MAT183/ MAT421 : Calculus I 246 Outer radius = outer curve/line − axis of rotation Outer radius = 0 − 2 = −2 inner radius = inner curve/line − axis of rotation inner radius = 2 4 − x − 2 = 2 4 − x − 4 = 2 − x Vwasher = ( ) ( ) = = − x b x a outer radius inner radius dx 2 2 Vwasher = ( ) = = − − − 4 0 2 2 2 2 x x dx x = − 4 0 2 4 4 dx x = 4 0 3 12 4 − x x = − − 0 12 64 16 = 3 32 units3 # Radius = equation of axis of rotation Radius = 2 − y Height = Right − left = (4 − 2y) − 0 = 4 − 2y VShell = ( )( ) = = y d y c 2 radius height dy VShell = ( )( ) = = − − 2 0 2 2 4 2 y y y y dy = ( )( ) − − 2 0 4 2 y 2 y dy = ( ) − + 2 0 2 4 4 4y y dy = 2 0 2 3 2 3 4 4 4 − + y y y = 2 0 3 2 3 4 4 2 − + y y y = − − + 0 3 8 4 8 8 = 3 32 units3 # y = 0 (outer curve/line) y = 2 (axis of rotation) line y = 2 4 − x (inner curve/line) r = 2 h = 4 y = 2 (axis of rotation) line x = 4 − 2y (R) line x = 0 (L)
Chapter 5 : Applications of Definite Integrals 247 d) = (r ) h 2 − (r ) h 2 3 = (4) (2) 2 − (4) (2) 3 2 = 32 − 3 32 = 3 64 units3 # (0, 0) x = 4 line x = 4 − 2y x = 4 (axis of rotation) y = 2 x = 4 (axis of rotation) line x = 4 − 2y = − x = 0 (axis of rotation) r = 4 h = 2 r = 4 h = 2
MAT183/ MAT421 : Calculus I 248 Outer radius = outer curve/line − axis of rotation Outer radius = 0 − 4 = −4 inner radius = inner curve/line − axis of rotation innter radius = (4 − 2y) − 4 = −2y Vwasher = ( ) ( ) = = − y d y c outer radius inner radius dy 2 2 Vwasher = ( ) ( ) = = − − − 2 0 2 2 4 2 y y y dy = ( ) − 2 0 2 16 4y dy = 2 0 3 3 4 16 − y y = − − 0 3 32 32 = 3 64 units3 # Radius = equation of axis of rotation Radius = 4 − x Height = top − bottom = 2 4 − x − 0 = 2 4 − x VShell = ( )( ) = = x b x a 2 radius height dx VShell = ( ) = = − − 4 0 2 4 2 4 x x dx x x = ( )( ) − − 4 0 4 x 4 x dx = ( ) − + 4 0 2 16 8x x dx = 4 0 2 3 2 3 8 16 − + x x x = 4 0 3 2 3 16 4 − + x x x = 4 0 0 3 64 64 64 − − + = 3 64 units3 # line x = 4 − 2y (inner curve/line) x = 4 (axis of rotation) line x = 0 (outer curve/line) r = 4 − x h = T − B x = 4 (axis of rotation) line y = (T) line y = 0 (B)
Chapter 5 : Applications of Definite Integrals 249 More Examples Example 3 (Disk Method – dx) Find the volume of the solid generated when the region enclosed by x = y 2 , y = 2 and y-axis is revolved about the line y = 2. Solution : Radius = curve − axis of rotation = ( x ) − (2) = x − 2 using Disk Method, Vdisk = ( ) = = x b x a radius dx 2 = ( ) = = − 4 0 2 2 x x x dx = ( ) − + 4 0 x 4 x 4 dx = − + 4 0 4 4 2 1 x x dx = 4 0 2 4 3 2 4 2 2 3 − • x + x x = 4 0 2 4 3 8 2 2 3 − x + x x = − − + 16 0 3 64 8 = 3 8 units3 # y = 2 (axis of rotation) y = x x = 0 x = 4 y x y = 2 x = y 2 Figure 11 Find the limits on x-axis (by solving simultaneous equation) y = 2 ---------- x = y 2 ---------- substitute into , x = (2)2 x = 4
MAT183/ MAT421 : Calculus I 250 Example 4 (Disk Method – dy) Use Disk Method to find the volume of the solid generated when the shaded region is revolved about the y-axis Solution : Radius = curve − axis of rotation = (2y − 2y 2 ) − (0) = 2y − 2y 2 using Disk Method, Vdisk = ( ) = = y d y c radius dy 2 = ( ) = = − 1 0 2 2 2 2 y y y y dy = ( ) − + 1 0 2 3 4 4y 8y 4y dy = ( ) − + 1 0 2 3 4 4 y 2y y dy = 1 0 3 4 5 4 5 2 3 4 − + y y y = 1 0 3 4 5 3 2 5 4 − + y y y = − − + 0 5 1 2 1 3 1 4 = 15 2 units3 # x = 2y − 2y 2 (curve) y x 0 1 line x = 0 (axis of rotation) y x x = 2y − 2y 2 Figure 12 Find the limits on y-axis, When x = 0 2y − 2y 2 = 0 2y(1 − y) = 0 y = 0, y = 1
Chapter 5 : Applications of Definite Integrals 251 Example 5 (Disk Method – dx) Figure 13 shows the shaded region bounded by the curve y = x 2 and horizontal line y = 1. Find the volume of the solid generated when the shaded region is revolved about the line y = 1. Solution : y = 1 y = x 2 y x 0 R Figure 13 Find the limits on x-axis (by solving simultaneous equation) = 2 − − − ① = 1 − − − ② substitute ① into ②, 2 = 1 = ±1 = 1 − 2 = ∫ () 2 =1 =−1 = ∫ (1 − 2 ) 2 1 −1 = ∫ (1 − 2 2 + 4 ) 1 −1 = [ − 2 3 3 + 5 5 ] −1 1 = [(1 − 2 3 + 1 5 ) − (−1 − −2 3 + −1 5 )] = [1 − 2 3 + 1 5 + 1 − 2 3 + 1 5 ] = 16 15 3 # = 1 (axis of rotation) radius = 2
