MTH101 Assignment # 3 Idea SolutionJune 2012

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MTH101 Assignment # 3 Idea SolutionJune 2012 Powered By Docstoc
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Question NO 1..
Find the area of the region confined by x y 2  12 and yx .
Solution
 y  y 2  12
y 2  y  12  0
y 2  4 y  3 y  12  0
y ( y  4)  3( y  4)  0
( y  3)( y  4)
y  3  0, y  3
y  4  0, y  4
4

 (y        y  12)
       2

3
                       4
   y3 y 2           
    12 y 
   3      2          3
   64 16            27 9        
    48                  36 
   3 2              3       2    
  64 16             27 9
        48              36
   3 2               3 2
  64                   9
       8  48  9   36
   3                   2
  64          9
       83 
   3          2
128  498  27          343
                   
        6                 6
  343
       ans
    6
Question NO 2
Find the volume of the solid that is obtained when the region under the curve
y 3 x3 over the interval [5, 24] is revolved about x-axis.
Solution………
     b
v     f ( x ) 2 dx
     a
                            2
    24
                  1
                     
v     ( x  3) 3  dx
    5               
    24
   ( x  3) 3 dx
                 2


     5
                       24
             2 
               1
    ( x  3) 3 
               
          5
                 
         3       5
                       24
    3        3
                
   ( x  3) 
              5

    5          5
                       24
 3                 5
                      
        ( x  3) 3 
 5                   5
  3       5     5
                   
   (27) 3 (8) 3 
  5               
  3
  (243  32)
  5
  3           633
  (211)           ans
  5              5
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