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.Surface Integrals.If the surface S is given by the graph z = z(x, y), where (x, y) lies inthe domain of D ⊂ R 2 , i.e. S√= { (x, y, z(x, y)) | (x, y) ∈ D }, then thesurface area element dS = 1 + z 2 x + z 2 y dxdy...If the surface S is given by parametric formr(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, where (u, v) lies in the domainof D of uv-plane. Surface area element dS = ∥N∥dudv=∂r∣∣∂u × ∂r∣∣ ∣∣ ∣∣∣ ∣∣∣ ∂v ∣∣ dudv = ∂(y, z) ∂(z, x) ∂(x, y) ∣∣∣ ∣∣∣i + j +∂(u, v) ∂(u, v) ∂(u, v) k dudv. = √ (y u z v − z u y v ) 2 + (x u z v − z u x v ) 2 + (x u y v − y u x v ) 2 du dv..Definition. Let g = g(x, y, z) be a continuous function defined ondomain ∫∫ containing∫∫S, the surface integral√of the function g on S isg(x, y, z) dS = g(x, y, z(x, y)) 1 + z 2 x + z 2 y dxdy, orS∫∫D= g(x(u, v), y(u, v), z(u, v))∂r∣∣. D∂u × ∂r∂u ∣∣ du dv. . . . . .
.Definition. Let S be a parameterized surface or a graph, then thesurface ∫∫ area ∫∫of √S is given by1 dS = 1 + z 2 x + z 2 y dxdy, orS ∫∫ D= ∥r u × r v ∥ dudv..D.Example.Find the area of the part of the surface 2z = x 2 that liesdirectly above the triangle in the xy-plane with vertices at. A(0, 0), B(1, 0) and C(1, 1).Solution. The surface is a graph z(x, y) = x 2 /2 defined on the regionD = { (x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }, then z x = x and z y = 0. Thesurface ∫∫ √area is1 + z 2 x + x 2 y dxdy∆ABC. . . . . .
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Page 7 and 8: .Example. Let S be a parametric sur
Page 9: .Surface Integral of Vector Field.L
Page 13 and 14: .Example. Find the flux of F(x, y,
Page 15 and 16: .Example. Suppose that a uniform so
Page 17 and 18: .Example. Let S be the surface (wit
Page 19 and 20: .Example. Let S be the sphere x 2 +
Page 21 and 22: .Example. Show that the divergence
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Page 25 and 26: .Stokes’ Theorem.Definition. A pi
Page 27 and 28: .Example. ∮ Let F(x, y, z) = 3zi
Page 29 and 30: .∫∫Example. Evaluate the surfac
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Page 37: .Example. Let F(x, y, z) = f (x, y,

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