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Lecture 4 Multiple Integration 4.1 The integral for a single variable Recall that in the Calculus I lectures we considered a function f (x ) defined over some bounded region R of x , and divided up R into n subregions, where δx i denoted the width of the i th subregion. Letting f i be the associated function value, the single integral was the limit Z f (x )dx = lim n→∞ δx i 0 n X i =1 f i δx i . (4.1) i f(x ) x i δ x i x Region R Figure 4.1: The integral for one variable 4.2 Double Integral The extension to two variables is straightforward. Consider a function f (x,y ) which 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 i th subregion. Let f i be the function value associated with the i th subregion. If the sum exists and is finite as n→∞ and 1

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Transcript of Lecture4 MultipleIntegration - University of Oxforddwm/Courses/1PD_2017/1PD-N4.pdf · Lecture4...

  • Lecture 4

    Multiple Integration

    4.1 The integral for a single variableRecall that in the Calculus I lectures we considered a function f (x) defined over somebounded region R of x , and divided up R into n subregions, where δxi denoted the widthof the ith subregion. Letting fi be the associated function value, the single integral wasthe limit∫

    f (x)dx = limn→∞δxi→0

    n∑i=1

    fiδxi . (4.1)

    if(x )

    xi

    δ

    xi

    x

    Region R

    Figure 4.1: The integral for one variable

    4.2 Double IntegralThe extension to two variables is straightforward. Consider a function f (x, y) whichis defined in some bounded region R of the (x, y) plane. Let R be divided up into nsubregions, where δAi denotes the area of the ith subregion. Let fi be the functionvalue associated with the ith subregion. If the sum exists and is finite as n→∞ and

    1

  • 4/2 LECTURE 4. MULTIPLE INTEGRATION

    δAi→0, then its limit is the double integral:∫∫f (x, y)dA = lim

    n→∞δAi→0

    n∑i=1

    fiδAi . (4.2)

    We assume here that it is irrelevant how the region is subdivided and the the xi , yichosen. This is always true for a continuous function f (x, y). Later we indicate howto extend integration to discontinuous functions.

    Ad i

    Function

    surface

    f i

    x

    y

    Region Rin x,y plane

    Figure 4.2: The double integral

    ♣ ExampleQ: Consider a thin plate, whose surface density (mass per unit area) is a functionσ(x, y). What is the mass of region R of the plate?

    A: The infinitesimal element of mass in area dA at (x, y) must be

    dM = σ(x, y)dA ⇒ Total M =∫∫

    R

    dM =∫∫

    R

    σ(x, y)dA . (4.3)

    4.2.1 Tiling the region

    To progress further we have to (i) define the region of integration, and (ii) define dA.You can think of placing the dA’s as generating a tiling across the region.

    In Cartesian coordinates, the xy plane is divided into a cake rack of x-constant, y -constant lines. The tiles are rectangles, so that in the limit

    dA = dxdy . (4.4)

    ♣ Example /ctd The integral for the mass is written

    Mass =

    ∫∫R

    σ(x, y)dxdy . (4.5)

  • 4.2. DOUBLE INTEGRAL 4/3

    x

    y

    Region in

    the x,y plane

    Tile dxdy in general position

    Figure 4.3: Placing a tile in general position

    You start by placing a tile in general position — at an (x, y) value that is not specialin any way — as shown in Figure 4.3.

    4.2.2 Summing tiles into strips

    Suppose we first summed the dxdy tiles into a strip parallel to the y -axis. For eachstrip this involves holding x constant and summing up the length along y from oneboundary intersection to the other.

    in a strip

    Sum over tiles

    x

    Region in

    the x,y plane

    x+dxx

    y2(x)

    y1(x)

    Figure 4.4: Summing tiles into a strip parallel to the y -axis.

    Note that we are assuming that the x =constant line intersects the boundary just twice.More on this later.