MAT183/ MAT421 : Calculus I 252 Example 6 (Washer Method – dx) Figure 14 shows the region R bounded by the curve y = 9 − x 2 and the line y = x + 3. Find the volume of the solid generated by rotating R about the x-axis. Solution : Outer Radius = outer curve − axis of rotation = (9 − x 2 ) − (0) = 9 − x 2 inner Radius = inner curve − axis of rotation = (x + 3) − (0) = x + 3 using Washer Method, Vwasher = ( ) ( ) = = − x b x a outer radius inner radius dx 2 2 = ( ) ( ) = =− − − + 2 3 2 2 2 9 3 x x x x dx = ( ) ( ) − − + − + + 2 3 2 4 2 81 18x x x 6x 9 dx = ( ) − − − + 2 3 2 4 72 6x 19x x dx = 2 3 3 5 2 3 5 19 72 3 − − − + x x x x = − − − + − − − + 5 243 216 27 171 5 32 3 152 144 12 = − − 5 603 15 1316 = 3 625 units3 # y y = 9 − x 2 0 Figure 14 x y = x + 3 R (2, 5) −3 y = x + 3 (inner curve/line) y = 9 − x 2 (outer curve/line) line y = 0 (axis of rotation)
Chapter 5 : Applications of Definite Integrals 253 Example 7 (Washer Method – dy) Consider the shaded region below which is bounded by the curve y = x 2 and the straight line y = 2x. Find the volume of the solid generated by revolving the shaded region about the line x = −1 using the Washer method. Solution : Outer Radius = outer curve − axis of rotation = y − (−1) = y + 1 inner Radius = inner curve − axis of rotation = y 2 1 −(−1) = y 2 1 + 1 using Washer Method, Vwasher = ( ) ( ) = = − y d y c outer radius inner radius dy 2 2 = ( ) = = + − + 4 0 2 2 1 2 1 1 y y y y dy = ( ) + + − + + 4 0 2 1 4 1 y 2 y 1 y y dy = − 4 0 2 4 1 2 y y dy = − 4 0 2 4 1 2 2 1 y y dy = 4 0 3 4 3 1 3 2 2 2 3 • − • y y = 4 0 3 3 12 4 2 3 − y y = − − 0 12 64 3 32 = 3 16 units3 # x = (outer curve) x = −1 (axis of rotation) x = (inner line) y x y = 2x y = x 2 (2, 4) Figure 15
MAT183/ MAT421 : Calculus I 254 Example 8 (Washer Method – dx) Figure 16 shows the shaded region R bounded by the curve y = 9 − x 2 and the line y = x + 3. Find the volume of the solid generated by rotating R about the x-axis. Solution : Outer Radius = outer curve − axis of rotation = (9 − x 2 ) − (0) = 9 − x 2 inner Radius = inner curve − axis of rotation = (x + 3) − (0) = x + 3 using Washer Method, Vwasher = ( ) ( ) = = − x b x a outer radius inner radius dx 2 2 = ( ) ( ) = =− − − + 2 3 2 2 2 9 3 x x x x dx = ( ) ( ) − − + − + + 2 3 2 4 2 81 18x x x 6x 9 dx = ( ) − − − + 2 3 2 4 72 6x 19x x dx = 2 3 3 5 2 3 5 19 72 3 − − − + x x x x = − − − + − − − + 5 243 216 27 171 5 32 3 152 144 12 = − − 5 603 15 1316 = 3 625 units3 # y y = 9 − x 2 0 Figure 16 x y = x + 3 R (2, 5) −3 y = x + 3 (inner curve/line) y = 9 − x 2 (outer curve/line) line y = 0 (axis of rotation)
Chapter 5 : Applications of Definite Integrals 255 Example 9 (Shell Method – dy) Consider the shaded region below which is bounded by the curve y = x 2 and the straight line y = 2x. Find the volume of the solid generated by revolving the shaded region about the line y = 4 using the Shell method. Solution : Radius = equation of axis of rotation = 4 − y height = Right − Left (dy) = y y 2 1 − using Shell Method, Vshell = ( )( ) = = y d y c 2 radius height dy = ( ) = = − − 4 0 2 1 2 4 2 1 y y y y y dy = − − + 4 0 2 2 1 2 4 2 2 3 2 1 y y y y dy = 4 0 2 3 2 3 1 5 2 2 2 3 2 2 4 2 5 2 3 • − − + • y y y y = 4 0 3 2 5 6 2 3 8 2 2 5 2 3 − − + y y y y = − − − + 0 6 64 5 64 16 3 64 2 = 5 32 units3 # y x y = 2x y = x 2 (2, 4) Figure 17 y = 4 (axis of rotation) x = (L) x = (R) y − 4 = 0 0 = 4 − y
