21c_02f_fs

# 21c_02f_fs - Math 21C Final Exam Solutions Fall 2002 1. Let...

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Math 21C Final Exam Solutions Fall 2002 1. Let f ( x, y )=4 - x 2 - y 2 . The xy -plane is given by z = 0. Thus the intersection with z =4 - x 2 - y 2 is given by x 2 + y 2 = 4. Since f x = - 2 x and f y = - 2 y , the area is ZZ D p 1+4 x 2 +4 y 2 dA where D = { ( x, y ) | x 2 + y 2 4 } . We must convert this to an iterated integral. This can be done in Cartesian coordinates in two ways: Z 2 - 2 Z 4 - y 2 - 4 - y 2 p 1+4 x 2 +4 y 2 dx dy = Z 2 - 2 Z 4 - x 2 - 4 - x 2 p 1+4 x 2 +4 y 2 dy dx and in polar coordinates in two ways: Z α +2 π α Z 2 0 p 1+4 r 2 rdrdθ = Z 2 0 Z α +2 π α p 1+4 r 2 rdθdr , where your answer can have any value for α ; e.g., 0 or - π . 2. (a) r = 1 2 +3=2, θ = π/ 3 and z =2. (b) ρ = 1 2 +3+2 2 = 8, θ = π/ 3 and φ = π/ 4. 3. (a) Any vector c h 1 , 1 , 0 i×h 0 , 1 , 2 i = c h 2 , - 2 , 1 i with c 6 =0. (b) Since h 0 , 0 , 0 i is on the ﬁrst line and h 1 , 1 , 1 i is on the second, the closest distance

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## This note was uploaded on 06/17/2009 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.

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21c_02f_fs - Math 21C Final Exam Solutions Fall 2002 1. Let...

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