    ♣ Example /ctd The mass of the thin strip is

    dM =

    [∫ y=y2(x)y=y1(x)

    σ(x, y)dy

    ]dx (4.6)

  • 4/4 LECTURE 4. MULTIPLE INTEGRATION

    4.2.3 Summing strips into the region

    Then we want to sum the strip values up from the smallest to the largest values of x ,

    Sum over stripsin the region

    Region in

    the x,y plane

    x x1 2

    Figure 4.5: Summing strips into the complete region

    ♣ Example /ctd The total mass of the region is

    Mass of region =

    ∫ x2x=x1

    [∫ y=y2(x)y=y1(x)

    σ(x, y)dy

    ]dx (4.7)

    Note that, because the inner limits are function of x , the length of the strip variesautomatically as x changes.

    4.2.4 Evaluating the integral

    This is done in two stages, as a repeated integral.

    • First, consider the integral[∫ y=y2(x)y=y1(x)

    σ(x, y)dy]. Because x is held constant, simply

    treat it no differently from any other constant. (Recall that we already used thismethod in Lecture 1 when integrating total or perfect differentials.) We will thenbe left with some new function of (x, y) which is evaluated between the limitsy = y1(x) and y = y2(x). This yields some function of x alone, g(x) say.

    • Then integrate the function g(x) over x and evaluate it between the limits x1 andx2.

    Thus the double integral is broken down into two single integrals.

  • 4.2. DOUBLE INTEGRAL 4/5

    ♣ ExampleQ: Derive the area of the shaded triangular region of the xy−plane by integration.

    1

    1 2

    y

    xy (x)

    y (x)

    1

    2

    A: Obviously the result should be 1.

    A =

    ∫∫R

    dxdy =∫ 2x=0

    [∫ y=1y=x/2

    dy]dx =

    ∫ 2x=0

    [1−

    1

    2x

    ]dx (4.8)

    =

    [x −

    1

    4x2∣∣∣∣2

    0

    = 2−1

    44 = 1 (4.9)

    4.2.5 The order of integration does not matter

    You can just as well sum tiles into strips along the x-direction, then sum strips in they -direction. You must take care to redefine the region.

    x (y)2

    1

    1 2

    y

    x

    1x (y)

    A =

    ∫∫R

    dxdy =∫ 1y=0

    [∫ x=2yx=0

    dx]dy =

    ∫ 1y=0

    2ydy (4.10)

    =[y 2∣∣1

    0= 1 (4.11)

  • 4/6 LECTURE 4. MULTIPLE INTEGRATION

    4.3 A standard recipe for success ...The recipe for success is:

    1. Sketch the region of integration.2. Draw the tile in general position3. 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 finally6. Do it.

    ♣ExampleQ: The mass per unit area of the thin triangular plate varies as σ(x, y) = k xy , wherek is constant. Find the plate’s mass.

    A:Step 1 and 2: Sketch region and place tile

    1

    1 2

    y

    xy (x)

    y (x)

    1

    2

    Step 3: Think about limitsTile into strip: y goes from x/2 to 1. Strip into region: x goes from 0 to 2.

    Step 4: Delta quantity. This is the physics

    dM = σ(x, y)dA = k xy dxdy . (4.12)

    Step 5: Write down integral

    M = k

    ∫ x=2x=0

    x

    [∫ y=1y=x/2

    ydy]dx . (4.13)

  • 4.4. COMPLICATED REGIONS 4/7

    Step 6: Do it

    M = k

    ∫ x=2x=0

    x

    [∫ y=1y=x/2

    ydy]dx (4.14)

    = k

    ∫ x=2x=0

    x1

    2

    [1−

    x2

    4

    ]dx =

    k

    2

    ∫ x=2x=0

    [x −

    x3

    4

    ]dx

    =k

    2

    [x2

    2−x4

    16

    ∣∣∣∣20

    =k

    2[2− 1] =

    k

    2

    4.4 Complicated regionsEarlier on it was noted that for a continuous function it does not matter how yousubdivide the region. There are several types of complicated regions where care mustbe taken.