MAT183/ MAT421 : Calculus I 256 Examples from previous semester papers Example 10/ MAR 2013/ MAT183/ Q5c Consider the shaded region W enclosed by the curve y = x 2 − 2x and the line y = x as shown in Figure 18 below i) Find the volume of the solid generated when the shaded region W is revolved about the y-axis (5 marks) ii) Find the volume of the solid generated when the shaded region W is revolved about the line y = 3 (6 marks) Solution : y y = x 2 − 2x 0 Figure 18 W x y = x (3, 3) ) = ℎℎ = − = () − ( 2 − 2) = − 2 + 2 = 3 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =3 =0 ℎ = 2 ∫ ()(3 − 2 ) 3 0 ℎ = 2 ∫ (3 2 − 3 ) 3 0 ℎ = 2 [ 3 3 3 − 4 4 ] 0 3 ℎ = 2 [ 3 − 4 4 ] 0 3 ℎ = 2 [(27 − 81 4 ) − 0] ℎ = 27 2 3 # ) = 3 − ( 2 − 2) = 3 − 2 + 2 = 3 − Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =3 =0 ℎ = ∫ (3 − 2 + 2) 2 − (3 − ) 2 3 0 ℎ = ∫ ( 4 − 4 3 − 2 2 + 12 + 9) − (9 − 6 + 2 ) 3 0 ℎ = ∫ ( 4 − 4 3 − 3 2 + 18) 3 0 ℎ = [ 5 5 − 4 4 4 − 3 3 3 + 18 2 2 ] 0 3 ℎ = [ 5 5 − 4 − 3 + 9 2 ] 0 3 ℎ = [( 243 5 − 81 − 27 + 81) − 0] ℎ = 108 5 3 # = 3 (axis of rotation) = 2 − 2 (outer) = 0 ( − ) (axis of rotation) = 2 − 2 height
Chapter 5 : Applications of Definite Integrals 257 Example 11/ OCT 2012/ MAT183/ Q5c The shaded region W is bounded by the curve y = x 2 +1 and the line y = x as shown in Figure 19. i) Find the volume of the solid generated when the shaded region W is revolved about the y-axis by using Shell method. (6 marks) ii) Find the volume of the solid generated when the shaded region W is revolved about the line y = −1 by using Washer method. 6 marks) Solution : y y = x y = x 2 + 1 W x 0 3 Figure 19 ) = ℎℎ = − = ( 2 + 1) − () = 2 + 1 − Using Shell method ℎ = 2 ∫ ()(ℎℎ) =3 =0 ℎ = 2 ∫ ()( 2 + 1 − ) 3 0 ℎ = 2 ∫ ( 3 + − 2 ) 3 0 ℎ = 2 [ 4 4 + 2 2 − 3 3 ] 0 3 ℎ = 2 [( 81 4 + 9 2 − 9) − 0] ℎ = 63 2 3 # ) = 2 + 1 − (−1) = 2 + 2 = − (−1) = + 1 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =3 =0 ℎ = ∫ ( 2 + 2) 2 − ( + 1) 2 3 0 ℎ = ∫ ( 4 + 4 2 + 4) − ( 2 + 2 + 1) 3 0 ℎ = ∫ ( 4 + 3 2 − 2 + 3) 3 0 ℎ = [ 5 5 + 3 3 3 − 2 2 2 + 3] 0 3 ℎ = [ 5 5 + 3 − 2 + 3] 0 3 ℎ = [( 243 5 + 27 − 9 + 9) − 0] ℎ = 378 5 3 # height = 0 ( − ) (axis of rotation) = −1 (axis of rotation) = 2 + 1 (outer)
MAT183/ MAT421 : Calculus I 258 Example 12/ MAR 2004/ MAT193/ BQ11 Figure 20 shows the curve x y 2 = and the straight line y = 3 − x which meet at the points A and B. a) Find the coordinates of A and B. (3 marks) b) Determine the area of the shaded region. (4 marks) c) The shaded region is rotated about the x-axis. Determine the volume of the solid by using the Washer method. (8 marks) Solution : Figure 20 x y = 3 − = 2 A B ) = 3 − − − − ① = 2 − − − ② ② into ① 2 = 3 − 2 = 3 − 2 2 − 3 + 2 = 0 ( − 1)( − 2) = 0 = 1, = 2 When = 1 into ① → = 3 − 1 = 2 When = 2 into ② → = 3 − 2 = 1 (1, 2) (2, 1) ) = ∫ ( − ) =2 =1 = ∫ (3 − ) − ( 2 ) 2 1 = ∫ (3 − − 2 ) 2 1 = [3 − 2 2 − 2 ] 1 2 = (6 − 2 − 2 2) − (3 − 1 2 − 2 1) = 3 2 − 2 2 = 0.1137 2 # ) = 3 − = 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =2 =1 ℎ = ∫ (3 − ) 2 − ( 2 ) 2 2 1 ℎ = ∫ (9 − 6 + 2 − 4 2 ) 2 1 ℎ = ∫ (9 − 6 + 2 − 4 −2 ) 2 1 ℎ = [9 − 6 2 2 + 3 3 − 4 −1 −1 ] 1 2 ℎ = [9 − 3 2 + 3 3 + 4 ] 1 2 ℎ = [(18 − 12 + 8 3 + 2) − (9 − 3 + 1 3 + 4)] ℎ = 1 3 3 # = 2 = 0 ( − ) (axis of rotation)
Chapter 5 : Applications of Definite Integrals 259 Example 13/ MAR 2012/ MAT183/ Q5b Consider the region R, enclosed by the curve y = x 3 + 1, the line x = 3 and x-axis as shown in Figure 21. i) Use the Disk Method to find the volume of the solid obtained by revolving the region R about the x-axis. (6 marks) ii) Using the Shell method, set up and simplify the integral to find the volume of the solid obtained by revolving the region R about the line x = −2. Do not evaluate the integral. (4 marks) Solution : Figure 21 y = x 3 + 1 y = 0 (x-axis) x = −1 x = 3 y = x 3 + 1 y = 0 (x-axis) x = −1 x = 3 y x y = x 3 + 1 x = 3 −1 R ) = 3 + 1 = ∫ () 2 =3 =−1 = ∫ ( 3 + 1) 2 3 −1 = ∫ ( 6 + 2 3 + 1) 3 −1 = [ 7 7 + 2 4 4 + ] −1 3 = [ 7 7 + 4 2 +] −1 3 = [( 2187 7 + 81 2 + 3) −(− 1 7 + 1 2 − 1)] = 2496 7 3 # radius s ) = + 2 ℎℎ = − = ( 3 +1) −(0) = 3 + 1 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =3 =−1 ℎ = 2 ∫ ( + 2)( 3 +1) 3 −1 ℎ = 2 ∫ ( 4 + 2 3 + + 2) 3 −1 ℎ = 2 [ 5 5 + 2 4 4 + 2 2 + 2] −1 3 ℎ = 2 [ 5 5 + 4 2 + 2 2 +2] −1 3 ℎ = 2 [( 243 5 + 81 2 + 9 2 + 6) −(− 1 5 + 1 2 + 1 2 − 2)] ℎ = 1008 5 3 #
