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# SurfaceIntgrls.pdf

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School
Department
Mathematics
Course
MATH209
Professor
Dragos Hrimiuc
Semester
Fall

Description
Surface Area of Parametric Surfaces. De▯nition of Surface area. Suppose that a smooth parametric curve S is given by ~ ~ ~ 2 r(u;v) = x(u;v) i + y(u;v) j + z(u;v) k; (u;v) in a domain D in R ; and S is covered once as (u;v) ranges over D. Then the surface area of S is ZZ ▯ ▯ A(S) = ▯ru▯ ~rv▯ dA: (1) D Example (from class): Evaluate the surface area of the upper part of the sphere of radius b which is inside the cylinder x + y = a (0 < a < b). Solution. First ▯nd a parametrization for the surface S. To do that, use spherical coordinates: ▯1 ▯ a x(▯;▯) = bsin▯cos▯; y(▯;▯) = bsin▯sin▯; z(▯;▯) = bcos▯; 0 ▯ ▯ ▯ sin ; 0 ▯ ▯ ▯ 2▯; b or in vector form: ▯ ▯ ~ ~ ~ ▯1 a r(▯;▯) = bsin▯cos▯ i + bsin▯sin▯ j + bcos▯ k; 0 ▯ ▯ ▯ sin b ; 0 ▯ ▯ ▯ 2▯: Find r▯and ~ r▯: @x ~ @y ~ @z ~ ~ ~ ~ r▯= i + j + k = bcos▯cos▯ i + bcos▯sin▯ j ▯ bsin▯ k; @▯ @▯ @▯ @x @y @z r▯= i + j + k = ▯bsin▯sin▯ i + bsin▯cos▯ j ▯ 0 k;~ @▯ @▯ @▯ Next, compute the cross product: ▯ ▯ ▯ ~ ~ ~ ▯ ▯ i j k ▯ 2 2 2 2 2 r▯▯ ~r▯= ▯ bcos▯cos▯ bcos▯sin▯ ▯bsin▯ ▯= b sin ▯cos▯ i + b sin ▯sin▯ j + b sin▯cos▯ k: ~ ▯ ▯bsin▯sin▯ bsin▯cos▯ 0 ▯ Next, compute the length of the cross product: ▯ ▯ q ▯ ▯2 ▯ ▯2 q ▯ru▯ ~rv▯= b sin ▯cos▯ + b sin ▯sin▯ + (b sin▯cos▯) = b 2 sin ▯cos ▯ + sin ▯sin ▯ + cos ▯sin ▯ 2 q q 2 4 2 2 2 2 = b sin ▯ + cos ▯sin ▯ = b sin ▯ = b sin▯; p where in the last equality sin ▯ = +sin▯ because 0 ▯ ▯ ▯ ▯ and so sin▯ is positive. Altogether, using formula (1), ZZ Z 2▯ Z sin1( ) Z sin1(a) h i ▯1 a ▯ ▯ b 2 2 b 2 sin (b) A(S) = r▯▯ ~r▯ dA = b sin▯ d▯d▯ = 2▯b sin▯ d▯ = 2▯b ▯ cos▯ ▯=0 ▯=0 ▯=0 ▯=0 D " p # h ▯ ▯a▯▯ i b ▯ a 2 = 2▯b 2 ▯cos sin ▯1 + 1 = 2▯b 2 1 ▯ : b b Example (not from class): Find the area of the surface S which is given by r(u;v) = 1 v i + u j + uv k; 0 ▯ u ▯ 2; 0 ▯ v ▯ 1: 2 Solution. Now the parametrization of the surface is already given. Thus, one can move directly to computing ruand ~rv: ru= 0 i + 2u j + v k; rv= v i + 0 j + u k: Next, compute the cross product: ▯ ▯ ▯~ ~ ~ ▯ ▯ i j k ▯ 2 2 ru▯ ~rv= ▯0 2u v ▯= 2u i + v j ▯ 2uv k:~ ▯v 0 u ▯ Next, compute the length of the cross product: ▯ ▯ q p q ~ru▯ rv ▯= (2u ) + (v ) + (▯2uv) = 2 4u + v + 4u v = 2 (2u + v ) = 2u + v : 2 Now, using formula (1) ZZ Z 2 Z 1 Z 2 ▯ 3▯1 Z 2 ▯ ▯ ▯ ▯ ▯ 2 2▯ 2 v 2 1 A(S) = ru▯ rv dA = 2u + v dvdu = 2u v + 3 du = 2u + 3 du D u=0 v=0 u=0 v=0 u=0 ▯ ▯ 2u 3 u 2 16 2 = + = + = 6: 3 3 u=0 3 3 Surface Integrals of Scalar Functions. De▯nition of Surface Integrals. Let f be a (scalar) function of three variables de▯ned on a surface S, and S is 2 given by r(u;v), where (u;v) is in a domain D in R . Then the surface integral of f over S is ZZ ZZ ▯ ▯▯ ▯ f(x;y;z)dS = f r(u;v) r~u▯ r~v▯dA: (2) S D ZZ ▯ Example (from class): Evaluate x + y + z) dA, where S is given by the parametric equations x(u;v) = u + v, S y(u;v) = u ▯ v, z(u;v) = 2u + v + 1, 0 ▯ u ▯ 2, 0 ▯ v ▯ ▯. ▯ Solution. In order to use formula (2), one needs to know ru▯ ~rv. Thus, start by ▯nding ruand ~rv: @x @y @z ru= i + j + k = 1 ▯ i + 1 ▯ j + 2 ▯ k = i + j + 2k: @u @u @u @x @y @z rv= i + j + k = 1 ▯ i + (▯1) ▯ j + 1 ▯ k = i ▯ j + k: @v @v @v Thus, ▯ ▯ ▯ i j ~k ▯ ▯ ▯ ▯ ▯ ▯ ▯ ru▯ ~rv= ▯ 1 1 2 ▯= 1 ▯ 1 ▯ 2 ▯ (▯1) i ▯ 1 ▯ 1 ▯ 2 ▯ 1 j + 1 ▯ (▯1) ▯ 1 ▯ 1 k = 3i + j ▯ 2k: ▯ ▯ 1 ▯1 1 ▯ ▯ p 2 2 2 p =) ~ ru▯ ~rv = 3 + 1 + (▯2) = 14: Also, ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ f r(u;v) = x(u;v) + y(u;v) + z(u;v) = u + v + u ▯ v + 2u + v + 1 = 4u + v + 1: Altogether, using formula (2), ZZ ZZ Z 2 Z 1 Z 2 ▯ ▯1 ▯ ▯ ▯ ▯ ▯ ▯ ▯p p v2 x + y + z dS = f r(u;v) ru▯ ~rv dA = 4u + v + 1 14 dvdu = 14 4uv + + v du u=0 v=0 u=0 2 v=0 S Z ▯ ▯D ▯ ▯ p 2 3 p 3u 2 p = 14 4u + du = 14 2u + = 11 14: u=0 2 2 u=0 ZZ ▯ ▯ Example (not from class): Evaluate x + yz2 dS where S is the part of the cylinder x +y = 4 that lies between S the planes z = 0 and z = 3 in the ▯rst octant. Solution. First, need to ▯nd a parametrization for the surface S. Since S is a part of a cylinder, it is natural to use cylindrical coordinates: r(u;v) = 2cosu i + 2sinu j + v k; 0 ▯ u ▯ ▯ ; 0 ▯ v ▯ 3: 2 Note that the restriction on u follows from the assumption that we take the part of the surface which lies in the ▯rst octant, i.e., we must have x(u;v) ▯ 0 and y(u;v) ▯ 0. Now that the parametrization is given, the next step is to rud ~ and rv: ru= ▯2sinu i + 2cos j + 0 k; rv= 0 i + 0j + 1 k: Compute the cross product: ▯ ~ ~ ~ ▯ ▯ i j k ▯ ▯ ▯ p ru▯ ~rv= ▯▯2sinu 2cosu 0 ▯= 2cosu i + 2sinu j =) ~▯ru▯ r v▯= (2cosu) + (2sinu) = 2: ▯ 0 0 1 ▯ ▯ ▯ Also, to use formula (2), need to know fr(u;v) : ▯ ▯ ▯ ▯ f r(u;v) = x(u;v) + y(u;v) ▯ z(u;v) 2= 2cosu + 2sinu ▯ v : Finally, use formula (2): ZZ ▯ ▯ ZZ ▯ ▯▯ ▯ Z 2 Z 3 ▯ ▯ x + yz2 dS = f r(u;v) ▯r ▯ r ▯dA = 2cosu + 2sinu ▯ v ▯ 2 dvdu u v S D u=0 v=0 Z ▯ ▯ ▯3 Z ▯ ▯ 2 v3 2 ▯ ▯ ▯ ▯▯2 = 4 v cosu + sinu du = 4 3cosu + 9sinu du = 4 3sinu ▯ 9cosu ▯ = 4(3 + 9) = 48: u=0 3 v=0 u=0 u=0 Special case: S is a graph of a function. In the special case when z = g(x;y), one can write the surface as ~ ~ ~ r(x;y) = x i + y j + g(x;y) k; and then s ▯ ▯2 ▯ ▯2 ▯ ▯ @g @g ~x ▯ ry = 1 + + : @x @y Therefore, using formula (2), ZZ ZZ s ▯ ▯ ▯ ▯ ▯ ▯ @g 2 @g 2 f(x;y;z) dS = f x;y;g(x;y) 1 + + dA: (3) @x @y S D ZZ ▯ Example (from class). Evaluate siny dS where S is the graph of g(x;y) = x + siny, 0 ▯ x ▯ 2, 0 . y ▯ 2 S Solution. Use formula (3). s ▯ ▯ 2 ▯ ▯ 2 p p @g @g @g @g 2 2 2 @x = 1; @y = cosy =) 1 + @x + @y = 1 + 1 + cos y = 2 + cos y; and f(x;y;z) = siny: Thus, s ZZ ZZ ▯ ▯2 ▯ ▯2 Z 2 Z 2 p ▯ ▯ @g @g 2 siny dS = f x;y;g(x;y) 1 + @x + @y dA = siny 2 + cos y dydx S D x=0 y=0 Z ▯ ▯▯ 2 1▯ 3=2 2 2▯ ▯ = ▯ 2 + cos y dx = ▯ 23=2+ 33=2 :
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