    (a) Suppose we wish to find the mass of a plate covering a region R comprised oftwo sub-regions R1 having a density function σ1(x, y) and R2 having a density functionσ2(x, y). Now it does matter how you subdivide R. No little element δA should straddlethe boundary of R1 and R2. The obvious solution is to perform two separate doubleintegrations and add up the results:∫∫

    RσdA =∫∫

    R1σ1dA+

    ∫∫R2σ1dA .

    Note that the functions σ1 and σ2 can be discontinuous at the boundary.

    σ1 σ2

    R1 R2

    R1

    R2

    One region Needs tworegions

    (a) (b) (c)

    Figure 4.6: Three region problems discussed above

    (b). Any closed smooth contour has two points where the tangent is parallel to thex axis and another two parallel to the y axes which coincide with the extremal values

  • 4/8 LECTURE 4. MULTIPLE INTEGRATION

    of x and y . Problems are caused when there are more parallel tangent points whichdo not coincide with the extremal has tangents which are parallel to either the x or yaxes, but which do not coincide with the extremal values. In this case, there will somelines of constant x and/or constant y which cut the boundary more than twice (butnote that the number of crossings is always even). But integrals only have two limits!The solution is to subdivide the region R into subregions not troubled by this shapeproblem, integrate over the subregions and then sum the results.

    (c). Sometimes the boundary is not defined by a single function. In the exampleshown, summing tiles into strips in the y direction would be straightforward, as thetwo boundaries provide the lower and upper limits. However, summing tiles along thex-direction would require splitting the region into two.

    4.5 Change of variables in double integralsThe definition of the double integral involved a subdivision of the region in into arbitrarilyshapes tiles dAi . Then, for a function f (x, y), it was noted that the obvious tiling haddA = dxdy .

    However, there is nothing particularly special about chopping the region of integrationinto little rectangles. Indeed, as we discussed in Lecture 3, often the shape and sym-metry of a problem suggests a different tiling or mesh, introduced by a transformationto a new set of variables.

    Suppose we specify a transformation to u, v coordinates as

    x = x(u, v) ; y = y(u, v) . (4.15)

    Let us keep the function in xy space and consider the new tiling.

    x

    y

    δu

    δv

    x

    y

    (a) (b)

    Figure 4.7: A region (a) showing lines of constant x and y giving rise to a dxdy tiling and (b) showinglines of constant u and v giving rise to a dudv tiling. These tiles are NOT rectangles, and do NOT havearea dudv .

  • 4.5. CHANGE OF VARIABLES IN DOUBLE INTEGRALS 4/9

    The tiling is no longer one of lines of constant x separated by δx and constant yseparated by δy , but rather one of lines of constant u separated by δu and constantv separated by δv . Typically, these will curves in the xy plane. (Indeed (u, v) arecommonly called curvilinear coordinates.)Two key things to note are that (i) these tiles are NOT rectangular, and (ii) theydo NOT have area dudv .

    To transform the integral we require three things:

    1. The function value in (u, v) coordinates.2. The tile area3. 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 usingx = x(u, v) and y = y(u, v). Thus F (u, v) = f (x(u, v), y(u, v)).

    2. The area of the tile.

    δu

    δv

    x

    y

    A

    B

    CD

    Figure 4.8: The new tile area

    The tile is not a rectangle, but in the limit is a parallelogram. Consider theparallelogram ABCD. Its area is given by the modulus of the vector product:

    dA = |du× dv| (4.16)

    Using ı̂, ĵ and k̂ as unit vectors in the x , y and z directions, we have

    δu = (xB − xA)̂ı + (yB − yA)̂j and δv = (xC − xA)̂ı + (yC − yA)̂j . (4.17)

  • 4/10 LECTURE 4. MULTIPLE INTEGRATION

    But

    (xB − xA) =∂x

    ∂uδu and (xC − xA) =

    ∂x

    ∂vδv (4.18)

    and similarly for the y terms, so that in the limit

    du =∂x

    ∂udûı +

    ∂y

    ∂udu ĵ and dv =

    ∂x

    ∂vdv ı̂ +

    ∂y

    ∂vdv ĵ . (4.19)