MAT183/ MAT421 : Calculus I 260 Example 14/ APR 2010/ MAT183/ Q5b The shaded region in Figure 25 below is bounded by the curve y = x , the y-axis and the line y = 2. i) Using the Disk method, find the volume of the solid generated by revolving the shaded region about the y-axis. (4 marks) ii) Using the Washer Method, find the volume of the revolution obtained when the shaded region is revolved about the line x = 4. (7 marks) iii) Using the Shell method, set up and simplify the integral to find the volume of the revolution obtained when the shaded region is revolved about the line x = −1. (3½ marks) Solution : y Figure 25 x y = 2 2 0 y = ) = 2 = ∫ ( 2 ) 2 =2 =0 = ∫ 4 2 0 = [ 5 5 ] 0 2 = [ 2 5 5 − 0] = 32 5 3 # ) = 4 − 0 = 4 = 4 − 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =2 =0 ℎ = ∫ (4) 2 − (4 − 2 ) 2 2 0 ℎ = ∫ (16 − (16 − 8 2 + 4 )) 2 0 ℎ = ∫ (8 2 − 4 ) 2 0 ℎ = [ 8 3 3 − 5 5 ] 0 2 ℎ = [( 64 3 − 32 5 ) − 0] ℎ = 224 15 3 #
Chapter 5 : Applications of Definite Integrals 261 ) = + 1 ℎℎ = − = 2 −√ Using Shell method ℎ = 2 ∫ ()(ℎℎ) =4 =0 ℎ = 2 ∫ ( + 1)(2 − √) 4 0 ℎ = 2 ∫ (2 +2 − 3 2 − 1 2) 4 0 ℎ = 2 [ 2 2 2 + 2 − 2 5 5 2 − 2 3 3 2] 0 4 ℎ = 2 [ 2 + 2 − 2 5 5 2 − 2 3 3 2] 0 4 ℎ = 2 [(16 + 8− 64 5 − 16 3 ) −0] ℎ = 176 15 3 #
MAT183/ MAT421 : Calculus I 262 Example 15/ JUN 2013/ MAT421/ Q5b The shaded region in Figure 27 below is bounded by the curve y = 4 2 x , the line x = 2 and the x-axis. Find the volume of the solid obtained when the shaded region is revolved about i) the line x = 2, using Disk Method. (6 marks) ii) the line y = −2, using Shell Method. (8 marks) Solution : ) = 2− 2√ = ∫ () 2 =1 =0 = ∫ (2 −2√) 2 1 0 = ∫ (4 −8√ + 4) 1 0 = ∫ (4 − 8 1 2 + 4) 1 0 = [4 − 8 3 2 ( 3 2 ) + 4 2 2 ] 0 1 = [4 − 8 ( 2 3 ) 3 2 + 2 2 ] 0 1 = [4 − 16 3 3 2 + 2 2 ] 0 1 = [(4 − 16 3 +2) − 0] = 2 3 3 # ) = + 2 ℎℎ = ℎ − = 2− 2√ Using Shell method ℎ = 2 ∫ ()(ℎℎ) =1 =0 ℎ = 2 ∫ ( + 2)(2 − 2√) 1 0 ℎ = 2 ∫ (2 +4 − 2 3 2 − 4 1 2) 1 0 ℎ = 2 [ 2 2 2 + 4 − 2 5 2 ( 5 2 ) − 4 3 2 ( 3 2 ) ] 0 1 ℎ = 2 [ 2 + 4 − 2 ( 2 5 ) 5 2 − 4 ( 2 3 ) 3 2] 0 1 ℎ = 2 [ 2 + 4 − 4 5 5 2 − 8 3 3 2] 0 1 ℎ = 2 [(1 +4 − 4 5 − 8 3 ) − 0] ℎ = 46 15 3 # y Figure 27 x 0 y = x = 2 = 2) (axis of rotation) = 2) (axis of rotation) = 2 4 2 = 4 = √4 = 2√ = 2 4 2 = 4 = √4 = 2√ = 2√ = 2√ ℎ = 2 → = 2 4 = 2 2 4 = 1 = 0 = 1 = −2 (axis of rotation) = 2√ = 2 = 0 = 1 = −2 (axis of rotation) = 2√ = 2
Chapter 5 : Applications of Definite Integrals 263 Example 16/ SEP 2011/ MAT183/ Q5 Figure 22 below shows the region R bounded by the curve y = x 3 , the line y = 2x + 4 and the y-axis. i) Find the area of the region R. (Ans : Area = 8 units2 ) (3 marks) ii) By using the Washer method, find the volume generated when the region R is rotated about the x-axis. (6 marks) iii) By using the Shell method, find the volume generated when the region R is rotated about the y-axis. (5 marks) Solution : y Figure 22 x y = x 3 y = 2x + 4 (2, 8) R ) = ∫ ( − ) =2 =0 = ∫ ((2 + 4) − ( 3 )) 2 0 = ∫ (2 + 4 − 3 ) 2 0 = [ 2 2 2 + 4 − 4 4 ] 0 2 = [ 2 + 4 − 4 4 ] 0 2 = [(4 + 8 − 4) − 0] = 8 2 = 0 = 2 = 0 ( ) (axis of rotation) ) = 2 + 4 = 3 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =2 =0 ℎ = ∫ (2 + 4) 2 − ( 3 ) 2 2 0 ℎ = ∫ (4 2 + 16 + 16) − ( 6 ) 2 0 ℎ = ∫ (4 2 + 16 + 16 − 6 ) 2 0 ℎ = [ 4 3 3 + 16 2 2 + 16 − 7 7 ] 0 2 ℎ = [ 4 3 3 + 8 2 + 16 − 7 7 ] 0 2 ℎ = [( 32 3 + 32 + 32 − 128 7 ) − 0] ℎ = 1184 21 3 #
MAT183/ MAT421 : Calculus I 264 Example 16 (iii) ) = ℎℎ = − = (2 + 4) − ( 3 ) = 2 + 4 − 3 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =2 =0 ℎ = 2 ∫ ()(2 + 4 − 3 ) 2 0 ℎ = 2 ∫ (2 2 + 4 − 4 ) 2 0 ℎ = 2 [ 2 3 3 + 4 2 2 − 5 5 ] 0 2 ℎ = 2 [( 16 3 + 16 2 − 32 5 ) − 0] ℎ = 2 [ 16 3 + 8 − 32 5 ] ℎ = 2 [ 104 15 ] ℎ = 208 15 3 # = 0 ( ) (axis of rotation) height = 0 = 2 How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232)
