VECTOR%20CALCULUS

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VECTOR CALCULUS
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VECTOR CALCULUS

• So far, we have considered special types of
  surfaces
• Cylinders
• Quadric surfaces
• Graphs of functions of two variables
• Level surfaces of functions of three variables

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VECTOR CALCULUS

• Here, we use vector functions to describe more
  general surfaces, called parametric surfaces, and
  compute their areas.

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VECTOR CALCULUS

• Then, we take the general surface area formula
  and see how it applies to special surfaces.

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VECTOR CALCULUS
17.6 Parametric Surfaces and their Areas
In this section, we will learn about Various
types of parametric surfaces and computing their
areas using vector functions.
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INTRODUCTION

• We describe a space curve by a vector function
  r(t) of a single parameter t.
• Similarly, we can describe a surface by a
  vector function r(u, v) of two parameters u and
  v.

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INTRODUCTION
Equation 1

• We suppose that r(u, v) x(u, v) i
  y(u, v) j z (u, v) kis a vector-valued
  function defined on a region D in the uv-plane.

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INTRODUCTION

• So x, y, and zthe component functions of rare
  functions of the two variables u and v with
  domain D.

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PARAMETRIC SURFACE
Equations 2

• The set of all points (x, y, z) in such that
  x x(u, v) y y(u, v) z
  z(u, v)and (u, v) varies throughout D, is
  called a parametric surface S.
• Equations 2 are called parametric equations of S.

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PARAMETRIC SURFACES

• Each choice of u and v gives a point on S.
• By making all choices, we get all of S.

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PARAMETRIC SURFACES

• In other words, the surface S is traced out by
  the tip of the position vector r(u, v) as (u, v)
  moves throughout the region D.

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PARAMETRIC SURFACES
Example 1

• Identify and sketch the surface with vector
  equation r(u, v) 2 cos u i v j 2 sin u k
• The parametric equations for this surface are
  x 2 cos u y v z 2 sin
  u

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PARAMETRIC SURFACES
Example 1

• So, for any point (x, y, z) on the surface, we
  have x2 z2 4 cos2u 4 sin2u
  4
• This means that vertical cross-sections parallel
  to the xz-plane (that is, with y constant) are
  all circles with radius 2.

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PARAMETRIC SURFACES
Example 1

• Since y v and no restriction is placed on v,
  the surface is a circular cylinder with radius 2
  whose axis is the y-axis.

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PARAMETRIC SURFACES

• In Example 1, we placed no restrictions on the
  parameters u and v.
• So, we obtained the entire cylinder.

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PARAMETRIC SURFACES

• If, for instance, we restrict u and v by writing
  the parameter domain as 0
  u p/2 0 v 3then x 0
  z 0 0 y 3

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PARAMETRIC SURFACES

• In that case, we get the quarter-cylinder with
  length 3.

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PARAMETRIC SURFACES

• If a parametric surface S is given by a vector
  function r(u, v), then there are two useful
  families of curves that lie on Sone with u
  constant and the other with v constant.
• These correspond to vertical and horizontal
  lines in the uv-plane.

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PARAMETRIC SURFACES

• Keeping u constant by putting u u0, r(u0, v)
  becomes a vector function of the single parameter
  v and defines a curve C1 lying on S.

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GRID CURVES

• Similarly, keeping v constant by putting v v0,
  we get a curve C2 given by r(u, v0) that lies on
  S.
• We call these curves grid curves.

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GRID CURVES

• In Example 1, for instance, the grid curves
  obtained by
• Letting u be constant are horizontal lines.
• Letting v be constant are circles.

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GRID CURVES

• In fact, when a computer graphs a parametric
  surface, it usually depicts the surface by
  plotting these grid curvesas we see in the
  following example.

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GRID CURVES
Example 2

• Use a computer algebra system to graph the
  surface
• r(u, v) lt(2 sin v) cos u, (2 sin v)
  sin u, u cos vgt
• Which grid curves have u constant?
• Which have v constant?

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GRID CURVES
Example 2

• We graph the portion of the surface with
  parameter domain 0 u 4p, 0 v 2p
• It has the appearance of a spiral tube.

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GRID CURVES
Example 2

• To identify the grid curves, we write the
  corresponding parametric equations
  x (2 sin v) cos u y (2 sin v) sin
  u z u cos v

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GRID CURVES
Example 2

• If v is constant, then sin v and cos v are
  constant.
• So, the parametric equations resemble those of
  the helix in Example 4 in Section 13.1

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GRID CURVES
Example 2

• So, the grid curves with v constant are the
  spiral curves.
• We deduce that the grid curves with u constant
  must be the curves that look like circles.