    The vector product is thus

    dv × dv = k̂(∂x

    ∂udu∂y

    ∂vdv −

    ∂x

    ∂vdu∂y

    ∂udv)

    (4.20)

    So that taking the modulus (k̂ is a unit vector!):

    dA =∣∣∣∣ (∂x∂u ∂y∂v − ∂x∂v ∂y∂u

    ) ∣∣∣∣dudv . (4.21)But the | . . . | term is nothing other than the modulus of the Jacobian, so that

    dA = dxdy = mod∂(x, y)

    ∂(u, v)dudv . (4.22)

    3. The last issue we have to consider is the region of integration. We have to expressthe region R′ in uv space using the limits of integration of u and v to exactly coverthe physical region R previously expressed in terms of x and y limits.

    v

    u

    uv

    spaceR’ in

    Figure 4.9: The new region R′ in uv space.

  • 4.5. CHANGE OF VARIABLES IN DOUBLE INTEGRALS 4/11

    ♣ Example #1Q: Consider a semicircle of radius a centred at the origin of xy space which lies in thex > 0 half space.

    i) Work out its area using integration in Cartesian coordinates.

    ii) Do the same by transforming to plane polars.

    (The answer should be πa2/2 of course!)

    a 2−x2

    a 2−x2

    y

    x0 a

    y

    x0 a

    r

    φ

    Figure 4.10: Semicircle region in (a) Cartesians and (b) Plane Polars

    A: (i) Place the dxdy tile in general position x, y . Integrate tiles into strips along they -direction, so that that limits are ylower(x) = −

    √a2 − x2 and yhigher(x) = +

    √a2 − x2.

    Then strips are integrated over x from x = 0 to x = a.

    So the integral is

    A =

    ∫ x=ax=0

    ∫ y=+√a2−x2y=−

    √a2−x2

    dydx =∫ x=ax=0

    2√a2 − x2dx . (4.23)

    This is now a standard 1-variable integral. We need a trig substitution, eg x = a sin pand dx = cos pdp∫ x=a

    x=0

    2√a2 − x2dx = 2a2

    ∫ π/2p=0

    cos2 pdp =a2

    2

    ∫ π/2p=0

    (1 + cos 2p)dp =a2π

    2. (4.24)

    A: (ii) The transformation is x = rcosφ, y = rsinφ.

    • The function is f (x, y) = 1, so that F (r, φ) = 1.

  • 4/12 LECTURE 4. MULTIPLE INTEGRATION

    • Find the MODULUS of the Jacobian.∂x

    ∂r= cosφ,

    ∂x

    ∂φ= −rsinφ

    ∂y

    ∂r= sinφ,

    ∂y

    ∂φ= rcosφ (4.25)

    so∂(x, y)

    ∂(r, φ)= r cos2 φ− (−r sin2 φ) = r . (4.26)

    Now r > 0, so |r | = r , and dA = rdrdφ.

    • Find region. This is 0 ≤ r ≤ a and −π/2 ≤ φ ≤ π/2. Note that the new regionis a rectangle in r, φ space. As we have noted before this is the sort of convenientshape we should hope for when make a “good” transformation.

    So the new integral is entirely separable

    A =

    ∫ ar=0

    ∫ +π/2φ=−π/2

    rdrdφ =

    ∫ +π/2φ=−π/2

    ∫ ar=0

    rdr =1

    2πa2. (4.27)

    ♣Example #2If you were not convinced about the merits of transformation by the previous example,try this.

    Q: The number of dopant atoms per unit area in a flat semicircular semiconductingwafer is α(x2 + 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. In Cartesiansthe total number of atoms is

    N = α

    ∫ ax=0

    ∫ +√a2−x2y=−

    √a2−x2

    (x2 + y 2)3/2dydx (4.28)

    which is a mess.

    Transforming ...

    • Find function in terms of r and φ: f (x, y) = α(x2 + y 2)3/2 = αr 3.

    • Modulus of the Jacobian, as before, is r , and dA = rdrdφ.