Chapter 5 : Applications of Definite Integrals 265 Example 17/ APR 2011/ MAT183/ Q5b Consider the region W, enclosed by the curve y = x 2 , the line y = 2x + 8 and y-axis as shown in Figure 23. i) Find the area of the region W. (Ans : Area = 3 28 units2 ) (4 marks) ii) Use the Washer Method to find the volume of the solid obtained by revolving the region W about the x-axis. (5 marks) iii) Find the volume of the solid obtained by revolving the region W about the y-axis by using the Shell Method. (6 marks) Solution : y Figure 23 x y = x 2 W y = 2x + 8 −2 ) = ∫ ( − ) =0 =−2 = ∫ ((2 + 8) − ( 2 )) 0 −2 = ∫ (2 + 8 − 2 ) 0 −2 = [ 2 2 2 + 8 − 3 3 ] −2 0 = [ 2 + 8 − 3 3 ] −2 0 = [0 − (4 − 16 + 8 3 )] = 28 3 2 top bottom = 2 = −2 = 0
MAT183/ MAT421 : Calculus I 266 Example 17 (ii), (iii) = 0 ( ) (axis of rotation) = −2 = 0 = 0 ( ) (axis of rotation) ) = 2 + 8 = 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =0 =−2 ℎ = ∫ (2 + 8) 2 − ( 2 ) 2 0 −2 ℎ = ∫ (4 2 + 32 + 64) − ( 4 ) 0 −2 ℎ = ∫ (4 2 + 32 + 64 − 4 ) 0 −2 ℎ = [ 4 3 3 + 32 2 2 + 64 − 5 5 ] −2 0 ℎ = [ 4 3 3 + 16 2 + 64 − 5 5 ] −2 0 ℎ = [0 − (− 32 3 + 64 − 128 + 32 5 )] ℎ = 1024 15 3 # ) = − ℎℎ = − = (2 + 8) − ( 2 ) = 2 + 8 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =0 =−2 ℎ = 2 ∫ (−)(2 + 8 − 2 ) 0 −2 ℎ = 2 ∫ (−2 2 − 8 + 3 ) 0 −2 ℎ = 2 [ −2 3 3 − 8 2 2 + 4 4 ] −2 0 ℎ = 2 [ −2 3 3 − 4 2 + 4 4 ] −2 0 ℎ = 2 [0 − ( 16 3 − 16 + 4)] ℎ = 2 [ 20 3 ] ℎ = 40 3 3 # How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232) = 0 ( ) (axis of rotation) = 0 ( ) (axis of rotation) = 0 = −2 height top bottom
Chapter 5 : Applications of Definite Integrals 267 Example 18/ OCT 2010/ MAT183/ Q5 The region R is bounded by the curve y = 2(x − 3)2 and y = 2x − 2. i) Find the area of R (Ans : Area = 9 units2 ) (4 marks) ii) Use the Washer Method to find the volume of the solid obtained by revolving the region R about the x-axis. (6 marks) iii) Find the volume of the solid obtained by revolving the region R about the line x = −1 by using the Shell Method. (5 marks) Solution : y Figure 24 x y = 2(x − 3)2 R y = 2x − 2 2 5 ) = ∫ ( − ) =5 =2 = ∫ ((2 − 2) − (2( − 3) 2 )) 5 2 = ∫ (2 − 2 − 2( − 3) 2 ) 5 2 = ∫ (2 − 2 − 2( 2 − 6 + 9)) 5 2 = ∫ (2 − 2 − 2 2 + 12 − 18) 5 2 = ∫ (14 − 20 − 2 2 ) 5 2 = [ 14 2 2 − 20 − 2 3 3 ] 2 5 = [7 2 − 20 − 2 3 3 ] 2 5 = [(175 − 100 − 250 3 ) − (28 − 40 − 16 3 )] = 9 2 top bottom = 2( − 3) 2 = 2 = 5
MAT183/ MAT421 : Calculus I 268 Example 18 (ii) Method I ) = 2 − 2 = 2( − 1) = 2( − 3) 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =5 =2 ℎ = ∫ (2( − 1)) 2 − (2( − 3) 2 ) 2 5 2 ℎ = ∫ 4( − 1) 2 − 4( − 3) 4 5 2 ℎ = 4 ∫ ( − 1) 2 − ( − 3) 4 5 2 ℎ = 4 ∫ ( − 1) 2 5 2 − 4 ∫ ( − 3) 4 5 2 ℎ = 4 [ ( − 1) 3 3 ] 2 5 − 4 [ ( − 3) 5 5 ] 2 5 ℎ = 4 3 [( − 1) 3]2 5 − 4 5 [( − 3) 5]2 5 ℎ = 4 3 [(5 − 1) 3 − (2 − 1) 3] − 4 5 [(5 − 3) 5 − (2 − 3) 5] ℎ = 4 3 [(4) 3 − (1) 3] − 4 5 [(2) 5 − (−1) 5] ℎ = 4 3 [64 − 1] − 4 5 [32 + 1] ℎ = 4 3 [63] − 4 5 [33] ℎ = 288 5 3 # = 2( − 3) 2 = 2 = 5 = 0 ( ) (axis of rotation)
Chapter 5 : Applications of Definite Integrals 269 Example 18 (ii) Method II ) = 2 − 2 = 2( − 3) 2 = 2( 2 − 6 + 9) = 2 2 − 12 + 18 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =5 =2 ℎ = ∫ (2 − 2) 2 − (2 2 − 12 + 18) 2 5 2 ℎ = ∫ (4 2 − 8 + 4) − (4 4 − 48 3 + 216 2 − 432 + 324) 5 2 ℎ = ∫ (−4 4 + 48 3 − 212 2 + 424 − 320) 5 2 ℎ = 4 ∫ (− 4 + 12 3 − 53 2 + 106 − 80) 5 2 ℎ = 4 [− 5 5 + 12 4 4 − 53 3 3 + 106 2 2 − 80] 2 5 ℎ = 4 [− 5 5 + 3 4 − 53 3 3 + 53 2 − 80] 2 5 ℎ = 4 [(−625 + 1875 − 6625 3 + 1325 − 400) − (− 32 5 + 48 − 424 3 + 212 − 160)] ℎ = 4 [(− 100 3 ) − (− 716 15 )] ℎ = 4 [ 72 5 ] ℎ = 288 5 3 #
MAT183/ MAT421 : Calculus I 270 Example 18 (iii) top bottom = 2( − 3) 2 = 2 = 5 ) = + 1 ℎℎ = − = (2 − 2) − 2( − 3) 2 = 2 − 2 − 2( 2 − 6 + 9) = 2 − 2 − 2 2 + 12 − 18 = 14 − 20 − 2 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =5 =2 ℎ = 2 ∫ ( + 1)(14 − 20 − 2 2 ) 5 2 ℎ = 2 ∫ (14 2 − 20 − 2 3 + 14 − 20 − 2 2 ) 5 2 ℎ = 2 ∫ (12 2 − 6 − 2 3 − 20) 5 2 ℎ = 2 [ 12 3 3 − 6 2 2 − 2 4 4 − 20] 2 5 ℎ = 2 [4 3 − 3 2 − 4 2 − 20] 2 5 ℎ = 2 [(500 − 75 − 625 2 − 100) − (32 − 12 − 8 − 40)] ℎ = 2 [ 81 2 ] ℎ = 81 3 # How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232) = −1 (axis of rotation) = −1 (axis of rotation) height
