Integration, Long Questions (Question 4)


Question 4:
Diagram below shows a curve x = y2 – 1 which intersects the straight line 3y = 2x at point Q.
Calculate the volume generated when the shaded region is revolved 360o about the y-axis.

Solution:

x= y 2 1( 1 ) 3y=2x x= 3 2 y( 2 ) Substitute (2) into (1), 3 2 y= y 2 1 2 y 2 3y2=0 ( 2y+1 )( y2 )=0 y= 1 2    or   y=2

when y=2,x= 3 2 ( 2 )=3, Q=( 3, 2 ) I 1 ( Volume of cone ) = 1 3 π r 2 h= 1 3 π ( 3 ) 2 ( 2 ) =6π  unit 3 I 2 ( Volume of the curve ) π 1 2 x 2 dy π 1 2 ( y 2 1 ) 2 dy π 1 2 ( y 4 2 y 2 +1 )dy =π [ y 5 5 2 y 3 3 +y ] 1 2 =π[ ( 2 5 5 2 ( 2 ) 3 3 +2 )( 1 5 5 2 ( 1 ) 3 3 +1 ) ] =π( 46 15 8 15 ) = 38 15 π  unit 3  Volume generated = I 1 I 2                                  =6π 38 15 π                                  = 52 15 π  unit 3