    • Semicircle in xy space transforms to rectangle 0 ≤ r ≤ a, −π/2 ≤ φ ≤ π/2.

    So

    N = α

    ∫ π/2φ=−π/2

    ∫ ar=0

    r 3 r dr dφ = απa5

    5. (4.29)

    Now divide by the area, A = πa2/2, and the average number per unit area is T/A =α2a3/5.

  • 4.6. TRIPLE (AND HIGHER) INTEGRALS 4/13

    4.6 Triple (and higher) integralsThe method we have discussed for double integrals is, straightforwardly extensible totriple and higher integrals, and is best explored now through examples. However, hereare the formalities.

    Definition: The triple integral is defined as∫∫∫R

    f (x, y , z)dV = limn→∞

    n∑i=1

    fiδVi (4.30)

    Volume element: In Cartesians, we can write dV = dxdydz and evaluate as a repeatedintegral.

    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 (a cuboid in Cartesians) ingeneral position (x, y , z), integrate the element first into a “rod”, then integrate therod 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 exactlythe same. 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) (4.31)

    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)) . (4.32)

    2. Finding the modulus of the Jacobian

    mod∂(x, y , z)

    ∂(u, v , w)= mod

    ∂x∂u

    ∂y∂u

    ∂z∂u

    ∂x∂v

    ∂y∂v

    ∂z∂v

    ∂x∂w

    ∂y∂w

    ∂z∂w

    ⇒ dV =∣∣∣∣ ∂(x, y , z)∂(u, v , w)

    ∣∣∣∣ dudvdw (4.33)3. Defining the new region R′

    4. And doing the integral

    I =

    ∫∫∫R′F (u, v , w)

    ∣∣∣∣ ∂(x, y , z)∂(u, v , w)∣∣∣∣ dudvdw. (4.34)

  • 4/14 LECTURE 4. MULTIPLE INTEGRATION

    ♣ Example #1Q: Find the mass of that part of the solid sphere x2 + y 2 + z2 ≤ a which occupies theoctant 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 firstin Cartesian coordinates.

    Volume element

    dxdydz

    x

    ymax

    max

    z

    x

    y

    dM = ρ(x, y , z)dV ⇒ M =∫∫∫

    R

    xyz dxdydz . (4.35)

    Place the volume element in general position (x, y , z), integrate into a rod over x , withy and z constant, then integrate over y into a plate, then finally over z .

    Integrating into a rod along x , variable x changes from xmin = 0 to xmax =√a2 − y 2 − z2.

    Note that xmax is not 1. The extreme position for the volume element is on the surfaceof the sphere, so x2max + y

    2 + z2 = a2.

    Integrating the rod along y , y changes from 0 to ymax, where (x = 0)2 +y 2max +z2 = a2,

    so ymax =√a2 − z2.

    The plate is integrated over z going from 0 to a.

  • 4.7. THE ROLE OF MOD[JACOBIAN] 4/15

    Thus the mass is

    M =

    ∫ az=0

    z

    ∫ √a2−z2y=0

    y

    ∫ √a2−y2−z2x=0

    x dxdydz (4.36)

    =1

    2

    ∫ az=0

    z

    ∫ √a2−z2y=0

    y[x2∣∣√a2−y2−z2

    0dydz (4.37)

    =1

    2

    ∫ az=0

    z

    ∫ √a2−z2y=0

    y(a2 − y 2 − z2) dydz (4.38)

    =1

    2

    ∫ az=0

    z

    ∫ √a2−z2y=0

    y(a2 − z2)− y 3 dydz (4.39)

    =1

    2

    ∫ az=0

    z

    [y 2

    2(a2 − z2)−

    y 4

    4

    ∣∣∣∣√a2−z2

    y=0

    dz (4.40)

    =1

    2

    ∫ az=0

    z

    [1

    2(a2 − z2)2 −

    1

    4(a2 − z2)2

    ]dz (4.41)

    =1

    8

    ∫ az=0

    z(a2 − z2)2dz (4.42)

    =1

    8

    ∫ az=0

    za4 − 2z3a2 + z5dz (4.43)

    =1

    8

    [1

    2z2a4 −

    1

    2z4a2 +

    1

    6z6∣∣∣∣az=0

    (4.44)

    =1

    48a6 (4.45)

    We will return to perform this integral in spherical polar coordinates after reviewing thestandard transformations introduced in Lecture 3.

