Vector Calculus Part II: Line integrals

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Vector Calculus Part II Line integrals Surface
integrals

• BET2533 Eng. Math. III
• R. M. Taufika R. Ismail
• FKEE, UMP

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Line integrals

• Suppose that a positive charge moves along
  direction in an electric field .
• If is uniform, then the energy dissipated by
  the electric field is equal to
• However, if the direction of varies, then the
  total energy becomes

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Let
Since
then
Hence we have
Notes Line integral differs with partial
integral. Along the curve C, x, y and z are
dependent of each others. Therefore, we cannot,
for example, when integrating w.r.t. x, we regard
y and z as constants.
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Example 1

• Evaluate the line integral
• where
• and C is given by

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Solution
Since
we have
then
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Example 2

• Evaluate
• where
• and C is the line segment from (1,1,1) to
• (1,2,4).

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Solution
In this example, C is not given directly like in
the previous example. Hence, we have to find C
Let and . Then
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Example 3

• Find the total work done by a force
• in moving a particle from (1,0,0) to (2,2,1)
• (i) along the curve
  .
• (ii) along the straight line segment connecting
  the two points.
• Given Total work .

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Solution
(i)
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(ii) The parametric eqn of the line
segment connecting the two points is
Then in this case the total work done is
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Example 4

• Find if and
• C is a closed loop formed by the semicircle
  
• from (1,0) to (-1,0) and the line
  
• segment from (-1,0) to (1,0).

Notes denotes integration along a
closed loop C.
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Solution
The loop C is formed by C1 and C2 as shown below
where
Hence,
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Conservative fields

• A vector field F is said to be conservative if
  there can be found some scalar field f such that
• In other words, if F is a gradient of some scalar
  field, then it is a conservative field
• Then the curl of F is

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• The line integral of F is
• but
• Therefore
• The above result show that, the exact form of the
  curve C does not come into play, that is how we
  go from P to Q does not matter

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• In other words, a line integral of a conservative
  field is independent of path
• Therefore, an alternative way of evaluating a
  line integral of a conservative field is to use
  any convenient and simple path, for example a
  line segment from P to Q
• Moreover, the integral of a conservative field
  along a closed loop is equal to zero, i.e.
• since the simplest path from a point to the same
  point is no need to move

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Example 5

• Show that
• is a conservative field, and find its potential
• function f.
• With the initial point P(1,0,2) and the terminal
• point Q(2,p,3), evaluate by
• (i) using the potential function f
• (ii) way of a line segment joining P and Q.

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Solution
Since , the vector field F is
conservative
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Since F is conservative, there is exist a
potential function such that
, that is
Integrating (1), (2) and (3) w.r.t. x, y and
z respectively, we obtain
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Comparing (4) with (5) and (6) we get
Therefore
Then, to find
(i) using theorem
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(ii) using the line segment from P to Q
The parametric eqn of the line segment is
Thus,
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Greens theorem

• Under certain conditions, we can evaluate a
  closed line integral on the xy-plane through a
  double integral by using Greens theorem

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Example 6

• Given
• and C is the boundary of the region in the first
  quadrant bounded by y-axis, x-axis and the circle
  . Evaluate
• (i) directly
• (ii) by using Greens theorem

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Solution
(i) Since , we perform the
line integral separately
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Surface integrals

• Suppose we are given a surface s as shown below,
  and a function y (x,y,z) defined and continuous
  on s.

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• Then, the surface integral of a scalar field y
  (x,y,z) over s is denoted by
• To evaluate a surface integral, we transform it
  into a double integral
• Let the surface s is given by z f (x,y) and R
  is the projection of the surface s on the
  xy-plane, then which implies

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• Then, we have the following theorem
• where R is the projection of the surface s on
  the xy-plane, z f (x,y) is the surface equation
  and is the unit normal in outward direction
  to the surface
• If , the value of the surface
  integral is equal to the area of the surface

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Example 7

• Evaluate
• where s is the part of the curved surface of the
  cylinder in the first octant
  between and .

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Solution
R is the projection of s on the xy-plane. Then
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To find the unit normal of s, let then
Therefore
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Example 8

• Find the surface area of the cone
• under the plane .

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Solution
R is the projection of s on the xy-plane. Then
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To find the unit normal of s, let then
Therefore
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• In most situations, the function y arises from
  scalar product involving a vector field
• For example, the flux that passes through a cross
  section (surface s) is
• ,
• where F is the vector field and S is the surface
  vector with direction normal to the surface

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Example 9

• If , compute
• where s is the surface of the paraboloid
• above the xy-plane with
  normal in upward direction.

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Solution
We project the surface s onto the xy-plane
as shown above. Then
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Let . Then
Therefore
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Gauss theorem

• Gauss theorem relates a surface integral to a
  triple integral
• Gauss theorem state that if G is a solid region
  bounded by a closed surface s and F is a vector
  field, then

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Example 10

• Use Gauss theorem to evaluate
• where
• and s is the closed surface formed by the
  cylinder , and the planes
• and .

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Solution
The cylinder is illustrated as
Using Gauss theorem,
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Stokes theorem

• Stokes theorem provides a relation between a
  line integral in 3-D and a surface integral
• Stokes theorem state that if s is a smooth open
  surface bounded by a piecewise smooth, simple and
  closed curve C traverse in positive direction,
  then

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Example 11

• Use Stokes theorem to evaluate
• where
• and C is the ellipse obtained from the
  intersection between the cylinder
  and the plane with the orientation
  counterclockwise looking from above

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Solution
The ellipse is illustrated as
Using Stokes theorem,
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R
Let . Then
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Hence
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Example 12

• Use Stokes theorem to evaluate
• where
• and s consists of three of the surfaces of the
  tetrahedron, that the plane makes with
  the coordinate planes excluding the plane
  .

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Solution
The tetrahedron without the plane is
illustrated as
Using Stokes theorem,
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C1 line segment (0,0,0) to (0,6,0)
C2 line segment (0,6,0) to (0,0,2)
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C3 line segment (0,0,2) to (0,0,0)
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