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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 2 4 2 x y y y = + = + , i.e. 2 6 ( 3)( 2) 0 y y 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: 2 2 2 2 3 3 2 3 2 3 [( 2) ( 2 4)] (6 ) 1 1 125 (6 ) 3 2 6 A y y y dy y y dy y 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 1 1 ( 2) ( 2) ( 2) ( 2) ( 4 ) 2 4 out in x x y y y y y π π π π π + + = + + = + Therefore, the volume is 4 2 0 1 4 8 ( 4 ) ( 8 8) 12 4 3 3 V y y y dy π π π = + = + = 3. (a) The area is 2 2 2 2 0 0 1 1 1 3 (1 cos ) 2 2 2 2 A r d d π π θ θ θ π π π = = = + = (b) The arc length is [using 2 1 1 cos 2sin ( ) 2 θ θ = ] 2 2 2 2 2 2 0 0 2 2 2 0 0 1 2cos cos sin
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