    4.7 The role of mod[Jacobian]Looking at the double and triple integrals

    I =

    ∫∫R′F (u, v)

    ∣∣∣∣∂(x, y)∂(u, v)∣∣∣∣ dudv and I = ∫∫∫

    R′F (u, v , w)

    ∣∣∣∣ ∂(x, y , z)∂(u, v , w)∣∣∣∣ dudvdw

    (4.46)

    will remind you that inserting the modulus of the Jacobian was the method of turningdudv into a proper area, and dudvdw into a proper volume. That is in our generalcurvilinear coordinate system

    dA =∣∣∣∣∂(x, y)∂(u, v)

    ∣∣∣∣ dudv and dV = ∣∣∣∣ ∂(x, y , z)∂(u, v , w)∣∣∣∣ dudvdw . (4.47)

  • 4/16 LECTURE 4. MULTIPLE INTEGRATION

    In plane, cylindrical and spherical polars, although the lines of constant r, φ, etc, arenot straight, they always meet each other at right angles, and it turns out then to bequite easy to visual the area and volume elements. Being familiar with them will helpyou define regions of integration after transformations.

    4.8 Standard transformations revisitedHere we specialize the general theory to look at the shapes of the area and volumeelements generated by the standard transformations.

    1. 2D Cartesian to plane polars. The Jacobian is r which is always positive so thatits modulus is r . Thus the element of area is dA = r dr dφ. This is readily seen fromthe diagram.

    dφ rd

    dA =rd rdφ

    φr

    y

    x

    Figure 4.11: dA = r dr dφ.

    2. 3D Cartesian to cylindrical polars. The Jacobian is r , and, as r > 0 so is themodulus of the Jacobian. Thus the volume element of area is dV = rdrdφdz . Againthis is evident from the diagram. Note also that the volume element is dimensionallycorrect.

    φd

    r φd

    φ

    y

    x

    zr

    dz

    dr

    Figure 4.12:

  • 4.8. STANDARD TRANSFORMATIONS REVISITED 4/17

    3. 3D Cartesian to spherical polars.The Jacobian is r 2 sin θ (see Lecture 3). Now theta ranges from 0 to π, so that theJacobian is always positive. Thus the element of area is dV = r 2 sin θ dr dθ dφ.

    One side of the volume element is dr , the second is rdθ, and the third is r sin θdφ.Note that the third side is generated by swing the arm of length r sin θ through thechange in azimuthal angle dθ.

    θdr

    sinr θ dφdr

    rsinθ

    x

    θd

    z

    φ

    θ

    y

    r

    Figure 4.13: Spherical polars: dV = r 2 sin θ dr, dθ dφ

    4.8.1 Earlier example, now in Spherical polars

    Recall the problem to find the mass of the positive octant of a sphere with density xyz .We convert the integral to spherical polars as follows:

    M =∫∫∫

    R { x y z } dxdydz (4.48)

    =∫∫∫

    R {(r sin θ cosφ)(r sin θ sinφ)(r cos θ)} mod∂(x, y , z)

    ∂(r, θ, φ)drdθdφ (4.49)

    =∫∫∫

    R

    {r 3(sin2 θ cos θ)(sinφ cosφ)

    }r 2 sin θdrdθdφ (4.50)

    =∫∫∫

    Rr5(sin3 θ cos θ)(sinφ cosφ)drdθdφ (4.51)

    =

    ∫ ar=0

    r 5dr∫ π/2θ=0

    sin3 θ cos θdθ∫ π/2φ=0

    sinφ cosφdφ (4.52)

    =

    [1

    6r 6∣∣∣∣ar=0

    [1

    4sin4 θ

    ∣∣∣∣π/2θ=0

    [1

    2sin2 φ

    ∣∣∣∣π/2φ=0

    (4.53)

    =1

    48a6 (4.54)

  • 4/18 LECTURE 4. MULTIPLE INTEGRATION

    Notice that the region of integration is nowdefined by limits of integration which are allconstants — which means that the integration is separable into the product of threesingle-variable integrations.

