max a So z=0 dm = ρ 0 a (3a/2 z)rdrdφdz M = ρ 0 a R (3a/2 z)rdrdφdz. Now think about the region and the limits this places on the integrals. M = ρ 0 a z=3a/2 φ=2π r=rrmax z=0 φ=0 r=0 (3a/2 z)rdrdφdz The only limit requiring thought is r max. From the inset diagram it is obvious that r 2 max = a 2 (z a) 2 = (2az z 2 ).

56 P / 58 Example M = ρ 0 a = 2π ρ 0 a = π ρ 0 a = π ρ 0 a = π ρ 0 a z=3a/2 z=0 z=3a/2 z=0 z=3a/2 z=0 z=3a/2 z=0 (3a/2 z) (3a/2 z) 1 2 φ=2π φ=0 dφ r= 2az z 2 r=0 [ r 2 2az z 2 0 dz (3a/2 z)(2az z 2 )dz [ z 3 7a ] 2 z2 + 3a 2 z dz [ 1 4 z4 7a 6 z3 + 3a2 z=3a/2 2 z2 z=0 [ ( ) 9a 1 9a 2 = πρ 0 7a ( ) ] 3a + 3a a 3 [ 9 = πρ ] 9a 3 [ ] 5 = πρ rdr dz [ ] 45 = πρ 0 a 3 64 It is encouraging that this is (i) positive, and (ii) dimensionally correct.

57 P / 58 Example #4: Q: Find the volume bounded by the cylinder x 2 + y 2 = 4 and the planes y + z = 4; z = 1. A: Place the volume element in general position, and integrate to a rod along the z-direction. Then integrate the rod over the circle: V = = = z=4 y Circle z= 1 2π 2 φ=0 2π φ=0 r=0 dzdxdy = (5 r sin φ)rdrdφ = Circle 2π (5 y)dxdy φ=0 [5r r 2 2 sin φ [10 2 sin φ] dφ = [10φ + 2 cos φ 2π 0 = 20π r=2 r=0 dφ

58 P / 58 You ll be glad to hear...