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GRID CURVES
Example 2

• Further evidence for this assertion is that, if
  u is kept constant, u u0, then the equation
  z u0
  cos v shows that the z-values vary from u0
  1 to u0 1.

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PARAMETRIC REPRESENTATION

• In Examples 1 and 2 we were given a vector
  equation and asked to graph the corresponding
  parametric surface.
• In the following examples, however, we are given
  the more challenging problem of finding a vector
  function to represent a given surface.
• In the rest of the chapter, we will often need
  to do exactly that.

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PARAMETRIC REPRESENTATIONS
Example 3

• Find a vector function that represents the plane
  that
• Passes through the point P0 with position vector
  r0.
• Contains two nonparallel vectors a and b.

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PARAMETRIC REPRESENTATIONS
Example 3

• If P is any point in the plane, we can get from
  P0 to P by moving a certain distance in the
  direction of a and another distance in the
  direction of b.
• So, there are scalars u and v such that
  ua vb

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PARAMETRIC REPRESENTATIONS
Example 3

• The figure illustrates how this works, by means
  of the Parallelogram Law, for the case where u
  and v are positive.
• See also Exercise 40 in Section 12.2

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PARAMETRIC REPRESENTATIONS
Example 3

• If r is the position vector of P, then
• So, the vector equation of the plane can be
  written as r(u, v) r0 ua
  vb where u and v are real numbers.

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PARAMETRIC REPRESENTATIONS
Example 3

• If we write r ltx, y, zgt r0 ltx0, y0, z0gt
  a lta1, a2, a3gt b ltb1, b2, b3gt we can
  write the parametric equations of the plane
  through the point (x0, y0, z0) as
• x x0 ua1 vb1
  y y0 ua2 vb2 z z0
  ua3 vb3

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PARAMETRIC REPRESENTATIONS
Example 4

• Find a parametric representation of the sphere
  x2 y2 z2 a2
• The sphere has a simple representation ? a in
  spherical coordinates.
• So, lets choose the angles F and ? in spherical
  coordinates as the parameters (Section 15.8).

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PARAMETRIC REPRESENTATIONS
Example 4

• Then, putting ? a in the equations for
  conversion from spherical to rectangular
  coordinates (Equations 1 in Section 15.8), we
  obtain
• x a sin F cos ? y a sin F sin ?
• z a cos F
• as the parametric equations of the sphere.

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PARAMETRIC REPRESENTATIONS
Example 4

• The corresponding vector equation is r(F, ?)
  a sin F cos ? i a sin F sin ? j a cos F k
• We have 0 F p and 0 ? 2p.
• So, the parameter domain is the rectangle
  D 0, p x 0,
  2p

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PARAMETRIC REPRESENTATIONS
Example 4

• The grid curves with
• F constant are the circles of constant latitude
  (including the equator).
• ? constant are the meridians (semicircles), which
  connect the north and south poles.

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APPLICATIONSCOMPUTER GRAPHICS

• One of the uses of parametric surfaces is in
  computer graphics.

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COMPUTER GRAPHICS

• The figure shows the result of trying to graph
  the sphere x2 y2 z2 1 by
• Solving the equation for z.
• Graphing the top and bottom hemispheres
  separately.

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COMPUTER GRAPHICS

• Part of the sphere appears to be missing because
  of the rectangular grid system used by the
  computer.

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COMPUTER GRAPHICS

• The much better picture here was produced by a
  computer using the parametric equations found in
  Example 4.

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PARAMETRIC REPRESENTATIONS
Example 5

• Find a parametric representation for the
  cylinder x2 y2 4 0
  z 1
• The cylinder has a simple representation r 2
  in cylindrical coordinates.
• So, we choose as parameters ? and z in
  cylindrical coordinates.

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PARAMETRIC REPRESENTATIONS
Example 5

• Then the parametric equations of the cylinder
  are x 2 cos ? y 2 sin ? z
  z
• where
• 0 ? 2p
• 0 z 1

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PARAMETRIC REPRESENTATIONS
Example 6

• Find a vector function that represents the
  elliptic paraboloid z x2 2y2
• If we regard x and y as parameters, then the
  parametric equations are simply x x
  y y z x2 2y2and
  the vector equation is r(x, y) x i
  y j (x2 2y2) k

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PARAMETRIC REPRESENTATIONS

• In general, a surface given as the graph of a
  function of x and yan equation of the form z
  f(x, y)can always be regarded as a parametric
  surface by
• Taking x and y as parameters.
• Writing the parametric equations as x x
  y y z f(x, y)

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PARAMETRIZATIONS

• Parametric representations (also called
  parametrizations) of surfaces are not unique.
• The next example shows two ways to parametrize a
  cone.