    4.9 The definition of several integral propertiesIt is important to be able to set up integrals from first principles. You should focus firston defining the the d(Quantity) first — the total quantity is then simply the integralover the region.

    4.9.1 In 2D ...

    Consider a lamina of uniform thickness and mass per unit area σ(x, y) occupying theregion 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

    2dxdy .(4) Polar moment of inertia about the origin is

    ∫∫R σ(x, y) (x

    2 + y 2)dxdy .(5) Centroid x-coordinate is

    ∫∫xdxdy/

    ∫∫dxdy .

    (6) Centre of mass x-coordinate is∫∫σ(x, y) x dxdy/

    ∫∫σ(x, y) dxdy .

    4.9.2 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 (4.55)

    (3) Centroid: Average value of x , y or z . For example x = (1/V )∫∫∫

    R x dxdydz

    (4) Centre of Mass: This is the point (xCOM, yCOM, zCOM) at which placing the entiremass of a body gives the same moment or torque Γ as integrating over the torqueof the distributed mass. Looking at Figure 4.14(a), and dealing with just xCOM, thecontribution of an element of mass dM at (x, y , z) (ie, in general posn) to the torqueabout x = 0 is (assuming an acceleration g)

    dΓ = g.xdM ⇒ Γ = g∫∫∫

    R

    xdM = gxCOMM ⇒ xCOM =1

    M

    ∫∫∫R

    xρ(x, y , z) dxdydz

    (4.56)

    (5) Moment of inertia about an axis:

  • 4.9. THE DEFINITION OF SEVERAL INTEGRAL PROPERTIES 4/19

    z

    y

    dMx

    z

    y

    dMr

    (a) Centre of Mass (b) Polar moment of inertia

    Figure 4.14:

    The value depends on the axis of rotation. Supposing this to the be the z-axis, as inFigure 4.14(b), the element of moment of inertia is

    dJz = r 2dM = (x2 + y 2)dM ⇒ Jz =∫∫∫

    R

    (x2 + y 2)ρ(x, y , z)dxdydz . (4.57)

    About x , the value is

    Jx =

    ∫∫∫R

    (y 2 + z2)ρ(x, y , z)dxdydz . (4.58)

  • 4/20 LECTURE 4. MULTIPLE INTEGRATION

    ♣ ExampleQ: Compute the moment of inertia Jx about the x-axis of the solid cylinder x2 +y 2 ≤ a2bounded by z = 0 and z = b. Assume uniform volume density ρ.

    z

    x

    ba

    Figure 4.15:

    A: The moment of inertia of a little element dxdydz about the x-axis is

    dJx = ρ(y 2 + z2)dxdydz (4.59)

    So

    Jx = ρ∫∫∫

    R(y2 + z2)dxdydz . (4.60)

    The geometry suggests transforming to cylindrical polar coordinates. There are againthree tasks in making the transformation: (i) rewrite the function, (ii) derive the ModJacobian, (iii) rewrite the region.

    The transformation is given by x = rcosφ, y = rsinφ, z = z . Thus the function(y 2 + z2) is (r 2 sin2 φ+ z2).

