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    x

    yy = f(x)

    a b

    x

    y

    a b

    y = g x

    y

    x = f(y)

    a

    b

    y = g(x)

    x

    y

    y = f(x)b

    a

    sr

    X

    r=f()

    =1

    O

    =2

    X

    r=f()

    =1

    O

    =2r=g()

    97422441.doc Page 1 / 4

    Pure Math Geom etric Applications of DeIntegrals

    Areaof PlaneFiguresCartesian coordinate system:

    The area under the equationy =f(x) in the regionx[a, b] is given by: =b

    adxxfA )(

    The area under thex-axis is negative whereas above the axis is positive The area bounded by the equationsy =f(x) andy =g(x) in the regionx[a, b] is given by:

    =b

    adxxgxfA )]()([

    The area under the equationx =f(y) in the regiony[a, b] is given by: =b

    adyyfA )(

    The area at right-hand side of they-axis is positive whereas at left-hand side of the axis is

    negative The area bounded by the equationsx =f(y) andx =g(y) in the regiony[a, b] is given by:

    =b

    adyygyfA )]()([

    Parametric equations: When a curve is given by parametric equation:

    ==

    )()(

    tgytfx t

    The area at the rangex[a, b] is:

    ====

    =

    2

    1

    )(')()(')(t

    t

    bx

    ax

    b

    adttftgdttftgydxA

    a =f(t1) i.e.x = a when t= t1 b =f(t2) i.e.x = b when t= t2

    The absolute area at the rangex[a, b] is: ==2

    1

    )('|)(|||t

    t

    b

    adttftgdxyA

    Polar equations: Definite integrals can be used to find the area bounded by r=f()

    and radius vectors =1, =2

    For a sector with angle and radius rhas the area === 221

    21

    21 )( rrrrsA

    The total area ofr=f() from =1 to =2 is: =

    =

    =

    =

    =2

    1

    2

    1

    )( 22

    1

    rAA

    When taking the limit 0,

    ====

    =

    2

    1

    2

    1

    dfdrrA 2212

    212

    21

    0)]([)(lim

    2

    1

    The area enclosed by r=f(), r=g() and the radius vectors =1, =2:

    ( ) ==22

    11

    22

    212

    2

    2

    121 )]([)]([)( dgfdrrA

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    y = f(x)

    a

    x

    s

    x

    y

    a b

    x

    x2x1

    Page 2 / 4 97422441.doc

    ArcLengthCartesian coordinate system: In the curvey=f(x), the length of the curve inx[a, b] iss In any partition of line ab, whose width is x and difference

    inf(x) is

    y, the portion of the curve is approximately equal

    to: 22 )()( yxs +=

    The total length of curve: =

    =

    =

    =

    +=bx

    ax

    bx

    ax

    yxss 22 )()(

    By taking the limit x0,

    +=

    +=+=

    =

    =

    =

    =

    b

    a

    bx

    axx

    bx

    axx

    dxdx

    dy

    x

    yxyxs

    22

    0

    22

    011lim)()(lim

    Length of curve: +=

    +=

    b

    a

    b

    a

    dxxfdxdx

    dys 2

    2

    )]('[11

    Parametric equations:

    When a curve is given by parametric equation:

    ==

    )()(tgytfx

    t

    The length of curve at the rangex[a, b] is:

    ( )

    +=

    +=

    +=

    1

    2

    1

    2

    2222

    /

    /11

    t

    t

    t

    t

    b

    a

    dtdt

    dy

    dtdxdt

    dtdx

    dtdx

    dtdydx

    dx

    dys

    Length of curve: ( ) +=

    +=

    2

    1

    1

    2

    22

    22

    )]('[)]('[t

    t

    t

    t

    dttgtfdtdt

    dy

    dtdxs

    a =f(t1) i.e.x = a when t= t1 b =f(t2) i.e.x = b when t= t2

    Polar equations: The length of curve of the equation r=f() from =1 to =2:

    Transformed to Cartesian coordinate:

    ====

    sin)(sincos)(cos

    fryfrx

    Length of curve: ( ) +=

    +=

    2

    1

    1

    2

    22

    22

    )]('[

    drrdd

    dy

    ddxs

    Volumeof Solidsof RevolutionDisc method: Lety =f(x) be a cts function defined on [a, b] and the curve is

    revolved about thex-axis At any partition [x1,x2] in [a, b] which has the width x, the volume

    of revolution of the curve in this partition is like a disc Volume of a disc: V= r2h = y2(x) = [f(x1)]2x

    Total volume of revolution: =

    =

    =

    ==

    bx

    ax

    bx

    axxxfVV 2)]([

    When taking the limit x0,

    ====

    =

    b

    a

    b

    a

    bx

    axx

    dxxfdxyxxfV )()]([lim222

    0

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    y=f(x)

    a b

    h

    y=f(

    a

    xy=g(x)

    97422441.doc Page 3 / 4

    The volume of revolution ofy =f(x) about the liney = h in theregionx[a, b] is:

    ==b

    a

    b

    adxhxfdxhyV 22 ])([)(

    The volume of revolution ofx =f(y) about they-axis in the regiony[a, b] is:

    ==b

    a

    b

    adyyfdyxV 22 )]([

    The volume of revolution ofx =f(y) about the linex = kin the regiony[a, b] is:

    ==b

    a

    b

    adxkyfdykxV 22 ])([)(

    Shell method: Lety =f(x) andy =g(x) be cts functions defined on [a, b]The area bounded byy =f(x),y =g(x),x = a,x = b is on the

    xy-planeThe area is revolved about they-axis

    At any partition in [a, b] of thex-axis, which has the width x, thestrip revolved is like a hollow cylinder Volume of the cylinder: V= (2x)[f(x) g(x)](x)

    Total volume is: =

    =

    =

    ==

    bx

    ax

    bx

    ax

    xxgxfxVV )]()([2

    By taking the limit x0, ===

    =

    b

    a

    bx

    axx

    dxxgxfxxxgxfxV )]()([2)]()([lim20

    Volume of revolution: ==b

    a

    b

    adxyyxdxxgxfxV )(2)]()([2 21

    y1 =f(x) y2 =g(x)y1 y2

    Lety =f(x) andy =g(x) be cts functions defined on [a, b]The area bounded byy =f(x),y =g(x),x = a,x = b is on thexy-plane

    Area bounded by curvesx =f(y) andx =g(y) in the regiony[a, b] is revolved about thex-axis

    Volume of revolution: ==b

    a

    b

    adyxxydyygyfyV )(2)]()([2 21

    Solids w/known parallel cross section: For a solid which the cross-sectional area perpendicular to thex-axis is known The area at any positionx0 on thex-axis isA(x0)

    The volume of the solid between positionsx = a andx = b is: =b

    adxxAV )(

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    Areaof Surfaceof RevolutionArea of surface of revolution: A curvey=f(x) in the regionx[a, b] is revolved about thex-axis In any partition in [a, b] on thex-axis, whose width is x, the curve is: 22 )()( yxs +=

    When revolved, the area of the surface is A = 2ys =2

    1)(2

    +x

    yxxf

    Total area: =

    =

    +bx

    ax

    xx

    yxfA

    2

    1)(2

    By taking the limit x0

    Total area:

    +=

    +=

    =

    =

    b

    a

    bx

    axx

    dxdx

    dyyx

    x

    yxfA

    22

    0121)(lim2

    Area of the surface of revolution: +=+=b

    a

    b

    a dxxfxfdxdxdyyA 2

    2

    )]('[1)(212