VECTOR CALCULUS
In this section, we will learn about: Integration of different types of surfaces. PARAMETRIC SURFACES Suppose a surface S has a vector equation
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
(u, v) D PARAMETRIC SURFACES •We first assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions ∆u and ∆v. •Then, the surface S is divided into corresponding patches Sij.
•We evaluate f at a point Pij* in each patch, multiply by the area ∆Sij of the patch, and form the Riemann sum
mn * f() Pij S ij ij11 SURFACE INTEGRAL Equation 1 Then, we take the limit as the number of patches increases and define the surface integral of f over the surface S as: mn * f( x , y , z ) dS lim f ( Pij ) S ij mn, S ij11 . Analogues to: The definition of a line integral (Definition 2 in Section 16.2);The definition of a double integral (Definition 5 in Section 15.1) . To evaluate the surface integral in Equation 1, we
approximate the patch area ∆Sij by the area of an approximating parallelogram in the tangent plane.
SURFACE INTEGRALS In our discussion of surface area in Section 16.6, we made the approximation
∆Sij ≈ |ru x rv| ∆u ∆v where: x y z x y z r i j k r i j k uvu u u v v v are the tangent vectors at a corner of Sij. SURFACE INTEGRALS Formula 2
If the components are continuous and ru and rv are nonzero and nonparallel in the interior of D, it can be shown from Definition 1—even when D is not a rectangle—that:
fxyzdS(,,) f ((,))|r uv r r | dA uv SD SURFACE INTEGRALS This should be compared with the formula for a line integral: b fxyzds(,,) f (())|'()|rr t tdt Ca
Observe also that:
1dS |rr | dA A ( S ) uv SD SURFACE INTEGRALS Example 1 Compute the surface integral x2 dS , where S is the unit sphere S x2 + y2 + z2 = 1. SURFACE INTEGRALS Example 1 As in Example 4 in Section 16.6, we use the parametric representation
x = sin Φ cos θ, y = sin Φ sin θ, z = cos Φ 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π . That is, r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
. we can compute: |rΦ x rθ| = sin Φ
SURFACE INTEGRALS Example 1 Therefore, by Formula 2, x2 dS S (sin cos )2 |rr |dA D 2 (sin22 cos sin dd 00
2 cos23dd sin 00 2 1 (1 cos 2 )dd (sin sin cos2 ) 002 2 1 1sin 2 cos 1 cos3 2 20 3 0 4 3 APPLICATIONS For example, suppose a thin sheet (say, of aluminum foil) has:
. The shape of a surface S.
. The density (mass per unit area) at the point (x, y, z) as ρ(x, y, z). CENTER OF MASS Then, the total mass of the sheet is: m (,,) x y z dS S The center of mass is: x,, y z 1 x x(,,) x y z dS m S 1 where y y(,,) x y z dS m S 1 z z(,,) x y z dS m S GRAPHS Any surface S with equation z = g(x, y) can be regarded as a parametric surface with parametric equations x = x y = y z = g(x, y)
. So, we have: gg rxy i k r j k xy GRAPHS Equation 3 •Thus, gg r r i j k xy xx
2 2 zz and |rrxy | 1 xy
•Formula 2 becomes: f(,,) x y z dS S
2 2 zz f( x , y , g ( x , y )) 1 dA D xy GRAPHS Example 2 Evaluate y dS where S is the surface S z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
z . 1 x and z 2y y GRAPHS Example 2 So, Formula 4 gives:
2 2 zz y dS y1 dA SDxy 12 y1 1 4 y2 dy dx 00 12 dx2 y 1 2 y2 dy 00 2 13 2 12 2 3/ 2 2 43 (1 2y ) 0 3 GRAPHS If S is a piecewise-smooth surface—a finite union of smooth surfaces S1, S2, . . . , Sn that intersect only along their boundaries—then the surface integral of f over S is defined by: f(,,) x y z dS S fxyzdS(,,)(,,) fxyzdS SS1 n GRAPHS Example 3 Evaluate z dS , where S is the surfaceS whose:
2 2 . Sides S1 are given by the cylinder x + y = 1.