Chapter 5 : Applications of Definite Integrals 271 Example 19/ MAR2004/ MAT149/ Q15b The shaded region in Figure 26 below is bounded by the curve y = 9 − x 2 and the line y = 9 − 2x. i) Using the Washer Method, find the volume of the revolution obtained when the shaded region is revolved about the x-axis. ii) Using the Shell method, set up and simplify the integral to find the volume of the revolution obtained when the shaded region is revolved about the line x = 3. Solution : y Figure 26 x y = 9 − 2x 9 0 y = 9 − x 2 −3 3 ) = 9 − 2 = 9 − 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =3 =0 ℎ = ∫ (9 − 2 ) 2 − (9 − 2) 2 3 0 ℎ = ∫ (81 − 18 2 + 4 ) − (81 − 36 + 4 2 ) 3 0 ℎ = ∫ (36 − 22 2 + 4 ) 3 0 ℎ = [ 36 2 2 − 22 3 3 + 5 5 ] 0 3 ℎ = [18 2 − 22 3 3 + 5 5 ] 0 3 ℎ = [(162 − 198 + 243 5 ) − 0] ℎ = 63 5 3 # = 0 (axis of rotation) = 0 = 3 outer = 9 − 2 = 3 (axis of rotation) height ) = 3 − ℎℎ = − = (9 − 2 ) − (9 − 2) = 2 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =3 =0 ℎ = 2 ∫ (3 − )(2 − 2 ) 3 0 ℎ = 2 ∫ (6 − 3 2 − 2 2 + 3 ) 3 0 ℎ = 2 ∫ (6 − 5 2 + 3 ) 3 0 ℎ = 2 [ 6 2 2 − 5 3 3 + 4 4 ] 0 3 ℎ = 2 [3 2 − 5 3 3 + 4 4 ] 0 3 ℎ = 2 [(27 − 45 + 81 4 ) − 0] ℎ = 2 [ 9 4 ] ℎ = 9 2 3 # How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232) = 3 (axis of rotation) = 9 − 2 = 9 − 2 top bottom
MAT183/ MAT421 : Calculus I 272 Example 20/ MAR 2004/ MAT183/ Q4b The region R is bounded by the curve x = y 2 + 1 and line y = 3 − x. Find the volume of the solid obtained by revolving R about the line i) x = 0 ii) y = 3 (10 marks) Solution : /. y x y = 3 − x R x = y 2 + 1 0 (5, −2) (2, 1) Figure 28 ) = 3 − = 2 + 1 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =1 =−2 ℎ = ∫ (3 − ) 2 − ( 2 + 1) 2 1 −2 ℎ = ∫ (9 − 6 + 2 ) − ( 4 + 2 2 + 1) 1 −2 ℎ = ∫ (8 − 6 − 2 − 4 ) 1 −2 ℎ = [8 − 6 2 2 − 3 3 − 5 5 ] −2 1 ℎ = [8 − 3 2 − 3 3 − 5 5 ] −2 1 ℎ = [(8 − 3 − 1 3 − 1 5 ) − (−16 − 12 + 8 3 + 32 5 )] ℎ = 117 5 3 # = 0 (axis of rotation) = 3 − = 2 + 1 (washer) = 0 (washer) outer inner = −2 = 1 ) = 3 − ℎℎ = ℎ − = (3 − ) − ( 2 + 1) = 2 − − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =1 =−2 ℎ = 2 ∫ (3 − )(2 − − 2 ) 1 −2 ℎ = 2 ∫ (6 − 3 − 3 2 − 2 + 2 + 3 ) 1 −2 ℎ = 2 ∫ (6 − 5 − 2 2 + 3 ) 1 −2 ℎ = 2 [6 − 5 2 2 − 2 3 3 + 4 4 ] −2 1 ℎ = 2 [(6 − 5 2 − 2 3 + 1 4 ) − (−12 − 10 + 16 3 + 4)] ℎ = 2 [ 63 4 ] ℎ = 63 2 3 # How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232) = 3 (axis of rotation) (shell) = 3 (washer) = 3 − right = 2 + 1 left height
Chapter 5 : Applications of Definite Integrals 273 Example 21/ APR 2006/ MAT183/ Q5b The curve y 2 = 6 − 2x intersects the line y = 2x at the point A and cuts the x-axis at the point B as shown in Figure 29. i) Find the coordinates of A and B. ii) Find the area of the shaded region. iii) Using the Washer method, set up the integral to find the volume of the solid obtained when the shaded region is revolved about the line x = −1. Do not evaluate the integral. (11 marks) Solution : y Figure 29 y 2 = 6 − 2x y = 2x 0 x B A 2 = 6 − 2 2 = 6 − 2 = 3 − 2 2 ) = − − − ① 2 = 6 − − − − ② ① into ② 2 = 6 − 2 + − 6 = 0 ( − 2)( + 3) = 0 = 2, = −3 Since A above the x-axis, so choose = 2 When = 2 ( ①) 2 = 2 → = 1 (, ) = 0 − − − ① 2 = 6 − 2 − − − ② ① into ② 0 = 6 − 2 2 = 6 = 3 (, )
MAT183/ MAT421 : Calculus I 274 Example 21 (ii) Solution : Example 21 (iii) left right = 3 − 2 2 = 2 = 0 ) = (3 − 2 2 ) − (−1) = 4 − 2 2 == ( 2 ) − (−1) = 2 + 1 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =2 =0 ℎ = ∫ (4 − 2 2 ) 2 − ( 2 + 1) 2 2 0 ℎ = ∫ (16 − 4 2 + 4 4 ) − ( 2 4 + + 1) 2 0 ℎ = ∫ (15 − − 17 2 4 + 4 4 ) 2 0 ℎ = [15 − 2 2 − 17 3 12 + 5 20] 0 2 ℎ = [(30 − 2 − 136 12 + 32 20) − 0] ℎ = 274 15 3 # ) = ∫ (ℎ − ) =2 =0 = ∫ ((3 − 2 2 ) − ( 2 )) 2 0 = ∫ (3 − 2 2 − 2 ) 2 0 = [3 − 3 6 − 2 4 ] 0 2 = (6 − 4 3 − 1) − 0 = 11 3 2 = −1 (axis of rotation) outer = 3 − 2 2 = 0 = −1 (axis of rotation) = 2
