Vector Calculus

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Vector Calculus

• CHAPTER 9.109.17

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Contents

• 9.10 Double Integrals
• 9.11 Double Integrals in Polar Coordinates
• 9.12 Greens Theorem
• 9.13 Surface Integrals
• 9.14 Stokes Theorem
• 9.15 Triple Integrals
• 9.16 Divergence Theorem
• 9.17 Change of Variables in Multiple Integrals

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9.10 Double Integrals

• Definition 9.10 Let f be a function of two
  variables defined on a closed region R. Then the
  double integral of f over R is given by
• 
  (1)
• Integrability If the limit in (1) exists, we
  say that f is integrable over R, and R is the
  region of integration.
• Area When f(x,y)1 on R.
• Volume When f(x,y) 0 on R.

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Properties of Double Integrals

• Theorem 9.11 Let f and g be functions of two
  variables that are integrable over a region R.
  Then
• (i) ,
  where k is any constant
• (ii)
  
• (iii)
  where
• and

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Regions of Type I and II

• Region of Type ISee the region in Fig 9.71(a)
  R a ? x ? b, g1(x) ? y ? g2(x)
• Region of Type IISee the region in Fig 9.71(b)
  R c ? y ? d, h1(y) ? x ? h2(y)

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Fig 9.71
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Iterated Integral

• For Type I (4)
• For Type II (5)

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Note

• Volume where z f(x, y) is the surface.

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Example 1

• Evaluate over the region
  bounded by y 1, y 2, y x, y -x 5. See
  Fig 9.73.
• SolutionThe region is Type II

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Fig 9.73
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Example 2

• Evaluate over the region in
  the first quadrant bounded by y x2, x 0, y
  4.
• SolutionFrom Fig 9.75(a) , it is of Type
  IHowever, this integral can not be computed.

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Fig 9.75(a) Fig 9.75(b)
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Example 2 (2)

• Trying Fig 9.75(b), it is of Type II

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9.11 Double Integrals in Polar Coordinates

• Double Integral Refer to the figure.The
  double integral is

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• Refer to the figure.The double integral is

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Change of Variables

• Sometimes we would like to change the rectangular
  coordinates to polar coordinates for simplifying
  the question. If and
  then (3)Recall x2 y2 r2
  and

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Example 2

• Evaluate
• SolutionFrom
  the graph is shown in Fig 9.84.Using x2
  y2 r2, then 1/(5 x2 y2 ) 1/(5 r2)

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Fig 9.84
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Example 2 (2)

• Thus the integral becomes

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Example 3

• Find the volume of the solid that is under
  and above the region bounded by x2 y2 y
  0. See Fig 9.85.
• SolutionFig 9.85

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Example 3 (2)

• We find that and the equations becomeand r
  sin ?. Now

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Example 3 (3)
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Area

• If f(r, ?) 1, then the area is

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9.12 Greens Theorem

• Along Simple Closed Curves For different
  orientations of simple closed curves, please
  refer to Fig 9.88.Fig 9.88(a) Fig
  9.88(b) Fig 9.88(c)

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Notations for Integrals Along Simple Closed
Curves

• We usually write them as the following
  formswhere and represents in the
  positive and negative directions, respectively.

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THEOREM 9.13

• IF P, Q, ?P/?y, ?Q/?x are continuous on R, which
  is
• bounded by a simply closed curve C, then

Greens Theorem in the Plane

• Partial ProofFor a region R is simultaneously of
  Type I and Type II,

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Fig 9.89(a) Fig 9.89(b)
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Partial Proof

• Using Fig 9.89(a), we have

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Partial Proof

• Similarly, from Fig 9.89(b),From (2)
  (3), we get (1).

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Note

• If the curves are more complicated such as Fig
  9.90, we can still decompose R into a finite
  number of subregions which we can deal with.
• Fig 9.90

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Example 2

• Evaluate
  where C is the circle (x 1)2 (y
  5)2 4 shown in Fig 9.92.

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Example 2 (2)

• SolutionWe have P(x, y) x5 3y andthenHence
  Since the area of this circle is 4?, we have

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Example 3

• Find the work done by F ( 16y sin x2)i
  (4ey 3x2)j along C shown in Fig 9.93.