2 2 . Bottom S2 is the disk x + y ≤ 1 in the plane z = 0.
. Top S3 is the part of the plane z = 1 + x that lies above S2. GRAPHS Example 3
For S1, we use θ and z as parameters (Example 5 in Section 16.6) and write its parametric equations as: x = cos θ y = sin θ z = z where: . 0 ≤ θ ≤ 2π . 0 ≤ z ≤ 1 + x = 1 + cos θ GRAPHS Example 3 Therefore, i j k
r rz sin cos 0 cos i sin j 0 0 1 and
22 |rr z | cos sin 1 GRAPHS Example 3
Thus, the surface integral over S1 is: z dS z||rr dA z SD1 2 1 cos z dz d 00 2 1 (1 cos )2 d 0 2 2 111 2cos (1 cos 2 ) d 220 3 113 2sin sin 2 2 2 2 4 0 2 GRAPHS Example 3
Since S2 lies in the plane z = 0, we have: z dS S2 0 dS S2 0 GRAPHS Example 3
S3 lies above the unit disk D and is part of the plane z = 1 + x.
. So, taking g(x, y) = 1 + x in Formula 4 and converting to polar coordinates, we have the following result. GRAPHS Example 3 2 2 zz z dS(1 x ) 1 dA xy SD3 21 (1 r cos ) 1 1 0 r dr d 00 21 2 (r r2 cos ) dr d 00 2 2 11 cosd 0 23 2 sin 22 230 GRAPHS Example 3 Therefore,
z dS z dS z dS z dS SSSS1 2 3 3 02 2 3 2 2 SURFACE INTEGRALS OF VECTOR FIELDS Suppose that S is an oriented surface with unit normal vector n. Then, imagine a fluid with density ρ(x, y, z) and velocity field v(x, y, z) flowing through S.
. Think of S as an imaginary surface that doesn’t impede the fluid flow—like a fishing net across a stream. Then, the rate of flow (mass per unit time) per unit area is ρv.
SURFACE INTEGRALS OF VECTOR FIELDS
If we divide S into small patches Sij , then Sij is nearly planar. SURFACE INTEGRALS OF VECTOR FIELDS So, we can approximate the mass of fluid crossing Sij in the direction of the normal n per unit time by the quantity
(ρv · n)A(Sij) where ρ, v, and n are evaluated at some point on Sij. . Recall that the component of the vector ρv in the direction of the unit vector n is ρv · n. VECTOR FIELDS Equation 7 Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function ρv · n over S: vn dS S (,,)(,,)(,,)xyzvn xyz xyzdS S . This is interpreted physically as the rate of flow through S. VECTOR FIELDS If we write F = ρv, then F is also a vector field on R3. Then, the integral in Equation 7 becomes: Fn dS S A surface integral of this form occurs frequently in physics—even when F is not ρv. It is called the surface integral (or flux integral) of F over S.