Chapter 5 : Applications of Definite Integrals 275 Example 22 The shaded region in Figure 30 below is bounded by the curve y = x and the curve y = x 3 . Find the volume generated when the shaded region R is rotated through the line x = 1, using : a) Washers method b) Cylindrical shells method. Solution : ) = 1 − ℎℎ = − = √ − 3 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =1 =0 ℎ = 2 ∫ (1 − )(√ − 3 ) 1 0 ℎ = 2 ∫ (1 − ) ( 1 2 − 3) 1 0 ℎ = 2 ∫ ( 1 2 − 3 − 3 2 + 4) 1 0 ℎ = 2 [ 2 3 3 2 − 4 4 − 2 5 5 2 + 5 5 ] 0 1 ℎ = 2 [( 2 3 − 1 4 − 2 5 + 1 5 ) − 0] ℎ = 2 [ 13 60] ℎ = 13 30 3 # ) = 1 − 2 = 1 − 1 3 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =1 =0 ℎ = ∫ (1 − 2 ) 2 − (1 − 1 3) 2 1 0 ℎ = ∫ (1 − 2 2 + 4 ) − (1 − 2 1 3 + 2 3) 1 0 ℎ = ∫ (−2 2 + 4 + 2 1 3 − 2 3) 1 0 ℎ = [ −2 3 3 + 5 5 + 2 ∙ 3 4 4 3 − 3 5 5 3] 0 1 ℎ = [ −2 3 3 + 5 5 + 3 2 4 3 − 3 5 5 3] 0 1 ℎ = [(− 2 3 + 1 5 + 3 2 − 3 5 ) − 0] ℎ = 13 30 3 # Figure 30 How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232)
MAT183/ MAT421 : Calculus I 276 Example 22 The shaded region in Figure 30 below is bounded by the curve y = x and the curve y = x 3 . Find the volume generated when the shaded region R is rotated through the line x = 1, using : a) Washers method b) Cylindrical shells method. Solution : ) = ℎℎ = − = () − ( 2 − 2) = − 2 + 2 = 3 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =3 =0 ℎ = 2 ∫ ()(3 − 2 ) 3 0 ℎ = 2 ∫ (3 2 − 3 ) 3 0 ℎ = 2 [ 3 3 3 − 4 4 ] 0 3 ℎ = 2 [ 3 − 4 4 ] 0 3 ℎ = 2 [(27 − 81 4 ) − 0] ℎ = 27 2 3 # ) = 1 − 2 = 1 − 1 3 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =1 =0 ℎ = ∫ (1 − 2 ) 2 − (1 − 1 3) 2 1 0 ℎ = ∫ (1 − 2 2 + 4 ) − (1 − 2 1 3 + 2 3) 1 0 ℎ = ∫ (−2 2 + 4 + 2 1 3 − 2 3) 1 0 ℎ = [ −2 3 3 + 5 5 + 2 ∙ 3 4 4 3 − 3 5 5 3] 0 1 ℎ = [ −2 3 3 + 5 5 + 3 2 4 3 − 3 5 5 3] 0 1 ℎ = [(− 2 3 + 1 5 + 3 2 − 3 5 ) − 0] ℎ = 13 30 3 # = 1 (axis of rotation) = 1 = 0 Figure 30 = 0 = 1 = 1 (axis of rotation) height radius
Chapter 5 : Applications of Definite Integrals 277 Example 23 Figure 31 shows the shaded region bounded by the graphs of y = x and line y = 2. a) Determine the area of the shaded region. (3 marks) b) Find the volume generated when the shaded region is rotated through 360o about the i) y-axis (6 marks) ii) line y = 2 (6 marks) c) Find the volume generated when the shaded region is rotated through 360o about the line x = −1, using : i) Volume by washers. (6 marks) ii) Volume by cylindrical shells. (6 marks) Solution : y = y = 2 y x 0 R Figure 31 = 2 −− − ① = √ − −− ② ② into ① : √ = 2 = 4 = 4 ) = ∫ ( − ) =4 =0 = ∫ ((2) − (√)) 4 0 = ∫ (2 − 1 2) 4 0 = [2 − 2 3 3 2] 0 4 = [(8 − 16 3 ) − 0] = 8 3 2 y = 2 = 0 = 4 = √ top bottom
MAT183/ MAT421 : Calculus I 278 Example 23b) (i), (ii) b) ) = 2 = ∫ ( 2 ) 2 =2 =0 = ∫ 4 2 0 = [ 5 5 ] 0 2 = [ 2 5 5 − 0] = 32 5 3 # ) ) = 2 − √ Using Washer method = ∫ () 2 =4 =0 = ∫ (2 − √) 2 4 0 = ∫ (4 − 4√ + ) 4 0 = ∫ (4 − 4 1 2 + ) 4 0 = [4 − 4 ∙ 2 3 3 2 + 2 2 ] 0 4 = [4 − 8 3 3 2 + 2 2 ] 0 4 = [(16 − 64 3 + 8) − 0] = 8 3 3 # = 2 (axis of rotation) = 0 = 4 = √
Chapter 5 : Applications of Definite Integrals 279 Example 23c) (i) x = −1 x = y 2 y = 0 y = 2 x = −1 x = y 2 y = 0 y = 2 (outer) x = 0 (inner) ) ) = ( 2 ) − (−1) = 2 + 1 = (0) − (−1) = 1 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =2 =0 ℎ = ∫ (( 2 + 1) 2 − (1) 2 ) 2 0 ℎ = ∫ (( 4 + 2 2 + 1) − (1)) 2 0 ℎ = ∫ ( 4 + 2 2 ) 2 0 ℎ = [ 5 5 + 2 3 3 ] 0 2 ℎ = [( 32 5 + 16 3 ) − 0] ℎ = 176 15 3 #
MAT183/ MAT421 : Calculus I 280 Example 23c) (ii) ) ) = + 1 ℎℎ = − = 2 − √ Using Shell method ℎ = 2 ∫ ()(ℎℎ) =4 =0 ℎ = 2 ∫ ( + 1)(2 − √) 4 0 ℎ = 2 ∫ (2 +2 − 3 2 − 1 2) 4 0 ℎ = 2 [ 2 2 2 + 2 − 2 5 5 2 − 2 3 3 2] 0 4 ℎ = 2 [ 2 + 2 − 2 5 5 2 − 2 3 3 2] 0 4 ℎ = 2 [(16 + 8− 64 5 − 16 3 ) −0] ℎ = 176 15 3 #
Chapter 5 : Applications of Definite Integrals 281 Example 24 Figure 32 shows the shaded region bounded by the curve y = x 2 − x − 4 and line y = − x. Figure 32 Find the volume generated when the shaded region is rotated through 360o about the line x = −2. (8 marks) Solution : (−2, 2) (2, −2) y = x 2 − x − 4 y = − x y x 0 = −2 (axis of rotation) (shell) = 0 (washer) = +2 ℎℎ = − = (−)− ( 2 − − 4) = 4 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =2 =−2 ℎ = 2 ∫ ( + 2)(4 − 2 ) 2 −2 ℎ = 2 ∫ (4 − 3 + 8 − 2 2 ) 2 −2 ℎ = 2 [ 4 2 2 − 4 4 + 8 − 2 3 3 ] −2 2 ℎ = 2 [2 2 − 4 4 + 8 − 2 3 3 ] −2 2 ℎ = 2 [(8 −4 + 16 − 16 3 )− (8 − 4 −16 + 16 3 )] ℎ = 2 [ 64 3 ] ℎ = 128 3 3 # = 2 − − 4 = − height top bottom = −2 (axis of rotation) = 2