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PARAMETRIZATIONS
Example 7

• Find a parametric representation for the surface
  
  that is, the top half of
  the cone z2 4x2 4y2

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PARAMETRIZATIONS
E. g. 7Solution 1

• One possible representation is obtained by
  choosing x and y as parameters x x
  y y
• So, the vector equation is

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PARAMETRIZATIONS
E. g. 7Solution 2

• Another representation results from choosing as
  parameters the polar coordinates r and ?.
• A point (x, y, z) on the cone satisfies x r
  cos ? y r sin ?

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PARAMETRIZATIONS
E. g. 7Solution 2

• So, a vector equation for the cone is
  r(r, ?) r cos ? i r sin ? j 2r kwhere
• r 0
• 0 ? 2p

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PARAMETRIZATIONS

• For some purposes, the parametric representations
  in Solutions 1 and 2 are equally good.
• In certain situations, though, Solution 2 might
  be preferable.

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PARAMETRIZATIONS

• For instance, if we are interested only in the
  part of the cone that lies below the plane z
  1, all we have to do in Solution 2 is change the
  parameter domain to 0 r ½ 0 ? 2p

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SURFACES OF REVOLUTION

• Surfaces of revolution can be represented
  parametrically and thus graphed using a computer.

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SURFACES OF REVOLUTION

• For instance, lets consider the surface S
  obtained by rotating the curve
  y f(x) a x b about the x-axis,
  where f(x) 0.

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SURFACES OF REVOLUTION

• Let ? be the angle of rotation as shown.

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SURFACES OF REVOLUTION
Equations 3

• If (x, y, z) is a point on S, then x x y
  f(x) cos ? z f(x) sin ?

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SURFACES OF REVOLUTION

• Thus, we take x and ? as parameters and regard
  Equations 3 as parametric equations of S.
• The parameter domain is given by a x b
  0 ? 2p

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SURFACES OF REVOLUTION
Example 8

• Find parametric equations for the surface
  generated by rotating the curve y sin x, 0 x
  2p, about the x-axis.
• Use these equations to graph the surface of
  revolution.

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SURFACES OF REVOLUTION
Example 8

• From Equations 3,
• The parametric equations are x x y
  sin x cos ? z sin x sin ?
• The parameter domain is 0 x 2p 0 ?
  2p

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SURFACES OF REVOLUTION
Example 8

• Using a computer to plot these equations and
  rotate the image, we obtain this graph.

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SURFACES OF REVOLUTION

• We can adapt Equations 3 to represent a surface
  obtained through revolution about the y- or
  z-axis.
• See Exercise 30.

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TANGENT PLANES

• We now find the tangent plane to a parametric
  surface S traced out by a vector function
  r(u, v) x(u, v) i y(u, v) j z(u, v) k
  at a point P0 with position vector r(u0, v0).

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TANGENT PLANES

• Keeping u constant by putting u u0, r(u0, v)
  becomes a vector function of the single parameter
  v and defines a grid curve C1 lying on S.

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TANGENT PLANES
Equation 4

• The tangent vector to C1 at P0 is obtained by
  taking the partial derivative of r with respect
  to v

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TANGENT PLANES

• Similarly, keeping v constant by putting v v0,
  we get a grid curve C2 given by r(u, v0) that
  lies on S.

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TANGENT PLANES
Equation 5

• Its tangent vector at P0 is

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SMOOTH SURFACE

• If ru x rv is not 0, then the surface is called
  smooth (it has no corners).
• For a smooth surface, the tangent plane is the
  plane that contains the tangent vectors ru and
  rv , and the vector ru x rv is a normal vector
  to the tangent plane.

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TANGENT PLANES
Example 9

• Find the tangent plane to the surface with
  parametric equations x u2
  y v2 z u 2v at the point
  (1, 1, 3).

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TANGENT PLANES
Example 9

• We first compute the tangent vectors

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TANGENT PLANES
Example 9

• Thus, a normal vector to the tangent plane is

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TANGENT PLANES
Example 9

• Notice that the point (1, 1, 3) corresponds to
  the parameter values u 1 and v 1.
• So, the normal vector there is
  2 i 4 j 4 k

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TANGENT PLANES
Example 9

• Therefore, an equation of the tangent plane at
  (1, 1, 3) is 2(x 1) 4(y 1)
  4(z 3) 0or x 2y 2z
  3 0

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TANGENT PLANES

• The figure shows the self-intersecting surface in
  Example 9 and its tangent plane at (1, 1, 3).