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Example 3 (2)

• SolutionWe have Hence from Greens theorem
  In view of R, it is better handled in polar
  coordinates, since R

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Example 3 (3)
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Example 4

• The curve is shown in Fig 9.94. Greens Theorem
  is not applicable to the integral since P, Q,
  ?P/?x, ?Q/?y are not continuous at the region.

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Fig 9.94
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Region with Holes

• Greens theorem cal also apply to a region with
  holes. In Fig 9.95(a), we show C consisting of
  two curves C1 and C2. Now We introduce cross cuts
  as shown is Fig 9.95(b), R is divided into R1 and
  R2. By Greens theorem (4)

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Fig 9.95(a) Fig 9.95(b)

• The last result follows from that fact that the
  line integrals on the crosscuts cancel each
  other.

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Example 5

• Evaluate where C C1? C2 is shown in Fig 9.96.
  
• SolutionBecause

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Example 5 (2)

• are continuous on the region bounded by C, then

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Fig 9.96
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Conditions to Simplify the Curves

• As shown in Fig 9.97, C1 and C2 are two
  nonintersecting piecewise smooth simple closed
  curves that have the same orientation. Suppose
  that P and Q have continuous first partial
  derivatives such that ?P/?y ?Q/?x in the region
  R bounded between C1 and C2, then we have

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Fig 9.97
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Example 6

• Evaluate the line integral in Example 4.
• SolutionWe find P y / (x2 y2) and Q x /
  (x2 y2) have continuous first partial
  derivatives in the region bounded by C and C.
  See Fig 9.98.

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Fig 9.98
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Example 6 (2)

• Moreover,we have

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Example 6 (3)

• Using x cos t, y sin t, 0 ? t ? 2? ,
  then
• Note The above result is true for every
  piecewise smooth simple closed curve C with the
  origin in its interior.

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9.13 Surface Integrals
DEFINITION 9.11

• Let f be a function with continuous first
  derivatives
• fx, fy on a closed region. Then the area of the
  surface
• zf(x,y) over R is given by (2)

Surface Area
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Example 1

• Find the surface area of portion of x2 y2 z2
  a2 and is above the xy-plane and within x2 y2
  b2, where 0 lt b lt a.
• SolutionIf we definethen
• Thuswhere R is shown in Fig 9.103.

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Fig 9.103
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Example 1 (2)

• Change to polar coordinates

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Differential of Surface Area

• The functionis called the differential of
  surface area.

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DEFINITION 9.12
Let G be a function of three variables defined
over a region of space containing the surface S.
Then the surface integral of G over S is given
by (4)
Surface Integral
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Method of Evaluation

• (5)where we define z f(x, y) to
  be the equation of S projecting into a region R
  of the xy-plane.

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Projection of S Into Other Planes

• If we define y g(x, z) to be the equation of S
  projecting into a region R of the xz-plane,
  then (6)
• Similarly, if x h(y, z) is the equation of S
  projecting into a region R of the yz-plane,
  then (7)

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Example 3

• Evaluate , where S is the portion
  of y 2x2 1 in the first octant bounded by x
  0, x 2, z 4 and z 8.
• Solution The projection graph on the xz-plane
  is shown in Fig 9.105.

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Example 3 (2)

• Let y g(x, z) 2x2 1. Since gx(x, z) 4x
  and gz(x, z) 0, then

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Orientable Surface

• A surface is said to be orientable or an oriented
  surface if there exists a continuous unit normal
  vector function n, where n(x, y, z) is called the
  orientation of the surface. e.g S is defined by
  g(x, y, z) 0, then n ?g /
  ?g (9)where is the gradient.

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Fig. 9.106
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Fig 9.107
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Example 4

• Consider x2 y2 z2 a2, a gt 0. If we define
  g(x, y, z) x2 y2 z2 a2, then
• Thus the two orientations arewhere n defines
  outward orientation, n1 - n defines inward
  orientation. See Fig 9.108.