MAT183/ MAT421 : Calculus I 282 Example 25 Figure 33 shows the shaded region R bounded by the curve x = y 2 + 1 and line y = 3 − x. Figure 33 Find : a) the area of the shaded region R. (3 marks) b) the volume generated when the shaded region R is rotated through 360o about the line x = 1 (use Washer Method) (7 marks) Solution : (5, −2) x = y 2 +1 x 0 R y (2, 1) y = 3 − x ) = ∫ (ℎ − ) =1 =−2 = ∫ ((3 − ) − ( 2 + 1)) 1 −2 = ∫ (2 − − 2 ) 1 −2 = [2 − 2 2 − 3 3 ] −2 1 = (2 − 1 2 − 1 3 ) − (−4 − 2 + 8 3 ) = 9 2 2 ) = (3 − ) − 1 = 2 − = ( 2 + 1) − 1 = 2 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =1 =−2 ℎ = ∫ (2 − ) 2 − ( 2 ) 2 1 −2 ℎ = ∫ (4 − 4 + 2 ) − ( 4 ) 1 −2 ℎ = ∫ (4 − 4 + 2 − 4 ) 1 −2 ℎ = [4 − 4 2 2 + 3 3 − 5 5 ] −2 1 ℎ = [4 − 2 2 + 3 3 − 5 5 ] −2 1 ℎ = [(4 − 2 + 1 3 − 1 5 ) − (−8 − 8 − 8 3 + 32 5 )] ℎ = 72 5 3 # = 3 − right left = 2 + 1 = −2 = 1 = 1 (axis of rotation) = −2 = 1
Chapter 5 : Applications of Definite Integrals 283 Example 26 The figure 34 shows the shaded region bounded by the parabola 2 = and the line 5 = + 4. a) Determine the area of the shaded region. b) Find the volume generated when the shaded region is rotated 360o through the line y = −1. Solution : 5y = x + 4 y 2 = x x y (1, 1) (16, 4) Figure 34 = 5 − 4 right left = 2 = 4 = 1 ) = ∫ (ℎ − ) =4 =1 = ∫ ((5 − 4) − ( 2 )) 4 1 = ∫ (5 − 4 − 2 ) 4 1 = [ 5 2 2 − 4 − 3 3 ] 1 4 = (40 − 16 − 64 3 ) − ( 5 2 − 4 − 1 3 ) = 9 2 2
MAT183/ MAT421 : Calculus I 284 END OF CHAPTER 5 Example 26 (b) Solution : /. END OF SYLLABUS (CALCULUS 1) ALL THE BEST FOR YOUR COMING FINAL ☺ ☺ ) = + 1 ℎℎ = ℎ − = (5 − 4) − ( 2 ) = 5 − 4 − 2 Using Shell method ℎ = 2 ∫ ()(ℎℎ) =4 =1 ℎ = 2 ∫ ( + 1)(5 − 4 − 2 ) 4 1 ℎ = 2 ∫ (5 2 − 4 − 3 + 5 − 4 − 2 ) 4 1 ℎ = 2 ∫ ( + 4 2 − 3 − 4) 4 1 ℎ = 2 [ 2 2 + 4 3 3 − 4 4 − 4] 1 4 ℎ = 2 [(8 + 256 3 − 64 − 16) − ( 1 2 + 4 3 − 1 4 − 4)] ℎ = 2 [ 63 4 ] ℎ = 63 2 3 # How to find the radius for Shell Method?? - Refer video 5.4.1 – 5.4.5 - Refer note book (page 232) (shell) = −1 (washer) = 5 − 4 right = 2 left = −1 (axis of rotation) height (washer) = −1 (washer) = 1 5 ( + 4) inner = √ outer = −1 = 1 = 16 (axis of rotation) = 1 = 4 ) = (√) − (−1) = √ + 1 = ( 5 + 4 5 ) − (−1) = 5 + 9 5 Using Washer method ℎ = ∫ (. ) 2 − (. ) 2 =16 =1 ℎ = ∫ (√ + 1) 2 − ( 5 + 9 5 ) 2 16 1 ℎ = ∫ ( + 2√ + 1) − ( 2 25 + 18 25 + 81 25) 16 1 ℎ = ∫ ( 7 25 + 2 1 2 − 56 25 − 2 25) 16 1 ℎ = [ 7 2 50 + 2 ∙ 2 3 3 2 − 56 25 − 3 75] 1 16 ℎ = [ 7 2 50 + 4 3 3 2 − 56 25 − 3 75] 1 16 ℎ = [( 1792 50 + 256 3 − 896 25 − 4096 75 ) − ( 7 50 + 4 3 − 56 25 − 1 75)] ℎ = 63 2 3 #
CONFIDENTIAL 7 CS/JUL 2022/MAT421 © Hak Cipta Universiti Teknologi MARA CONFIDENTIAL APPENDIX 1 RULES OF DIFFERENTIATION 1. Product rule ( ( ) ( )) d ( ) ( ) ( ) ( ) d f x g x g x f x f x g x x = + 2. Quotient rule ( ) 2 d ( ) ( ) ( ) ( ) ( ) d ( ) ( ) f x g x f x f x g x x g x g x − = 3. Power rule ( ) d 1 ( ) ( ) ( ) d n n f x n f x f x x − = 4. Chain rule ( ) ( ) d ( ) ( ) ( ) d f g x f g x g x x = TABLE OF INTEGRALS 1. ( ) ( ) 1 ) ; 1 d ( 1) 1 ln ; 1 n n ax b C n ax b x a n ax b C n a + + + − + = + + + = 2. x x C x = + d ln 1 3. = − ax + C a ax x cos 1 sin d 4. = ax + C a ax x sin 1 cos d 5. = ax +C a ax x tan 1 sec d 2 6. = − ax + C a ax x cot 1 csc d 2
CONFIDENTIAL 8 CS/JUL 2022/MAT421 © Hak Cipta Universiti Teknologi MARA CONFIDENTIAL APPENDIX 2 DEFINITION OF DIFFERENTIATION ( ) ( ) h f x h f x f x h ( ) lim 0 + − = → LINEAR APPROXIMATION f x f x f x x x ( ) + − ( 0 0 0 ) ( ) ( ) THE SQUEEZING THEOREM 1. 1 sin lim 0 = → x x x 2. 0 1 cos lim 0 = − → x x x SIGMA NOTATION 1. n n k ak = a + a + a + + a = ... 1 2 3 1 2. ( ) 1 1 1 2 3 ... 2 n k n n k n = + = + + + + = 3. ( )( ) 2 2 2 2 2 1 1 2 1 1 2 3 ... 6 n k n n n k n = + + = + + + + = 4. ( ) 2 3 3 3 3 3 1 1 1 2 3 ... 2 n k n n k n = + = + + + + =