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SURFACE AREA

• Now, we define the surface area of a general
  parametric surface given by Equation 1.

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SURFACE AREAS

• For simplicity, we start by considering a
  surface whose parameter domain D is a rectangle,
  and we divide it into subrectangles Rij.

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SURFACE AREAS

• Lets choose (ui, vj) to be the lower left
  corner of Rij.

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PATCH

• The part Sij of the surface S that corresponds to
  Rij is called a patch and has the point Pij with
  position vector r(ui, vj) as one of its corners.

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SURFACE AREAS

• Let ru ru(ui, vj) and rv
  rv(ui, vj) be the tangent vectors at Pij as
  given by Equations 5 and 4.

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SURFACE AREAS

• The figure shows how the two edges of the patch
  that meet at Pij can be approximated by vectors.

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SURFACE AREAS

• These vectors, in turn, can be approximated by
  the vectors ?u ru and ?v rv because partial
  derivatives can be approximated by difference
  quotients.
• So, we approximate Sij by the parallelogram
  determined by the vectors ?u ru and ?v rv.

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SURFACE AREAS

• This parallelogram is shown here.
• It lies in the tangent plane to S at Pij.

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SURFACE AREAS

• The area of this parallelogram is So, an
  approximation to the area of S is

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SURFACE AREAS

• Our intuition tells us that this approximation
  gets better as we increase the number of
  subrectangles.
• Also, we recognize the double sum as a Riemann
  sum for the double integral
• This motivates the following definition.

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SURFACE AREAS
Definition 6

• Suppose a smooth parametric surface S is
• Given by r(u, v) x(u, v) i y(u, v) j
  z(u, v) k
  (u, v) D
• Covered just once as (u, v) ranges throughout
  the parameter domain D.

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SURFACE AREAS
Definition 6

• Then, the surface area of S iswhere

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SURFACE AREAS
Example 10

• Find the surface area of a sphere of radius a.
• In Example 4, we found x a sin F cos ?, y
  a sin F sin ?, z a cos F where the
  parameter domain is D (F, ?) 0
  F p, 0 ? 2p)

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SURFACE AREAS
Example 10

• We first compute the cross product of the
  tangent vectors

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SURFACE AREAS
Example 10

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SURFACE AREAS
Example 10

• Thus,
• since sin F 0 for 0 F p.

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SURFACE AREAS
Example 10

• Hence, by Definition 6, the area of the sphere is

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SURFACE AREA OF THE GRAPH OF A FUNCTION

• Now, consider the special case of a surface S
  with equation z f(x, y), where (x, y) lies in D
  and f has continuous partial derivatives.
• Here, we take x and y as parameters.
• The parametric equations are x x
  y y z f(x, y)

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GRAPH OF A FUNCTION
Equation 7

• Thus,and

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GRAPH OF A FUNCTION
Equation 8

• Thus, we have

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GRAPH OF A FUNCTION
Formula 9

• Then, the surface area formula in Definition 6
  becomes

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GRAPH OF A FUNCTION
Example 11

• Find the area of the part of the paraboloid z
  x2 y2 that lies under the plane z 9.
• The plane intersects the paraboloid in the
  circle x2 y2 9, z 9

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GRAPH OF A FUNCTION
Example 11

• Therefore, the given surface lies above the disk
  D with center the origin and radius 3.

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GRAPH OF A FUNCTION
Example 11

• Using Formula 9, we have

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GRAPH OF A FUNCTION
Example 11

• Converting to polar coordinates, we obtain

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SURFACE AREA

• The question remains
• Is our definition of surface area (Definition 6)
  consistent with the surface area formula from
  single-variable calculus (Formula 4 in Section
  8.2)?

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SURFACE AREA

• We consider the surface S obtained by rotating
  the curve y f(x), a x b about the
  x-axis, where
• f(x) 0.
• f is continuous.

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SURFACE AREA

• From Equations 3, we know that parametric
  equations of S are
• x x y f(x) cos ? z f(x)
  sin ? a x b 0
  ? 2p

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SURFACE AREA

• To compute the surface area of S, we need the
  tangent vectors

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SURFACE AREA

• Thus,

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SURFACE AREA

• Hence,because f(x) 0.

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SURFACE AREA

• Thus, the area of S is

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SURFACE AREA

• This is precisely the formula that was used to
  define the area of a surface of revolution in
  single-variable calculus (Formula 4 in Section
  8.2).