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Final Exam Solution

Final Exam Solution - Math 3010 Fall 2012 Final Exam Dec 8...

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Math 3010 Student Name: Fall 2012 Student Number: Final Exam Dec. 8, 2012 Time Limit: 180 Minutes This exam contains 14 pages (including this cover page) and 10 problems. Check to see if any pages are missing. Enter all the requested information on the top of this page. You may not use your books, notes, or calculator during the exam. Show your full solutions for each problem unless otherwise speciFed. If you need more space, use the back of the pages. Clearly indicate if you have done this. Do not write on the table below. Question Points Bonus Score 1 15 0 2 10 0 3 10 0 4 10 0 5 10 0 6 10 0 7 10 0 8 15 0 9 10 0 10 0 5 Total: 100 5
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Math 3010 Final Exam - Page 2 of 14 Dec. 8, 2012 FORMULA SHEET Hyperbolic functions cosh θ = e θ + e - θ 2 and sinh θ = e θ e - θ 2 d cosh θ = sinh θ and d sinh θ = cosh θ cosh 2 θ + sinh 2 θ = cosh 2 θ cosh 2 θ sinh 2 θ = 1 2 sinh θ cosh θ = sinh 2 θ Change of variables Polar coordinates: dA = r dr dθ Cylindrical coordinates: dV = r dr dθ dz Spherical coordinates: dV = ρ 2 sin ϕ dϕ dθ dρ Integral theorems Green’s theorem: i ∂D ( P dx + Q dy ) = ii D p ∂Q ∂x ∂P ∂y P dA Stokes’ theorem: i ∂S v F · dvs = ii S v ∇ × v F · d v S Divergence theorem: ii ∂W v F · d v S = iii W v ∇ · v F dV Generalized Stokes’ theorem: i ∂M ω = i M DiFerential forms Let f be a 0-form, ω be a k -form and η be an -form on K with 0 l k + l 3. Also let σ 1 , σ 2 and σ 3 be k 1 , k 2 and k 3 forms respectively with 0 l k 1 + k 2 + k 3 l 3. For each k there is a zero k -form with the property 0 + ω = ω and 0 η = 0 for all ω and η . If ω 1 and ω 2 are both k -forms then ( 1 + ω 2 ) η = f ( ω 1 η ) + ( ω 2 η ) ω η = ( 1) kℓ ( η ω ) ω 1 ( ω 2 ω 3 ) = ( ω 1 ω 2 ) ω 3 ω ( ) = ( ) η = f ( ω η ) dx dy = dxdy , dy dz = dy dz , dz dx = dz dx dx dx = dy dy = dz dz = 0 dx ( dy dz ) = dxdy dz = ( dx dy ) dz f ω = df = ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz . If ω 1 and ω 2 are k -forms then d ( ω 1 + ω 2 ) = 1 + 2 . d ( ω η ) = ( η ) + ( 1) k ( ω ) d ( dx ) = d ( dy ) = d ( dz ) = 0.
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Math 3010 Final Exam - Page 3 of 14 Dec. 8, 2012 1. Improper integrals (a) (6 points) Let D = { ( x, y ) | x [ a, b ] , ϕ 1 ( x ) l y l ϕ 2 ( x ) } where ϕ i are continuous. Let f : R 2 R be continuous in the interior of D but possibly discontinuous on points on ∂D . De±ne the following improper integrals: ii D f dA , i ϕ 2 ( x ) ϕ 1 ( x ) f ( x, y ) dy and i b a i ϕ 2 ( x ) ϕ 1 ( x ) f ( x, y ) dy dx. Solution: Let D η,δ = { ( x, y ) | x [ a + η, b η ] , y [ ϕ 1 ( x ) + δ, ϕ 2 ( x ) δ ] } . Let η, δ > 0. D f dA = lim ( η,δ ) (0 , 0) D η,δ f dA = lim ( η,δ ) (0 , 0) i b η a + η i ϕ 2 ( x ) δ ϕ 1 ( x )+ δ f ( x, y ) dy dx. i ϕ 2 ( x ) ϕ 1 ( x ) f ( x, y ) dy = lim δ 0 i ϕ 2 ( x ) δ ϕ 1 ( x )+ δ f ( x, y ) dy.
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Final Exam Solution - Math 3010 Fall 2012 Final Exam Dec 8...

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