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17 VECTOR CALCULUS
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So far, we have considered special types of surfaces:
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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Parametric Surfaces and their Areas
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) k is a vector-valued function defined on a region D in the uv-plane.
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INTRODUCTION So x, y, and z—the component functions of r—are 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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Each choice of u and v gives a point on S.
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. Fig , p. 1106
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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. Fig , p. 1106
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In Example 1, we placed no restrictions on the parameters u and v.
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 ≤ u ≤ π/ ≤ v ≤ 3 then x ≥ 0 z ≥ 0 0 ≤ y ≤ 3
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In that case, we get the quarter-cylinder with length 3.
PARAMETRIC SURFACES In that case, we get the quarter-cylinder with length 3. Fig , p. 1107
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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 S—one 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. Fig , p. 1107
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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. Fig , p. 1107
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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 curves—as we see in the following example.
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Use a computer algebra system to graph the surface
GRID CURVES Example 2 Use a computer algebra system to graph the surface r(u, v) = <(2 + sin v) cos u, (2 + sin v) sin u, u + cos v> 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 ≤ u ≤ 4π, 0 ≤ v ≤ 2π It has the appearance of a spiral tube. Fig , p. 1107
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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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If v is constant, then sin v and cos v are constant.
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 14.1
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So, the grid curves with v constant are the spiral 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. Fig , p. 1107
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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 13.2 Fig , p. 1108
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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 = <x, y, z> r0 = <x0, y0, z0> a = <a1, a2, a3> b = <b1, b2, b3> we can write the parametric equations of the plane through the point (x0, y0, z0) as: x = x0 + ua1 + vb y = y0 + ua2 + vb 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 ρ = a in spherical coordinates. So, let’s choose the angles Φ and θ in spherical coordinates as the parameters (Section 15.8).
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PARAMETRIC REPRESENTATIONS
Example 4 Then, putting ρ = a in the equations for conversion from spherical to rectangular coordinates (Equations 1 in Section 15.8), we obtain: x = a sin Φ cos θ y = a sin Φ sin θ z = a cos Φ as the parametric equations of the sphere.
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PARAMETRIC REPRESENTATIONS
Example 4 The corresponding vector equation is: r(Φ, θ) = a sin Φ cos θ i + a sin Φ sin θ j + a cos Φ k We have 0 ≤ Φ ≤ π and 0 ≤ θ ≤ 2π. So, the parameter domain is the rectangle D = [0, π] x [0, 2π]
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PARAMETRIC REPRESENTATIONS
Example 4 The grid curves with: Φ constant are the circles of constant latitude (including the equator). θ constant are the meridians (semicircles), which connect the north and south poles.
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APPLICATIONS—COMPUTER 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. Fig , p. 1108
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COMPUTER GRAPHICS Part of the sphere appears to be missing because of the rectangular grid system used by the computer. Fig , p. 1108
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COMPUTER GRAPHICS The much better picture here was produced by a computer using the parametric equations found in Example 4. Fig , p. 1108
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PARAMETRIC REPRESENTATIONS
Example 5 Find a parametric representation for the cylinder x2 + y2 = ≤ z ≤ 1 The cylinder has a simple representation r = 2 in cylindrical coordinates. So, we choose as parameters θ and z in cylindrical coordinates.
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PARAMETRIC REPRESENTATIONS
Example 5 Then the parametric equations of the cylinder are x = 2 cos θ y = 2 sin θ z = z where: 0 ≤ θ ≤ 2π 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 + 2y2 and 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 y—an 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. 7—Solution 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. 7—Solution 2 Another representation results from choosing as parameters the polar coordinates r and θ. A point (x, y, z) on the cone satisfies: x = r cos θ y = r sin θ
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PARAMETRIZATIONS E. g. 7—Solution 2 So, a vector equation for the cone is r(r, θ) = r cos θ i + r sin θ j + 2r k where: r ≥ 0 0 ≤ θ ≤ 2π
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In certain situations, though, Solution 2 might be preferable.
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: ≤ r ≤ ½ 0 ≤ θ ≤ 2π
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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, let’s 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 θ be the angle of rotation as shown. Fig , p. 1110
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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 θ z = f(x) sin θ Fig , p. 1110
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SURFACES OF REVOLUTION
Thus, we take x and θ as parameters and regard Equations 3 as parametric equations of S. The parameter domain is given by: a ≤ x ≤ b 0 ≤ θ ≤ 2π
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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 ≤ 2π, 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 θ z = sin x sin θ The parameter domain is: ≤ x ≤ 2π 0 ≤ θ ≤ 2π
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SURFACES OF REVOLUTION
Example 8 Using a computer to plot these equations and rotate the image, we obtain this graph. Fig , p. 1110
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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. Fig , p. 1110
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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. Fig , p. 1110
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Its tangent vector at P0 is:
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 = u y = v z = u + 2v at the point (1, 1, 3).
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We first compute the tangent vectors:
TANGENT PLANES Example 9 We first compute the tangent vectors:
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Thus, a normal vector to the tangent plane is:
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) = 0 or 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). Fig , p. 1111
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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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Let’s choose (ui*, vj*) to be the lower left corner of Rij.
SURFACE AREAS Let’s choose (ui*, vj*) to be the lower left corner of Rij. Fig , p. 1111
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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. Fig , p. 1111
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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. Fig a, p. 1112
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SURFACE AREAS These vectors, in turn, can be approximated by the vectors Δu ru* and Δv rv* because partial derivatives can be approximated by difference quotients. So, we approximate Sij by the parallelogram determined by the vectors Δu ru* and Δv rv*.
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This parallelogram is shown here.
SURFACE AREAS This parallelogram is shown here. It lies in the tangent plane to S at Pij. Fig b, p. 1112
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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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Suppose a smooth parametric surface S is:
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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Then, the surface area of S is where:
SURFACE AREAS Definition 6 Then, the surface area of S is where:
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Find the surface area of a sphere of radius a.
SURFACE AREAS Example 10 Find the surface area of a sphere of radius a. In Example 4, we found x = a sin Φ cos θ, y = a sin Φ sin θ, z = a cos Φ where the parameter domain is: D = {(Φ, θ) | 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π)
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We first compute the cross product of the tangent vectors:
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 Φ ≥ 0 for 0 ≤ Φ ≤ π.
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Hence, by Definition 6, the area of the sphere is:
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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Then, the surface area formula in Definition 6 becomes:
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. Fig , p. 1113
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GRAPH OF A FUNCTION Example 11 Using Formula 9, we have:
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Converting to polar coordinates, we obtain:
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 9.2)?
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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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From Equations 3, we know that parametric equations of S are:
SURFACE AREA From Equations 3, we know that parametric equations of S are: x = x y = f(x) cos θ z = f(x) sin θ a ≤ x ≤ b ≤ θ ≤ 2π
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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 9.2).
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