# MTH101 Assignment # 3 Idea SolutionJune 2012 by gonglu01

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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
0306-4424443

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