Tugas fisika untuk matematika 2

Tugas fisika untuk matematika 2

• 1.
  TUGAS FISIKA UNTUKMATEMATIKA 2 Penyelesaian Soal Soal Chapter 5 Dosen Pengampu : Drs. Pujayanto, MSi DISUSUN OLEH MAY NURHAYATI (K2315048) PROGRAM STUDI PENDIDIKAN FISIKA FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN UNIVERSITAS SEBELAS MARET SURAKARTA 2016
• 2.
  Section 2 2.1 ∫∫ 𝟑𝐱 𝐝𝐲𝐝𝐱 𝟒 𝐲=𝟐 𝟏 𝐱=𝟎 =∫ [3𝑥𝑦]2 4 𝑑𝑥 1 𝑥=0 = ∫ (3x. 4 − 3x. 2)𝑑𝑥 1 𝑥=0 = ∫ (12x − 6x)𝑑𝑥 1 𝑥=0 = ∫ (6x)𝑑𝑥 1 𝑥=0 = [3𝑥2]0 1 = 3 2.5 ∫ ∫ 𝐲 𝐝𝐲𝐝𝐱 𝐞 𝐱 𝐲=𝐱 𝟏 𝐱=𝟎 =∫ [ 1 2 y2 ] x ex dx 1 x=0 =∫ ( 1 2 (ex)2 − 1 2 x2 )dx 1 x=0 = 1 2 ∫ (e2x − x2)dx 1 x=0 = 1 2 [ 1 2 e2x − 1 3 x3 ] 0 1 = ( 1 4 e2 − 1 6 ) – ( 1 4 − 0) = ( 1 4 e2 − 1 6 ) – ( 1 4 − 0) = 1 4 𝑒2 − 2−3 12 = 1 4 𝑒2 − 5 12 2.11 𝒘𝒉𝒆𝒓𝒆 𝑨 𝒊𝒔 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒂 𝒚 = 𝒙 𝟐 𝒂𝒏𝒅 𝒕𝒉𝒆 𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆 𝟐𝒙 − 𝒚 + 𝟖 = 0.  Penyelesaian: Menentukanbatas y x = x 𝑥2 = 𝑥2 𝑦 = ( 𝑦 − 8 2 ) 2 𝑦 = 𝑦2 − 16𝑦 + 64 4 4𝑦 = 𝑦2 − 16𝑦 + 64 𝑦2 − 20𝑦 + 64 = 0 ( 𝑦 − 16) 𝑉( 𝑦 − 4) 𝑦 = 16 𝑉 𝑦 = 4 Menentukanbatas x
• 3.
  𝑦 = 𝑦 𝑥2= 2𝑥 + 8 𝑥2 − 2𝑥 − 8 = 0 ( 𝑥 − 4) 𝑉 (𝑥 + 2) 𝑥 = 4 𝑉 𝑥 = −2 Menentukanluas ∫ ∫ 𝑥𝑑𝑥𝑑𝑦 16 𝑦=4 4 𝑥=−2 = ∫ [ 𝑥𝑦]4 16 4 𝑥=−2 𝑑𝑥 = ∫ [16𝑥 − 4𝑥] 𝑑𝑥 4 −2 = ∫ [12𝑥] 𝑑𝑥 4 −2 = [6𝑥2]−2 4 = 96 − 24 = 72 2.39 ∫ ∫ ∫ 𝟔𝐲𝐝𝐱𝐝𝐳𝐝𝐲 = ∫ ∫ [6xy]y+z 2y+z2 z=1 dzdy 3 y=−2 𝟐𝐲+𝐳 𝐱=𝐲+𝐳 𝟐 𝐳=𝟏 𝟑 𝐲=−𝟐 = ∫ ∫ (6(2y+ z)y − 6(y + z)y 2 z=1 dzdy 3 y=−2 = ∫ ∫ 6((2y2 + yz) − (y2 + yz)) 2 z=1 dzdy 3 y=−2 = ∫ ∫ 6(y2 ) 2 z=1 dzdy 3 y=−2 = ∫ ∫ (6y2 ) 2 z=1 dzdy 3 y=−2 = ∫ [6y2 z]z=1 2 dy 3 y=−2 = ∫ (12y2 − 6y2 )dy 3 y=−2 = ∫ (6y2 )dy 3 y=−2 = [6 . 1 3 y3 ] −2 3 = [2y3]−2 3 = 54 − (−16) = 70
• 4.
