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# VECTOR CALCULUS - PowerPoint PPT Presentation

16. VECTOR CALCULUS. 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. VECTOR CALCULUS.

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VECTOR CALCULUS

• So far, we have considered special types of surfaces:

• Cylinders

• Graphs of functions of two variables

• Level surfaces of functions of three variables

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

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

16.6

Parametric Surfaces and their Areas

• In this section, we will learn about:

• Various types of parametric surfaces

• and computing their areas using vector functions.

• 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.

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.

• So x, y, and z—the component functions of r—are functions of the two variables u and v with domain D.

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 equationsof S.

• Each choice of u and v gives a point on S.

• By making all choices, we get all of S.

• 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.

Example 1

• Identify and sketch the surface with vector equation r(u, v) = 2 cos ui + vj + 2 sin uk

• The parametric equations for this surface are: x = 2 cos uy = vz = 2 sin u

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.

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.

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

• So,we obtained the entire cylinder.

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

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

• 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.

• 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.

• 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.

• In Example 1, for instance, the grid curves obtained by:

• Letting u be constant are horizontal lines.

• Letting v be constant are circles.

• 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.

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?

Example 2

• We graph the portion of the surface with parameter domain 0 ≤ u ≤ 4π, 0 ≤v≤ 2π

• It has the appearance of a spiral tube.

Example 2

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

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

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.

Example 2

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

• 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.

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.

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

Example 3

• The figure illustrates how this works, by means of the Parallelogram Law, for the case where u and v are positive.

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 + vbwhere u and v are real numbers.

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 + vb1y = y0 + ua2 + vb2z = z0 + ua3 + vb3

Example 4

• Find a parametric representation of the spherex2 + y2 + z2 = a2

• The sphere has a simple representation ρ = ain spherical coordinates.

• So, let’s choose the angles Φand θin spherical coordinates as the parameters (Section 15.8).

Example 4

• Then, putting ρ = ain the equations for conversion from spherical to rectangular coordinates (Equations 1 in Section 15.8), we obtain:

• x = a sin Φ cos θy = asin Φ sin θ

• z = a cos Φ

• as the parametric equations of the sphere.

Example 4

• The corresponding vector equation is: r(Φ, θ) = a sin Φ cos θi + a sin Φ sin θj + a cos Φk

• We have 0 ≤ Φ ≤ πand 0 ≤ θ≤ 2π.

• So,the parameter domain is the rectangle D = [0, π] x [0, 2π]

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.

• One of the uses of parametric surfaces is in 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.

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

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

Example 5

• Find a parametric representation for the cylinderx2 + y2 = 4 0 ≤ z≤ 1

• The cylinder has a simple representation r = 2 in cylindrical coordinates.

• So, we choose as parameters θand zin cylindrical coordinates.

Example 5

• Then the parametric equations of the cylinder are x = 2 cos θy = 2 sin θz = z

• where:

• 0 ≤ θ≤ 2π

• 0 ≤z≤ 1

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 = xy = yz = x2 + 2y2and the vector equation isr(x, y) = xi + yj + (x2 + 2y2) k

• 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 = xy = yz = f(x, y)

• Parametric representations (also called parametrizations) of surfaces are not unique.

• The next example shows two ways to parametrize a cone.

Example 7

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

E. g. 7—Solution 1

• One possible representation is obtained by choosing x and y as parameters:x = xy = y

• So, the vector equation is:

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 θy = r sin θ

E. g. 7—Solution 2

• So, a vector equation for the cone is r(r, θ) = r cos θi + r sin θj + 2rkwhere:

• r≥ 0

• 0 ≤ θ≤ 2π

• For some purposes, the parametric representations in Solutions 1 and 2 are equally good.

• In certain situations, though, Solution 2 might be preferable.

• 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 ≤θ≤ 2π

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

• For instance, let’s consider the surface S obtained by rotating the curve y = f(x) a≤ x ≤ babout the x-axis, where f(x) ≥ 0.

• Let θbe the angle of rotation as shown.

Equations 3

• If (x, y, z) is a point on S, then x = xy = f(x) cos θz = f(x) sin θ

• Thus, we take x and θas parameters and regard Equations 3 as parametric equations of S.

• The parameter domain is given by: a≤ x ≤ b 0 ≤ θ≤ 2π

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.

Example 8

• From Equations 3,

• The parametric equations are: x = xy = sin x cos θz = sin x sin θ

• The parameter domain is: 0 ≤ x ≤ 2π 0 ≤θ≤ 2π

Example 8

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

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

• See Exercise 30.

• 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) kat a point P0 with position vector r(u0, v0).

• Keeping u constant by putting u = u0, r(u0, v) becomes a vector function of the single parameter v and defines a grid curve C1lying on S.

Equation 4

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

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

Equation 5

• Its tangent vector at P0 is:

• 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.

Example 9

• Find the tangent plane to the surface with parametric equations x = u2y = v2z = u + 2vat the point (1, 1, 3).

Example 9

• We first compute the tangent vectors:

Example 9

• Thus, a normal vector to the tangent plane is:

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

Example 9

• Therefore, an equation of the tangent plane at (1, 1, 3) is: –2(x – 1) – 4(y – 1) + 4(z – 3) = 0orx + 2y – 2z + 3 = 0

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

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

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

• Let’s choose (ui*, vj*) to be the lower left corner of Rij.

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

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

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

• These vectors, in turn, can be approximated by the vectors Δuru* and Δvrv* because partial derivatives can be approximated by difference quotients.

• So, we approximate Sij by the parallelogram determined by the vectors Δuru* and Δvrv*.

• This parallelogram is shown here.

• It lies in the tangent plane to S at Pij.

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

• 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.

Definition 6

• Suppose a smooth parametric surface S is:

• Given byr(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.

Definition 6

• Then, the surface areaof S iswhere:

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 Φwhere the parameter domain is:D = {(Φ, θ) | 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π)

Example 10

• We first compute the cross product of the tangent vectors:

Example 10

Example 10

• Thus,

• since sin Φ ≥ 0 for 0 ≤ Φ≤π.

Example 10

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

• Now, consider the special case of a surface S with equation z = f(x, y), where (x, y) lies in Dand f has continuous partial derivatives.

• Here, we take x and y as parameters.

• The parametric equations are:x = xy = yz = f(x, y)

Equation 7

• Thus,and

Equation 8

• Thus, we have:

Formula 9

• Then, the surface area formula in Definition 6 becomes:

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

Example 11

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

Example 11

• Using Formula 9, we have:

Example 11

• Converting to polar coordinates, we obtain:

• 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)?

• 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.

• From Equations 3, we know that parametric equations of S are:

• x = xy = f(x) cos θz = f(x) sin θa≤ x ≤ b 0 ≤ θ≤ 2π

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

• Thus,

• Hence,because f(x) ≥ 0.

• Thus, the area of S is:

• 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).