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# Class Note for MATH 223 at UA 2

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COURSE
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TYPE
Class Notes
PAGES
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KARMA
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This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 02/06/15
Sample solutions to Sec 162 worksheet 1 a First 2 dy 2 1 2 1 1 i 1 7 1 z y z y z 2 2 1 if one makes a Change of variables u z y Then lt 1 1 d1n61n51n571114 Tx2 21 1 b As with the rst problem it is the easiest to do the problem in stages 5 2 2 4ltz 2ygtdz 3 2yz y 2127 4 25 777 2 274710 2 2yy 1 Now the rest of the problem is just algebra to write out the expression above as a polynomial in y and then to do the outer integral 1 C The inner integral is 1 2 2 y 1 1 z 3 2 1 d 1 i 1 77i AW M wb2 11 M 962 2 296 62 And so the outer integral is 1 zz1z11 11111 02 2 23 2 2 2quot 2 Notice that the points of intersection of z 0 and z 11 7 y are just the solutions to 0 11 7 y ie y i1 Therefore the region is 0 S x S 117y2 and 71 y S 1 This gives 1 11712 21sz zyzdxdy R 71 0 Now integrate starting from the inner integral 1722 1 2 2 1 2 1 y y y2dy2ltlt 2 gt0gty2 y2y4 0 Next the outer integral is 1 12 4 12 12 d 129 my 23 25 3 a If you sketch the region you will see that one can describe it by 0 S y S 48 and 112y S x S y3 therefore 4 12 48 1 y3 f967ydyd96 0 3z2 0 112y 3 b If you sketch the region you will see that it is the triangle ABC where A 00 B 12 and C 13 There is no uniform description of z in terms of y throughout the triangle But if we split it into two parts we get a uniform description for x on each part as follows if 0 S y S 2 then 13y if 2 S y S 3 then 13y Therefore we have 1 31 2 12y 3 1 f967ydyd f967yd96dy f967yd96dy 0 2m 0 13y 2 13y 3 c ls similar to a and b WM 1 l l l l 4 Sketch the region Observe that it can be described as 1 S x S e 0 S y S lnz therefore we change the order of integration as follows 1 5 5 d d d d o 1nltzgt 9 y 1nltzgt y 9 Ina 1mm 9m a5 71 5 I will just show how to set up the integral The limits of inte gration for z are the points of integration of the two curve ie the solutions to x2 z 2 These are 71 and 2 Therefore the area is 2 m2 1dydz 1 m2 6 The volume in question is equal to R faxW1 which is where f y 2x2 y2 1 and R is the region in the xy plane that is cut out by the coordinate planes and the plane z y 1 If we draw 3 R we see that this a triangle such that 0 S x S 1 and 0 S y S 1 7 x Thus we have to compute 1 17m 2x2 y2 1dydx 0 0 One can evaluate the integral in stages as parts 1a 10 of this handout 7 This is similar to 6

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