7B_2016-2017_H2_Maths_App_of_Integration_Assignment_2_Solutions

National Junior College Mathematics Department 2016 Applications of Integration (Definite Integrals) Page 1 of 3 National Junior College 2016 – 2017 H2 Mathematics Applications of Integration (Area and Volume) Assignment 2 Solutions 1(i)   2 2 1 1 1 1 0. y x xxy y xx y         Integrating with respect to y, area of region A 4 1 4 1 11 d (1)(1)2 12 2 5 .2 y y y y y           [M1] Formed an area integral. [A1] Correct expression for area of A. [A1] 5 .2 1(ii) Volume generated     4 2 1 4 1 11π(1) (1) π d3 π π ln3 π π ln 4 ln13 1π ln 4 .3 yy y         [M1] Formed a volume integral. [A1] Correct expression for volume of solid. [A1]   4 1ln y [A1] 1π ln 4 .3  

National Junior College Mathematics Department 2016 Applications of Integration (Definite Integrals) Page 2 of 3 2(i) When x = 0, 22 4 2 .y a y a   When y = 0,  4 0 .a a x x a        2 2 0 23 0 3 2 Area of d 4 12 22 12 4 .3 a a yR a y a yay a aaa a a      [M1] Formed area integral. [A1] 23 0 .12 a yay a   [A1] 24 .3 a 2(ii)     0 22 0 33 3 π 4 d π 4 2 π 4 2 2π. a x a V a a x x a x ax aa a       [M1] Formed volume integral w.r.t. x. [A1] 22 0 4 2 . a a x ax  [A1] 32π.a 2(iii)         2 2 22 2 4 2 2 00 235 2 2 0 35 2 2 3 3 π d π d4 2 16 π 6 80 22π2 6 80 16 π15 88 2π.15 15 a a y a x y y yV a y a y aa yyay a aaaa a a aV                   [M1] Formed volume integral w.r.t. y. [A1] 235 2 2 06 80 a yyay a   [A1] 8 15 yxVV (a.g.).

National Junior College Mathematics Department 2016 Applications of Integration (Definite Integrals) Page 3 of 3 2 (iv) Volume of solid formed when S is rotated completely about the y-axis     2 33 3 π2 162ππ 15 14 π.15 ya a V aa a    [M1] Expressing volume as a difference. [A1] 314 π.15 a 3(i) When 0,y cos 0 2    . Hence, π1 2sin 1 2x     or π1 2sin 3. 2x     When 0,x 1 π1 2sin 0 sin 26        . Hence π3cos 62y    . [B1] Shape of curve [B2,1,0] Axial intercepts 3(ii) When 2,x π2 1 2sin . 6    Area of region   π 6 π 6 π 26 π 6 π 6 π 6 π 6 π- 6 cos 2cos d 2 cos d cos 2 1 d 1 sin 22 1 π π 1 π πsin 2 sin 22 6 6 2 6 6 3 π .23                                                      [M1] Method of substitution. [A1] π 26 π 6 2 cos d .  [A1] π 6 π 6 1 sin 2 .2     [A1] 3 π .23