    The Jacobian iscosφ rsinφ 0

    sinφ −rcosφ 00 0 1

    = r

    Now as r > 0 always, |r | = r .The region is 0 ≤ r ≤ a, 0 ≤ φ ≤ 2π and 0 ≤ z ≤ b.So

    Jx = ρ

    ∫ ar=0

    rdr

    ∫ 2πφ=0

    ∫ bz=0

    (r 2 sin2 φ+ z2)dz (4.61)

    = ρ

    ∫ ar=0

    rdr

    ∫ 2πφ=0

    (br 2 sin2 φ+ b3/3)dφ (4.62)

  • 4.9. THE DEFINITION OF SEVERAL INTEGRAL PROPERTIES 4/21

    Now ∫ 2π0

    sin2 φdφ =1

    2

    ∫ 2π0

    (1− cos 2φ)dφ = π

    so that

    Jx = ρ

    ∫ ar=0

    rdr(r 2πb +b3

    32π) (4.63)

    = ρ

    [a4πb

    4+b32πa2

    6

    ](4.64)

    = ρa2bπ

    12(3a2 + 4b2). (4.65)

    ♣ ExampleQ: A spherical settling tank has radius a and, as shown in the diagram, is filled to adepth 3a/2 with a fluid suspension whose volume density is

    ρ0a

    (3a/2− z) where z ismeasured upwards from the bottom of the tank. Calculate the mass of fluid suspensionin the tank.

    rmax

    In general posn

    at r,φ,z

    rmax

    z

    z=3a/2

    z=0

    a

    az−a

    Figure 4.16:

    A: The spherical tank pulls you towards spherical polars, but the dependence of thedensity on z alone should be enough to make you realize that cylindrical polars areeasier!

    So

    dM =ρ0a

    (3a/2− z)rdrdφdz ⇒ M =ρ0a

    ∫∫∫R

    (3a/2− z)rdrdφdz . (4.66)

  • 4/22 LECTURE 4. MULTIPLE INTEGRATION

    Now think about the region and the limits this places on the integrals.

    M =ρ0a

    ∫ z=3a/2z=0

    ∫ φ=2πφ=0

    ∫ r=rrmaxr=0

    (3a/2− z)rdrdφdz (4.67)

    The only limit requiring thought is rmax. From the inset diagram it is obvious that

    r 2max = a2 − (z − a)2 = (2az − z2) . (4.68)

    M =ρ0a

    ∫ z=3a/2z=0

    (3a/2− z)∫ φ=2πφ=0

    dφ∫ r=√2az−z2r=0

    rdr dz

    = 2πρ0a

    ∫ z=3a/2z=0

    (3a/2− z)1

    2

    [r 2∣∣√2az−z2

    0dz

    = πρ0a

    ∫ z=3a/2z=0

    (3a/2− z)(2az − z2)dz

    = πρ0a

    ∫ z=3a/2z=0

    [z3 −

    7a

    2z2 + 3a2z

    ]dz

    = πρ0a

    [1

    4z4 −

    7a

    6z3 +

    3a2

    2z2∣∣∣∣z=3a/2z=0

    = πρ09a

    4

    [1

    4

    (9a2

    4

    )−

    7a

    6

    (3a

    2

    )+

    3a2

    2

    ]= πρ0

    9a3

    4

    [9

    16−

    7

    4+

    3

    2

    ]= πρ0

    9a3

    4

    [5

    16

    ]= πρ0a

    3

    [45

    64

    ]

    It is encouraging that this is (i) positive, and (ii) dimensionally correct.

  • 4.9. THE DEFINITION OF SEVERAL INTEGRAL PROPERTIES 4/23

    ♣ ExampleQ: Find the volume bounded by the cylinder x2 + y 2 = 4 and the planes y + z = 4;z = −1.

    Figure 4.17: Volume under a surface

    Place the volume element in general position, and integrate to a rod along the z-direction. Then integrate the rod over the circle:

    V =

    ∫∫Circle

    ∫ z=4−yz=−1

    dzdxdy (4.69)

    =

    ∫∫Circle

    (5− y)dxdy (4.70)

    =

    ∫ 2πφ=0

    ∫ 2r=0

    (5− r sinφ)rdrdφ (4.71)

    =

    ∫ 2πφ=0

    [5r −

    r 2

    2sinφ

    ∣∣∣∣r=2r=0

    dφ (4.72)

    =

    ∫ 2πφ=0

    [10− 2 sinφ] dφ (4.73)

    = [10φ+ 2 cosφ|2π0 = 20π (4.74)