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Fig 9.108

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Computing Flux

• We have (10) See Fig 9.109.
• Flux of F through S
• The total volume of a
• fluid passing through S
• per unit time.

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Example 5

• Let F(x, y, z) zj zk represent the flow of a
  liquid. Find the flux of F through the surface S
  given by that portion of the plane z 6 3x
  2y in the first octant oriented upward.
• SolutionRefer to the figure.

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Example 5 (2)

• We define g(x, y, z) 3x 2y z 6 0. Then
  a unit normal vector with a positive k component
  (it should be upward) is Thus With R the
  projection of the surface onto the xy-plane, we
  have

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9.14 Stokes Theorem

• Vector Form of Greens Theorem If F(x, y) P(x,
  y)i Q(x, y)j, then Thus, Greens Theorem
  can be written as

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THEOREM 9.14
Let S be a piecewise smooth orientable surface
bounded by a piecewise smooth simple closed
curve C. Let F(x, y, z) P(x, y, z)i Q(x, y,
z)j R(x, y, z)k be a vector field for which P,
Q, R, are continuous and have continuous first
partial derivatives in a region of 3-space
containing S. If C is traversed in the positive
direction, thenwhere n is a unit normal to S
in the direction of the orientation of S.
Stokes Theorem
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Example 1

• Let S be the part of the cylinder z 1 x2 for
  0 ? x ? 1, -2 ? y ? 2. Verify Stokes theorem if
  F xyi yzj xzk.
• Fig 9.116

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Example 1 (2)

• Solution See Fig 9.116. Surface Integral From
  F xyi yzj xzk, we find

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Example 1 (3)
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Example 1 (4)
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Example 1 (5)
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9.15 Triple Integrals
DEFINITION 9.13

• Let F be a function of three variables defined
  over a
• Closed region D of 3-space. Then the triple
  integral of F
• over D is given by (1)

The Triple Integral
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Evaluation by Iterated Integrals

• See Fig 9.123.

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Fig 9.123
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Example 1

• Find the volume of the solid in the first octant
  bounded by z 1 y2, y 2x and x 3.
• Fig 9.125(a) Fig 9.125(b)

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Example 1 (2)

• Solution Referring to Fig 9.125(a), the first
  integration with respect to z is from 0 to 1
  y2. From Fig 9.125(b), we see that the projection
  of D in the xy-plane is a region of Type II.
  Hence

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Cylindrical Coordinates

• Refer to Fig 9.127.

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Conversion of Cylindrical Coordinates to
Rectangular Coordinates

• The relationship between the cylindrical
  coordinates (r, ?, z) and rectangular coordinates
  (x, y, z) x r cos ?, y r sin ?, z z
  (3)

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Example 1

• Convert (8, ?/3, 7) in cylindrical coordinates to
  rectangular coordinates.
• SolutionFrom (3)

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Conversion of Rectangular Coordinates to
Cylindrical Coordinates

• Also we have

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Example 4

• Solution

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Fig 9.128
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Triple Integrals in Cylindrical Coordinates

• See Fig 9.129.
• We have

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Fig 9.129
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Spherical Coordinates

• See Fig 9.131.

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Conversion of Spherical Coordinates to
Rectangular and Cylindrical Coordinates

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Example 6

• Convert (6, ?/4, ?/3) in spherical coordinates to
  rectangular and cylindrical coordinates.
• Solution

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Inverse Conversion
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Triple Integrals in Spherical Coordinates

• See Fig 9.132.

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• We have

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9.16 Divergence Theorem

• Another Vector Form of Greens TheoremLet F(x,
  y) P(x, y)i Q(x, y)j be a vector field, and
  let T (dx/ds)i (dy/ds)j be a unit tangent to
  a simple closed plane curve C. If n (dy/ds)i
  (dx/ds)j is a unit normal to C, then

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• that is, The result in (1) is a special case
  of the divergence or Gauss theorem.

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THEOREM 9.15
Let D be a closed and bounded region in 3-space
with a piecewise smooth boundary S that is
oriented outward. Let F(x, y, z) P(x, y, z)i
Q(x, y, z)j R(x, y, z)k be a vector field for
which P, Q, and R are continuous and have
continuous first partial derivatives in a
region of 3-space containing D. Then (2)
Divergence Theorem
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Example 1

• Let D be the region bounded by the hemisphere
• SolutionThe closed region is shown in Fig 9.140.

