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17. VECTOR CALCULUS. VECTOR CALCULUS. 17.7 Surface Integrals. In this section, we will learn about: Integration of different types of surfaces. SURFACE INTEGRALS.

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## VECTOR CALCULUS

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**17**VECTOR CALCULUS**VECTOR CALCULUS**17.7 Surface Integrals • In this section, we will learn about: • Integration of different types of surfaces.**SURFACE INTEGRALS**• The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length.**SURFACE INTEGRALS**• Suppose f is a function of three variables whose domain includes a surface S. • We will define the surface integral of f over S such that, in the case where f(x, y, z) = 1, the value of the surface integral is equal to the surface area of S.**SURFACE INTEGRALS**• We start with parametric surfaces. • Then, we deal with the special case where S is the graph of a function of two variables.**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 Rijwith dimensions ∆u and ∆v. Fig. 17.7.1, p. 1117**PARAMETRIC SURFACES**• Then, the surface Sis divided into corresponding patches Sij. Fig. 17.7.1, p. 1117**PARAMETRIC SURFACES**• We evaluate f at a point Pij* in each patch, multiply by the area ∆Sijof the patch, and form the Riemann sum Fig. 17.7.1, p. 1117**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:**SURFACE INTEGRALS**• Notice the analogy with: • The definition of a line integral (Definition 2 in Section 17.2) • The definition of a double integral (Definition 5 in Section 16.1)**SURFACE INTEGRALS**• 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 17.6, we made the approximation ∆Sij ≈ |rux rv| ∆u ∆v • where: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:**SURFACE INTEGRALS**• This should be compared with the formula for a line integral: • Observe also that:**SURFACE INTEGRALS**• Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D. • When using this formula, remember that f(r(u, v) is evaluated by writing x =x(u, v), y =y(u, v), z =z(u, v) in the formula for f(x, y, z)**SURFACE INTEGRALS**Example 1 • Compute the surface integral , where S is the unit sphere x2 + y2 + z2 = 1.**SURFACE INTEGRALS**Example 1 • As in Example 4 in Section 17.6, we use the parametric representation x = sin Φ cos θ, y = sin Φ sin θ, z = cos Φ0 ≤ Φ ≤ π,0≤ θ ≤ 2π • That is, r(Φ,θ) = sin Φcos θi + sin Φ sin θj + cos Φk**SURFACE INTEGRALS**Example 1 • As in Example 10 in Section 17.6, we can compute that: • |rΦxrθ| = sin Φ**SURFACE INTEGRALS**Example 1 • Therefore, by Formula 2,**SURFACE INTEGRALS**Example 1**APPLICATIONS**• Surface integrals have applications similar to those for the integrals we have previously considered.**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).**MASS**• Then, the total mass of the sheet is:**CENTER OF MASS**• The center of mass is: where**MOMENTS OF INERTIA**• Moments of inertia can also be defined as before. • See Exercise 39.**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:**GRAPHS**Equation 3 • Thus, • and**GRAPHS**Formula 4 • Therefore, in this case, Formula 2 becomes:**GRAPHS**• Similar formulas apply when it is more convenient to project S onto the yz-plane or xy-plane.**GRAPHS**• For instance, if S is a surface with equation y =h(x, z) and D is its projection on the xz-plane, then**GRAPHS**Example 2 • Evaluate where S is the surface z =x +y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 Fig. 17.7.2, p. 1119**GRAPHS**Example 2 • So, Formula 4 gives:**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:**GRAPHS**Example 3 • Evaluate , where S is the surface whose: • Sides S1 are given by the cylinder x2 + y2 = 1. • Bottom S2 is the disk x2 + y2≤ 1 in the plane z = 0. • Top S3 is the part of the plane z = 1 + x that lies above S2.**GRAPHS**Example 3 • The surface S is shown. • We have changed the usual position of the axes to get a better look at S. Fig. 17.7.3, p. 1120**GRAPHS**Example 3 • For S1, we use θand z as parameters (Example 5 in Section 17.6) and write its parametric equations as: • x = cos θ y = sin θ z =zwhere: • 0 ≤θ≤ 2π • 0 ≤z ≤ 1 + x = 1 + cos θ Fig. 17.7.3, p. 1120**GRAPHS**Example 3 • Therefore, • and**GRAPHS**Example 3 • Thus, the surface integral over S1 is:**GRAPHS**Example 3 • Since S2 lies in the plane z = 0, we have: Fig. 17.7.3, p. 1120**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. Fig. 17.7.3, p. 1120**GRAPHS**Example 3**GRAPHS**Example 3 • Therefore,**ORIENTED SURFACES**• To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown. • It is named after the German geometer August Möbius (1790–1868). Fig. 17.7.4, p. 1121**MOBIUS STRIP**• You can construct one for yourself by: • Taking a long rectangular strip of paper. • Giving it a half-twist. • Taping the short edges together. Fig. 17.7.5, p. 1121**MOBIUS STRIP**• If an ant were to crawl along the Möbius strip starting at a point P, it would end up on the “other side” of the strip—that is, with its upper side pointing in the opposite direction. Fig. 17.7.4, p. 1121**MOBIUS STRIP**• Then, if it continued to crawl in the same direction, it would end up back at the same point P without ever having crossed an edge. • If you have constructed a Möbius strip, try drawing a pencil line down the middle. Fig. 17.7.4, p. 1121**MOBIUS STRIP**• Therefore, a Möbius strip really has only one side. • You can graph the Möbius strip using the parametric equations in Exercise 32 in Section 17.6. Fig. 17.7.4, p. 1121**ORIENTED SURFACES**• From now on, we consider only orientable (two-sided) surfaces.**ORIENTED SURFACES**• We start with a surface S that has a tangent plane at every point (x, y, z) on S (except at any boundary point). • There are two unit normal vectors n1 and n2 = –n1at (x, y, z). Fig. 17.7.6, p. 1122

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