FLUX INTEGRAL Definition 8 If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is: Fd S F n dS SS
. This integral is also called the flux of F across S. FLUX INTEGRAL If S is given by a vector function r(u, v), then n is given by Equation 6. . Then, from Definition 8 and Equation 2, we have (D is the parameters’ domain): rr FSFd uv dS SSrruv rruv F( r (u , v )) ruv r dA rr D uv . So, Fd S F () r r dA uv SD
FLUX INTEGRALS Example 4 •Find the flux of the vector field F(x, y, z) = z i + y j + x k across the unit sphere :x2 + y2 + z2 = 1 •Using the parametric representation: r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
0 ≤ Φ ≤ π 0 ≤ θ ≤ 2π F(r(Φ, θ)) = cos Φ i + sin Φ sin θ j + sin Φ cos θ k FLUX INTEGRALS Example 4 From Example 10 in Section 16.6,
2 2 rΦ x rθ = sin Φ cos θ i + sin Φ sin θ j + sin Φ cos Φ k 2 Therefore, F(r(Φ, θ)) · (rΦ x rθ) = cos Φ sin Φ cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ Then, by Formula 9, the flux is: FSd S F () r r dA D 2 (2sin2 cos cos sin 3 sin 2 ) dd 00 FLUX INTEGRALS Example 4 2 2 sin2 cos dd cos 00 2 sin32dd sin 00 2 0 sin32dd sin 00 4 3
. This is by the same calculation as in Example 1. FLUX INTEGRALS The figure shows the vector field F in Example 4 at points on the unit sphere. VECTOR FIELDS If, for instance, the vector field in Example 4 is a velocity field describing the flow of a fluid with density 1, then the answer, 4π/3, represents:
. The rate of flow through the unit sphere in units of mass per unit time. VECTOR FIELDS In the case of a surface S given by a graph z = g(x, y), we can think of x and y as parameters and use Equation 3 to write:
gg F()() rxy r PQR i j k i j k xy VECTOR FIELDS Formula 10 Thus, Formula 9 becomes:
gg FSd P Q R dA SDxy
. This formula assumes the upward orientation of S. . For a downward orientation, we multiply by –1. VECTOR FIELDS Example 5 Evaluate FSd S where:
. F(x, y, z) = y i + x j + z k . S is the boundary of the solid region E enclosed by the paraboloid z = 1 – x2 – y2 and the plane z = 0. VECTOR FIELDS Example 5 S consists of:
. A parabolic top surface S1.
. A circular bottom surface S2. VECTOR FIELDS Example 5 Since S is a closed surface, we use the convention of positive (outward) orientation.
. This means that S1 is oriented upward.
. So, we can use Equation 10 with D being
the projection of S1 on the xy-plane, namely, the disk x2 + y2 ≤ 1. VECTOR FIELDS Example 5
On S1, P(x, y, z) = y Q(x, y, z) = x R(x, y, z) = z = 1 – x2 – y2 Also, gg 22xy xy VECTOR FIELDS Example 5 So, we have: FSd S1 gg P Q R dA D xy [ y ( 2 x ) x ( 2 y ) 1 x22 y ] dA D (1 4xy x22 y ) dA D VECTOR FIELDS Example 5 21 (1 4r22 cos sin r ) r dr d 00 21 (r r33 4 r cos sin ) dr d 00 2 (1 cos sin ) d 0 4 1 4 (2 ) 0 2 VECTOR FIELDS Example 5
The disk S2 is oriented downward. So, its unit normal vector is n = –k and we have:
Fd S F ()() k dS z dA SSD22
00dA D since z = 0 on S2. VECTOR FIELDS Example 5 Finally, we compute, by definition, FSd as the sum of the surface integrals S of F over the pieces S1 and S2:
FSFSFSddd SSS12 0 22 APPLICATIONS Although we motivated the surface integral of a vector field using the example of fluid flow, this concept also arises in other physical situations. ELECTRIC FLUX For instance, if E is an electric field (Example 5 in Section 16.1), the surface integral ESd S is called the electric flux of E through the surface S. GAUSS’S LAW Equation 11 One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is:
Qd ES 0 S where ε0 is a constant (called the permittivity of free space) that depends on the units used. –12 2 2 . In the SI system, ε0 ≈ 8.8542 x 10 C /N · m GAUSS’S LAW Thus, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is:
Q = 4πε0/3 HEAT FLOW Another application occurs in the study of heat flow.
. Suppose the temperature at a point (x, y, z) in a body is u(x, y, z). HEAT FLOW •Then, the heat flow is defined as the vector field F = –K ∇u where K is an experimentally determined constant called the conductivity of the substance. •Then, the rate of heat flow across the surface S in the body is given by the surface integral FSSd K u d SS