MATH ajajoo/math2433

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1 MATH Aarti Jajoo Office : PGH 606 Lecture : MoWeFre 10-11am in SR 116 Office hours : MW 11:30-12:30pm and BY APPOINTMENT ajajoo/math2433 A. Jajoo, UH MATH / 31

2 Parametrized Surfaces; Surface Area A. Jajoo, UH MATH / 31

3 We can parametrized a surface S in space by a vector function r = r(u, v), where (u, v) ranges over some region Ω of the uv-plane. A. Jajoo, UH MATH / 31

4 The fundamental vector product Let the surface S be parametrized by a differentiable vector function We denote r = r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k. r u = x u i + y u j + z u k, r v = x v i + y v j + z v k. Definition The cross product N = r u r v is called the fundamental vector product of the surface. A. Jajoo, UH MATH / 31

5 The fundamental vector product The vector N(u, v) = r u(u, v) r v (u, v) is perpendicular to the surface S at the tip of r(u, v). A. Jajoo, UH MATH / 31

6 Example Find the fundamental vector product for r(u, v) = u cos(v)i + u sin(v)j + vk. A. Jajoo, UH MATH / 31

7 How to evaluate the area of S Let S be a surface parametrized by a continuously differentiable function r = r(u, v), (u, v) Ω. Definition If Ω is a basic region in the uv-plane and r is one-to-one on the interior of Ω, we call S a smooth surface. If N(u, v) is never zero on the interior of Ω, then we have area(s) = N(u, v) dudv. Ω A. Jajoo, UH MATH / 31

8 How to evaluate the area of S (particular case) If S is the graph of z = f (x, y) for (x, y) Ω, we can parametrize it like this: S : r(x, y) = xi + yj + f (x, y)k, (x, y) Ω. Then, we have area(s) = Ω [f x (x, y)] 2 + [f y (x, y)] dxdy. A. Jajoo, UH MATH / 31

9 Example Find the area of the surface z = y 2, for 0 x y, 0 y 1. A. Jajoo, UH MATH / 31

10 (continue) Find the area of the surface z = y 2, for 0 x y, 0 y 1. A. Jajoo, UH MATH / 31

11 Surface Integrals A. Jajoo, UH MATH / 31

12 Surface integrals Let H = H(x, y, z) be a scalar field continuous over S : r = r(u, v), with (u, v) Ω. The surface integral of H over S is: S H(x, y, z) dσ = Ω H(x(u, v), y(u, v), z(u, v)) N(u, v) dudv Notice that if H(x, y, z) = 1, we have dσ = S Ω N(u, v) dudv = area(s). A. Jajoo, UH MATH / 31

13 Example Evaluate S 2y dσ where S is the surface S : z = 1 2 y 2, 0 x 2, 0 y 1. A. Jajoo, UH MATH / 31

14 (continue) Evaluate S 2y dσ where S is the surface S : z = 1 2 y 2, 0 x 2, 0 y 1. A. Jajoo, UH MATH / 31

15 Example Evaluate S 2xy dσ where S is the surface x + 2y + 2z = 4 in the first octant. A. Jajoo, UH MATH / 31

16 (continue) Evaluate S 2xy dσ where S is the surface x + 2y + 2z = 4 in the first octant. A. Jajoo, UH MATH / 31

17

18 The Vector Differential Operator A. Jajoo, UH MATH / 31

19 Formal definition The vector differential operator is formally defined by = x i + y j + z k. So far, we have seen applied to a differentiable scalar field, e.g. f, but it can also be applied to vector fields. A. Jajoo, UH MATH / 31

20 Divergence and curl Consider vector field v = v 1 (x, y, z)i + v 2 (x, y, z)j + v 3 (x, y, z)k. Divergence v = v 1 x + v 2 y + v 3 z. Curl v = Det i j k x y z v 1 v 2 v 3 A. Jajoo, UH MATH / 31

21 The curl of a gradient is zero. A. Jajoo, UH MATH / 31

22 Example Given v(x, y) = calculate v and v. ( ) ( ) 1 1 x 2 + y 2 i + x 2 + y 2 j, A. Jajoo, UH MATH / 31

23 (continue) Given v(x, y) = calculate v and v. ( ) ( ) 1 1 x 2 + y 2 i + x 2 + y 2 j, A. Jajoo, UH MATH / 31

24 Example Given calculate v and v. v(x, y, z) = (3yz) i + (4xz) j + (3xy)k, A. Jajoo, UH MATH / 31

25 (continue) Given calculate v and v. v(x, y, z) = (3yz) i + (4xz) j + (3xy)k, A. Jajoo, UH MATH / 31

26 The Divergence Theorem Section 17.9 A. Jajoo, UH MATH / 31

27 Divergence theorem Set v = Qi Pj. Green s theorem can be written as: ( v)dxdy = (v )ds, where is the outer unit normal for Ω. In 3D, we have: Divergence theorem Ω Let T be a solid bounded by a closed oriented surface S which is piecewise smooth. If the vector field v = v(x, y, z) is continuously differentiable throughout T, then ( v)dxdydz = (v )dσ. where is the outer unit normal. T C S A. Jajoo, UH MATH / 31

28 Flux Definition The quantity is called flux of v out of S. S (v )dσ The divergence theorem tells you that you can evaluate the flux of v out of S through a triple integral on T, which is the solid bounded by S flux of v out of S = ( v)dxdydz. T A. Jajoo, UH MATH / 31

29 Example Use the divergence theorem to find the total flux out of the solid x 2 + y 2 4, 0 z 1, given v(x, y, z) = xi + 3y 2 j + 2z 2 k. A. Jajoo, UH MATH / 31

30 (continue) Use the divergence theorem to find the total flux out of the solid x 2 + y 2 4, 0 z 1, given v(x, y, z) = xi + 3y 2 j + 2z 2 k. A. Jajoo, UH MATH / 31

31 Example Use the divergence theorem to find the total flux out of the solid 0 x 1, 0 y 1 x, 0 z 1 x y, given v(x, y, z) = 2x 2 i + 4xyj 4xzk. A. Jajoo, UH MATH / 31

32 (continue) Use the divergence theorem to find the total flux out of the solid 0 x 1, 0 y 1 x, 0 z 1 x y, given v(x, y, z) = 2x 2 i + 4xyj 4xzk. A. Jajoo, UH MATH / 31