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Fig 9.140

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Example 1 (2)

• Triple Integral Since F xi yj (z-1)k, we
  see div F 3. Hence (10)
• Surface Integral We write ??S ??S1 ??S2,
  where S1 is the hemisphere and S2 is the plane z
  1. If S1 is a level surfaces of g(x, y) x2
  y2 (z 1)2, then a unit outer normal is

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Example 1 (3)
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Example 1 (4)
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9.17 Change of Variables in Multiple Integrals

• Introduction If f (x) is continuous on a, b,
  then if x g(u) and dx g?(u) du, we have
  where c g(a), d g(b).If we write
  J(u) dx/du, then we have

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Double Integrals

• If we have x f(u, v), y g(u, v)
  (3)we expect that a change of variables would
  take the form where S is the region in
  the uv-plane, and R is the region in the
  xy-plane. J(u,v) is some function obtained from
  the partial derivatives of the equation in (3).

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Example 1

• Find the image of the region S shown in Fig
  9.146(a) under the transformations x u2 v2, y
  u2 - v2.
• SolutionFig 9.146(a) Fig 9.146(b)

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Example 1 (2)
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Some of the Assumptions

1. The functions f, g have continuous first partial
   derivatives on S.
2. The transformation is one-to-one.
3. Each of region R and S consists of a piecewise
   smooth simple closed curve and its interior.
4. The following determinant is not zero on S.

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• Equation (7) is called the Jacobian of the
  transformation TS R and is denoted by ?(x,
  y)/?(u, v).Similarly, the inverse transformation
  of T is denoted by T-1. See Fig 9.147.

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• If it is possible to solve (3) for u, v in terms
  of x, y, then we have
• u h(x,y), v k(x,y) (8)The Jacobian of
  T-1 is

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Example 2

• The Jacobian of the transformation x r cos
  ?, y r sin ?is

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THEOREM 9.6
If F is continuous on R, then (11)
Change of Variables in a Double Integral
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Example 3

• Evaluate over the region R in Fig
  9.148(a).
• Fig 9.148(a) Fig 9.148(b)

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Example 3 (2)

• Solution We start by letting u x 2y, v x
  2y.

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Example 3 (3)

• The Jacobian matrix is

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Example 3 (4)

• Thus

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Example 4

• Evaluate over the region R in Fig
  9.149(a).Fig 9.149(a) Fig
  9.149(b)

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Example 4 (2)

• Solution The equations of the boundaries of R
  suggest u y/x2, v xy (12)The four
  boundaries of the region R become u 1, u 4, v
  1, v 5. See Fig 9.149(b).The Jacobian matrix
  is

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Example 4 (3)

• Hence

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Triple Integrals

• Let x f(u, v, w), y g(u, v, w), z h(u, v,
  w)be a one-to-one transformation T from a region
  E in the uvw-space to a region in D in xyz-space.
  If F is continuous in D, then

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Thank You !
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Quiz (2)

• 1. (40) (i) Find the level surface of
  passing through (1,1,1). (ii) Derive the
  equation of the tangent plane to the above level
  surface at (1,1,1). (iii) Find the unit vector
  such that the directional derivative of
  in at (1,1,1) achieves the
  minimum value among all possible .
• 2. (30) Let the curve C be represented by
  i j,
• , where
  and . Let , please
  compute (i) , (ii)
  , and (iii)

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• 1.(10)(i) F(1,1,1)1- 4 1 -2
• (15)(ii)
  2xi - 8yj 2zk
• 2(x-1)(-8)(y-1)2(z-1)0
• 2x-2-8y82z-20
• The tangent plane equation is
• 2x-8y2z40

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• (iii)(15)

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• 2. r(t) x(t) i y(t) j , x(t) 2cost, y(t)
  2sint
• dx -2sint dt dy 2cost dt
• -21-(-1) -4
  (10)
• 20-0 0 (10)
• 
  
• 
  (10)