Integration, Long Questions (Question 6)


Question 6:
Diagram below shows part of the curve   y= 2 ( 3x2 ) 2  which passes through (1, 2).

 (a)   Find the equation of the tangent to the curve at the point B.
 (b)  A region is bounded by the curve, the x-axis and the straight lines x = 2 and x = 3.
       (i) Find the area of the region.
       (ii) The region is revolved through 360° about the x–axis. Find the volume generated,
             in terms of p.

Solution:
(a)
y= 2 ( 3x2 ) 2 =2 ( 3x2 ) 2 dy dx =4 ( 3x2 ) 3 ( 3 ) dy dx = 12 ( 3x2 ) 3 dy dx = 12 ( 3( 1 )2 ) 3  , x=1 dy dx =12 y2=12( x1 ) y2=12x+12 y=12x+14
(b)(i)
Area =  2 3 y dx = 2 3 2 ( 3x2 ) 2 dx= 2 3 2 ( 3x2 ) 2 dx = [ 2 ( 3x2 ) 1 1( 3 ) ] 2 3 = [ 2 3( 3x2 ) ] 2 3 =[ 2 3[ 3( 3 )2 ] ][ 2 3[ 3( 2 )2 ] ] = 2 21 + 1 6 = 1 14  unit 2

(b)(ii)
Volume generated  π y 2 dx =π 2 3 4 ( 3x2 ) 4   dx=π 2 3 4 ( 3x2 ) 4   dx =π [ 4 ( 3x2 ) 3 3( 3 ) ] 2 3 =π [ 4 9 ( 3x2 ) 3 ] 2 3 =π[ 4 9 [ 3( 3 )2 ] 3 ][ 4 9 [ 3( 2 )2 ] 3 ] =π( 4 3087 + 4 576 ) = 31 5488 π  unit 3