Math 2433 Section 26013 MW 1-2:30pm GAR 205 Bekki George bekki@math.uh.edu 639 PGH Office Hours: 11:00 - 11:45am MWF or by appointment Last week’s video went over 17.6 and some of 17.7 We will pick up today with one more topic from 17.7 – Flux of a Vector Field Suppose S is a smooth surface parameterized by r(u,v) with a unit normal of n(x,y,z) that is continuous on all of S. If v is a vector field that is continuous on S, then the surface integral: ∬ (v ⋅ n)dσ S gives the flux of v across S in the direction of n. Example: Calculate the flux out of the sphere x2 + y2 + z2 = 16 given v = y i − x j 17.8 The Vector Differential Operator ∇ ∂ ∂ ∂ ∇ = i + j+ k ∂x ∂y ∂z So, we are familiar with ∇ from our work with gradients but how does this work with vector fields? This is called the divergence of v: ∇ • v = div v = ∂v1 ∂v2 ∂v3 + + ∂x ∂y ∂z And the curl of v is: i ∂ ∇ × v = curl v = ∂x v1 j ∂ ∂y v2 k ∂ ∂z v3 Note that the curl of a gradient is zero And the divergence of a curl is zero Example: ( ) 2 2 v(x, y) = x + y i + ( xy ) j 1.Calculate ∇· v and ∇× v given: 2.Calculate ∇· v and ∇× v given: Popper 22 1 2 v = xyi − y j + zk 1. Find the divergence of 2 17.9 The Divergence Theorem Recall Green’s Theorem: ⎛ ∂Q ∂P ⎞ ∫C h(r) ⋅dr = ∫C P ( x, y ) dx + Q ( x, y ) dy = ∫∫Ω ⎜⎝ ∂x − ∂ y ⎟⎠ dx dy In vector terms, Green’s Theorem can be written as ∫∫ (∇ • v ) dxdy = ∫ ( v • n) ds Ω C (n is the outer normal) In higher dimensions, this gives: ∫∫∫ (∇ i v ) dxdydz = ∫∫ ( v i n) dσ T S And the Outward Flux per unit volume is (∇ i v ) dxdydz = flux of v out of N ∫∫∫ N ∈ Examples: 1.Use the divergence theorem to find the total flux out of the solid x2 + y2 ≤ 4, 0 ≤ z ≤ 1 given v(x, y, z) = xi + 3y2 j + 2z2 k. 2.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) = 2x2 i + 4xy j - 4xzk. 3. Use the divergence theorem to find ∫∫S (v i n)dσ for 2 2 v(x, y, z) = (z + 2x)i + (4 y + z 2 )j + (y + 2z)k on the surface S : x + y = 1, 0 ≤ z ≤ 4 2.Set up the integral that uses the divergence theorem to find where v = xyi − 1 2 y j + zk 2 ∫∫ (v i n)dσ S and S is the surface z = 4 − 3x 2 − 3y 2 ; z > 0 17.10 Stoke’s Theorem Let’s revisit Green’s Theorem: ⎛ ∂Q ∂P ⎞ h(r) ⋅dr = P x, y dx + Q x, y dy = ( ) ∫∫Ω ⎜⎝ ∂x − ∂ y ⎟⎠ dx dy ∫C ∫C ( ) If we let v = P i + Q j + R k, then (∇ × v ) i k = i j k ∂ ∂x ∂ ∂y ∂ ∂Q ∂P ik = − ∂z ∂x ∂ y P Q R So, Green’s Thm can be written as: ∫∫ ⎡⎣(∇ × v ) i k ⎤⎦ dxdy = ∫ v (r ) • dr Ω C And if S is a flat surface in space bounded by a Jordan curve and v is continuously differentiable on S, we have ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ = ∫ v (r ) • dr S C This is Stoke’s Theorem. Example: 1. Let S be the upper half of the unit sphere x2 + y2 + z2 = 1 and take n as the upper unit normal. Use Stoke's theorem to find ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ given v(x, y, z) = 2z2 i + 4x j - 4y3 k. S 2. Let S be the portion of z = 1- x2 - y2 above the xy-plane. Use Stoke's theorem to find ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ given v(x, y, z) = y i - x j + z k. S 3.Find ∫∫ ⎡⎣(∇ × v ) i n ⎤⎦ dσ where v = -3y i + 3x j + z4 k and S is the portion of S 1 the ellipsoid 2x + 2y + z = 1 that lies above the plane z = 2 and n is the upper unit normal. 2 2 2

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