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Lecture 35 : Surface Area; Surface Integrals

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Lecture 35 : Surface Area; Surface Integrals 1 Lecture 35 : Surface Area; Surface Integrals In the previous lecture we de¯ned the surface area a(S) of the parametric surface S, de¯ned by r(u; v) on T , by the double integral RR a(S) = k ru £ rv k dudv: (1) T We will now drive a formula for the area of a surface de¯ned by the graph of a function. Area of a surface de¯ned by a graph: Suppose a surface S is given by z = f(x; y); (x; y) 2 T , thatp is, S is the graph of the function f(x; y). (For example, S is the unit hemisphere de¯ned by z = 1 ¡ x2 ¡ y2 where (x; y) lies in the circular region T : x2 +y2 · 1.) Then S can be considered as a parametric surface de¯ned by: r(x; y) = xi + yj + f(x; y)k; (x; y) 2 T: In this case the surface area becomes RR q 2 2 a(S) = 1 + fx + fy dxdy: (2) T q 2 2 because k ru £ rv k = k ¡fxi ¡ fyj + k k = 1 + fx + fy . Example 1: Let us ¯nd the area of the surface of the portion of the sphere x2 + y2 + z2 = 4a2 2 2 that lies inside thep cylinder x + y = 2ax. Note that the sphere can be considered as a union of 2 2 2 two graphs: z = § 4ap¡ x ¡ y . We will use the formula given in (2) to evaluate the surface area. Let z = f(x; y) = 4a2 ¡ x2 ¡ y2. Then q q ¡x ¡y 2 2 4a2 fx = p ; fy = p and 1 + f + f = 2 2 2 : 4a2¡x2¡y2 4a2¡x2¡y2 x y 4a ¡x ¡y Let T be the projection of the surface z = f(x; y) on the xy-plane (see Figure 1). Then, because of the symmetry, the surface area is ¼ q 2 2a cos θ RR 2 R R a(S) = 2 4a dxdy = 2 £ 2 p2ardrdθ : 4a2¡x2¡y2 4a2¡r2 T 0 0 Remark: Since 2 2 2 2 2 2 2 2 2 2 k ru £ rv k = k ru k k rv k sin θ = k ru k k rv k (1 ¡ cos θ) = k ru k k rv k ¡(ru ¢ rv) ; the formula given in (1) can be written as RR p a(S) = EG ¡ F 2dudv (3) T where E = ru ¢ ru;G = rv ¢ rv and F = ru ¢ rv. 2 Example 2: Let us compute the area of the torus x = (a + b cos Á) cos θ; y = (a + b cos Á) sin θ; z = b sin Á where 0 · θ · 2¼; 0 · Á · 2¼, and a and b are constants such that 0 < b < a. Since the surface is given in the parametric form with the parameters θ and Á, we can either use the formula given in (1) or (3) and ¯nd the surface area. We do not have to know how the surface looks like. However the surface is given in Figure 2 for understanding. Note that rθ = ¡(a + b cos Á) sin θi + (a + b cos Á) cos θj + 0k; rv = ¡b sin Á cos θi ¡ b sin Á sin θj + b cos Ák: p 2 2 2 This implies that E = ru ¢ ru = (a + b cos Á) ;F = 0, G = b and hence EG ¡ F = b(a + b cos '). Therefore, by (3), the surface area is RR R R 2¼ 2¼ 2 a(S) = b(a + b cos ')dθdÁ = 0 0 b(a + b cos ')dθdÁ = 4¼ ab: T Note that this problem can also be solved using the Pappus theorem : a(S) = 2¼½L = 2¼ ¢ a ¢ 2¼b. Surface Integrals: We will de¯ne the concept of integrals, called surface integrals, to the scalar functions de¯ned on parametric surfaces. Surface integrals are used to de¯ne center of mass and moment of inertia of surfaces, and the surface integrals occur in several applications. We will not get in to the applications of the surface integrals in this course. We will de¯ne the surface integrals and see how to evaluate them. Let S be a parametric surface de¯ned by r(u; v); (u; v) 2 T . Suppose rRRu and rv are continuous. Let g : S ! R be bounded. The surface integral of g over S, denoted by gdσ, is de¯ned by S RR RR RR p 2 g dσ = g(r(u; v)) k ru £ rv k dudv = g(r(u; v)) EG ¡ F dudv (4) S T T provided the RHS double integral exists. If S is de¯ned by z = f(x; y), then RR RR q 2 2 g dσ = g[x; y; f(x; y)] 1 + fx + fy dxdy: (5) S T where T is the projection of the surface S over the xy-plane. RR Example 3: Let S be the hemispherical surface z = (a2¡x2¡y2)1=2. Let us evaluate dσ : [x2+y2+(z+a)2]1=2 S We ¯rst parameterize the surface S as follows: S := r(θ; Á) = (a sin Á cos θ; a sin Á sin θ; a cos Á); 0 · θ · 2¼; 0 · Á · ¼: p 2 2 2 2 2 1=2 Á Simple calculation shows that EG ¡ F = a sin Á and [x +y +(z+a) ] = 2a cos 2 . Therefore, by equation (4), the surface integral is RR R R 2 dσ = 2¼ ¼=2 a sin Á dφdθ: [x2+y2+(z+a)2]1=2 0 0 2a cos Á S 2 RR Example 4: Let us evaluate the surface integral S g dσ where g(x; y; z) = x+y+z and the surface S is described by z = 2x+3y; x ¸ 0; y ¸ 0 and x+y · 2. We use the formula given in (5) to evaluate the surface integral. Note that the projection T of the surface is f(x; y): x ¸ 0; y ¸ 0; x + y · 2g. The surface integral is RR RR q R R p 2 2 2 2¡y g dσ = (x + y + z) 1 + fx + fy dxdy = 0 0 (x + y + 2x + 3y) 14 dxdy: S T Remark: Under certain general conditions (we deal with surfaces satisfying such conditions) the value of the surface integral is independent of the representation..
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