solution_practice_final_1

solution_practice_final_1 - Solutions Practice Final #1 1....

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Solutions – Practice Final #1 1. The two curves intersect when 2 24 2 xy y y = +− = + , i.e. 2 6( 3 ) ( 2 )0 yy y y + −= + − = , so y = -3 and y =2. Since the value of x when y =0 (which lies between the two crossing points) is -4 for the first curve and +2 for the 2 nd curve, the 2 nd curve is to the right of the 1 st throughout the region where we have to find the area. Thus, the area between the curves is: 22 33 2 32 3 [( 2) ( 2 4)] (6 ) 11 1 2 5 (6 ) 6 A y d y y y d y y −− =+ + = =− = ∫∫ 2. The curves intersect when y = 0 and y = 4. The volume can be computed by slicing the volume at fixed y’s – the area of each ‘washer’ (disk with a hole in it) is 2 2 2 ( 2 )(2 )( 2 2 4 ) out in x xy y y y y ππ πππ += + Therefore, the volume is 4 2 0 14 8 (4 ) ( 8 8 ) 1 2 43 3 Vy y y d y π =−− + = + = 3. (a) The area is 00 1 3 (1 cos ) 2 2 Ar d d θ θθ == = + = (b) The arc length is [using 2 1 1c o s 2 s i n( ) 2 −= ] 2 2 2 12 c o s c o s s i n 1 21 c o s 2s i n ( )8 2 dr sr d d d dd θθθ θ θ ⎛⎞ = + + ⎜⎟ ⎝⎠ = = 4.
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This note was uploaded on 05/26/2008 for the course MATH 115 taught by Professor Staff during the Spring '08 term at Yale.

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solution_practice_final_1 - Solutions Practice Final #1 1....

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