Spring 2006 - Nordgren's Class - Exam 1 (Version A)

# Spring 2006 - Nordgren's Class - Exam 1 (Version A) - plane...

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Solutions to Version A of Midterm 1 By H˚ akan Nordgren R EMARK : Please note that there were two versions of the midterm. If these questions look similar - but not the same - as your questions, apply the methods presented below to your question. Problem 1: Evaluate Z 1 0 Z 2 x 0 Z y 0 xzdzdydx. Solution: Z 1 0 Z 2 x 0 Z y 0 xzdzdydx. = Z 1 0 x Z 2 x 0 ± z 2 2 ² y 0 dydx = Z 1 0 x Z 2 x 0 y 2 2 dydx = Z 1 0 x ± y 3 6 ² 2 x 0 dx = Z 1 0 8 x 4 6 dx = ± 8 x 5 30 ² 1 0 = 8 30 = 4 15 . Problem 2: Interchange the order of integration in the following two integrals. 1. Z 1 0 Z 2 x 0 f ( x, y ) dydx. 2. Z 1 0 Z 4 - 2 y y - 1 g ( x, y ) dxdy. Solution: In each part of the question, begin by drawing the region of intergration. This will make it clear what the limits of integration will be when you change the order. 1. Z 1 0 Z 2 x 0 f ( x, y ) dydx = Z 2 0 Z 1 x 2 f ( x, y ) dxdy. 2. Z 1 0 Z 4 - 2 y y - 1 g ( x, y ) dxdy = Z 0 - 1 Z x +1 0 g ( x, y ) dydx + Z 2 0 Z 1 0 g ( x, y ) dydx + Z 4 2 Z 2 - x 2 0 g ( x, y ) dydx. 1

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Problem 3: Find the volume of the region above the paraboloid { ( x, y, z ) R 3 : z = x 2 + y 2 } and below the
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Unformatted text preview: plane { ( x, y, z ) ∈ R 3 : z = 4 } . Solution: Cylindrical polar co-ordinates are best for this problem. Thus we have Volume = Z 4 Z 2 π Z √ z rdrdθdz = Z 4 Z 2 π ± r 2 2 ² √ z dθdz = Z 4 Z 2 π z 2 dθdz = Z 4 2 π z 2 dz = 2 π ± z 2 4 ² 4 dz = 2 π 16 4 = 8 π Problem 4: Evaluate Z B ( x 2 + y 2 + z 2 ) dv, where B = { ( x, y, z ) ∈ R 3 : x 2 + y 2 + z 2 < 4 } . Solution: For this problem, spherical polars are best. Thus, Z B ( x 2 + y 2 + z 2 ) dv = Z 2 Z 2 π Z π r 4 sin( φ ) dφdθdr = ± r 5 5 ² 2 2 π [-cos( φ )] π = 2 7 5 π. 2...
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Spring 2006 - Nordgren's Class - Exam 1 (Version A) - plane...

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