Ying-Ying Tran 2016 May 10 Review

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1 MATH 920 Final review Ying-Ying Tran 206 Ma 0 Review hapter 3: Vector geometr vectors dot products cross products planes quadratic surfaces clindrical and spherical coordinates hapter 4: alculus of vector-valued functions vector functions arc length hapter 5: Di erentiation in several variables level sets partial derivatives di erentiabilit and tangent planes gradient and directional derivatives chain rule optimization in several variables Lagrange multipliers hapter 6: Multiple integration integration in two and three variables integration in polar, clindrical, and spherical coordinates applications hapter 7: Line and surface integrals scalar line integral vector line integral scalar surface integral vector surface integral conservative vector fields hapter 8: Fundamental theorems of vector analsis Green s theorem tokes theorem Divergence theorem

2 TA: M Hunh Discussion 27 - Final Review MATH 920 ections 209/24 December 0, 206. Justif the following statement: The flu of F =( 3, 3,z 3 )througheverclosed surface is positive. 2. Let be a unit sphere z 2 =andf(,, z) =(z 4 2, 3,z 2 ). Evaluate (r F) d. 3. Evaluate the flu F d if F(,, z) =( 2 2 z,3z cos, 4 2 )and is the boundar (surface) of the region bounded b 4+2 +z = 4 (first octant) and the coordinate planes. The orientation is given b the outward pointing normal.

3 MATH 920 Final review Ying-Ying Tran 206 Ma 0 Eercises. (8..26) Let F be the vorte vector field F(, ) = 2 + 2, In section 6.3, we verified that F dr =2, where R is the circle of radius R centered at the origin. I R Prove that F dr =2 for an simple closed curve whose interior contains the origin. Hint: Appl the general form of Green s Theorem to the domain between and R where R is so small that R is contained in. 2. (8..28) Estimate the circulation of a vector field F around a circle of radius R =0., assuming that curl z (F) takes the value 4 at the center of the circle. 3. (8..40) If v is the velocit field of a fluid, the flu of v across is equal to the flow rate (amount of fluid flowing across in square meters per second). Find the flow rate across the circle of radius 2 centered at the origin if div(v) = 2. D 4. (8.2.6) alculate curl(f) iff = +, z 2 4, p E 2 +. Then appl tokes theorem to compute the flu of curl(f) through the surface of the wedge-shaped bo below (bottom included, top ecluded) with outward-pointing normal. 2 z (0,, 2) (, 0, 2) + = 5. (8.2.28) uppose that F has a vector potential and that F(,, 0) = k. Find the flu of F through the surface pictured below, oriented with an upward-pointing normal. z Unit circle 6. (8.2.32) Assume that f and g have continuous partial derivatives of order 2. Prove that I frg dr = rf rg

4 7. (8.3.pre2) Justif the following statement: The flu of F = 3, 3,z 3 through ever closed surface is positive. 8. (8.3.36) Let F = r n e r,wheren is an number, r =( z 2 ) /2, and e r = h,, zi /r is the unit radial vector. (a) alculate div(f). (b) alculate the flu of F through the surface of a sphere of radius R centered at the origin. For which values of n is this flu independent of R? 9. (8.3.38) Assume that ' is harmonic. how that div('r') =kr'k 2 and conclude that 'D n ' d = kr'k 2 dv.

5 Partial answers:. Appl Divergence theorem! /3 2

6 MATH 920 Final review Ying-Ying Tran 206 Ma 0 Partial Eercise Answers. (8..26) Let F be the vorte vector field F(, ) = 2 + 2, In section 6.3, we verified that F dr =2, where R is the circle of radius R centered at the origin. I R Prove that F dr =2 for an simple closed curve whose interior contains the origin. Hint: Appl the general form of Green s Theorem to the domain between and R where R is so small that R is contained in. olution: Let R>0 be su cientl small so that the circle R is contained in. Let D denote the region between R and. e appl Green s Theorem to the region D, wherethecurves and R are both oriented counterclockwise. This gives Now, F dr F dr = @F = (2 + 2 ) (2) ( ) 2 = 2 2 ( = (2 + 2 )( ) + (2) ( ) 2 = 2 2 ( ) 2. ince we are given that R R F dr =2, substituting in () we obtain or F dr =2. F dr 2 = D 0 da da. 2. (8..28) Estimate the circulation of a vector field F around a circle of radius R =0., assuming that curl z (F) takes the value 4 at the center of the circle. Answer: /25 3. (8..40) If v is the velocit field of a fluid, the flu of v across is equal to the flow rate (amount of fluid flowing across in square meters per second). Find the flow rate across the circle of radius 2 centered at the origin if div(v) = 2. Answer: 4 4. (8.2.6) alculate curl(f) if F = D +, z 2 4, p E 2 +. Then appl tokes theorem to compute the flu of curl(f) through the surface of the wedge-shaped bo below (bottom included, top ecluded) with outward-pointing normal.

7 2 z (0,, 2) (, 0, 2) + = Answer: curl(f) = p 2 + 2z, p 2 +, ;flu/2 5. (8.2.28) uppose that F has a vector potential and that F(,, 0) = k. Find the flu of F through the surface pictured below, oriented with an upward-pointing normal. z Answer: Unit circle 6. (8.2.32) Assume that f and g have continuous partial derivatives of order 2. Prove that I frg dr = rf rg d. olution: Note hfg,fg,fg z i = h(fg z ) (fg ) z, (fg ) z (fg z ), (fg ) (fg ) i = hf g z f z g,f z g f g z,f g f g i = hf,f,f z i hg,g,g z i = rf rg. Thus b tokes theorem and the above equation, we have frg dr = curl(frg) d rf rg d. 7. (8.3.pre2) Justif the following statement: The flu of F = 3, 3,z 3 through ever closed surface is positive. olution: Note div(f) = 3( z 2 ) > 0 for all (,, z) 6= (0, 0, 0). Therefore b the Divergence theorem, the flu of F through a closed surface is F d = 3( z 2 ) dv > (8.3.36) Let F = r n e r,wheren is an number, r =( z 2 ) /2, and e r = h,, zi /r is the unit radial vector. (a) alculate div(f). Answer: (n + 2)r n

8 (b) alculate the flu of F through the surface of a sphere of radius R centered at the origin. For which values of n is this flu independent of R? Answer: flu 4 R n+2 ; independent if and onl if n = 2 9. (8.3.38) Assume that ' is harmonic. how that div('r') =kr'k 2 and conclude that 'D n ' d = kr'k 2 dv. olution: First notice div('r') =div('', '', '' z )=('' ) +('' ) +('' z ) z = ' 2 + '' + ' 2 + '' + ' 2 z + '' zz = kr'k 2 + ' '. ince ' is harmonic, ' = 0, so div('r') =kr'k 2 and 'D n ' d = 'r' n d = 'r' d = div('r') dv = kr'k 2 dv.

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