  Section 3 3.3 Athin rod 10 fr longhas a densitywhich variesuniformlyfrom 4 to 24 lb/ft.Find M, 𝒙̅, 𝑰 𝒎,𝒂𝒏𝒅 𝑰.  Penyelesaian: 4 2 24 (𝜆) 0 x 10 (x) 𝑥−0 10−0 = 𝜆−4 24−4 𝑥 10 = 𝜆−4 20 2𝑥 = 𝜆 − 4 𝜆 = 2𝑥 + 4 𝑎. 𝑀 = ∫ 𝜆 𝑑𝑥 = ∫ (2𝑥 + 4) 𝑑𝑥 = [ 𝑥2 + 4𝑥]0 10 10 𝑥=0 = 100 + 40 = 140 𝑏. 𝑥̅ = ∫ 𝑥 𝑑𝑚 ∫ 𝑑𝑚 = ∫ 𝑥 (2𝑥+4) 𝑑𝑥 ∫(2𝑥+4) = ∫ (2𝑥2+4𝑥) 𝑑𝑥 ∫(2𝑥+4) = [ ( 2 3 𝑥3+2𝑥2) ( 𝑥2+4𝑥) ] 0 10 = [ ( 2 3 (1000)+2(100)) (100+40) ] = 130 21 𝑐. 𝐼 𝑦0 = ∫ 𝑥2 𝑑𝑚 = ∫ 𝑥2(2𝑥 + 4) 𝑑𝑥 10 𝑥=0 = ∫ 2𝑥3 + 4𝑥210 𝑥=0 = [ 2 4 𝑥4 + 4 3 𝑥3] 0 10 = [ 2 4 (10000) + 4 3 (1000)] = 5000 + 4000 3 = 19000 3 𝑑. 𝐼 = 𝐼𝑐𝑚 + 𝑚𝑑2 𝐼 𝑐𝑚 = 𝐼 − 𝑚𝑑2 𝐼 𝑐𝑚 = 19000 3 − 140. 130 21 (6. 4 2 )
• 5.
  = 19000 3 − 1280 7 𝐼 = 𝐼𝑐𝑚+ 𝑚𝑑2 = ( 19000 3 − 1280 7 )+ 140. 80 24 = ( 19000 3 − 1280 7 )+ 140. 80 24 = ( 19000 3 − 1280 7 )+ 1400 3 = 133000−3840+9800 21 = 138960 21 = 46320 7 3.7 A rectangular lamina has vertices (0,0), (0,2), (3,0), (3,2) and density xy. Find: a) 𝒎 = ∫ 𝝆𝒅𝒎 = ∫ ∫ 𝑥𝑦𝑑𝑥𝑑𝑦 3 𝑥=0 2 𝑦=0 = ∫ ∫ 𝑥𝑑𝑥𝑦𝑑𝑦 3 𝑥=0 2 𝑦=0 = ∫ ( 1 2 𝑥2 ) 𝑥=0 32 𝑦=0 = ∫ ( 9 2 ) 2 𝑦=0 𝑦𝑑𝑦 = ( 9 4 𝑦2 ) 𝑦=0 2 = 9 b) 𝒙̅ = ∫ 𝒙𝒅𝒎 ∫ 𝒅𝒎 = ∫ ∫ 𝑥2 𝑦𝑑𝑥𝑑𝑦 9 = ∫ ∫ 𝑥2 𝑑𝑥𝑦𝑑𝑦 3 𝑥=0 2 𝑦=0 9 = ∫ ( 1 3 𝑥3) 0 3 𝑦𝑑𝑦 2 𝑦=0 9 = ∫ 9𝑦𝑑𝑦 2 𝑦=0 9 = ( 9 2 𝑦2) 0 2 9 = 18 9 = 2 𝒚̅ = ∫ 𝒚𝒅𝒎 ∫ 𝒅𝒎 = ∫ ∫ 𝑦2 𝑥𝑑𝑥𝑑𝑦 9
• 6.
  = ∫ ∫ 𝑥𝑑𝑥𝑦2 𝑑𝑦 3 𝑥=0 2 𝑦=0 9 = ∫( 1 2 𝑥2) 0 3 𝑦2 𝑑𝑦 2 𝑦=0 9 = ∫ 9 2 𝑦2 𝑑𝑦 2 𝑦=0 9 = ( 9 6 𝑦3) 0 2 9 = 12 9 = 4 3 c) 𝑰 𝒙 = ∫ 𝒚 𝟐 𝒅𝒎 = ∫ ∫ 𝑦3 𝑥𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑦3 𝑑𝑦𝑥𝑑𝑥 2 𝑦=0 3 𝑥=0 = ∫ ( 1 4 𝑦4 ) 0 2 𝑥𝑑𝑥 3 𝑥=0 = ∫ 4𝑥𝑑𝑥 3 𝑥=0 = (2𝑥2)0 3 = 18 = 2𝑚 𝐼 𝑦 = ∫ 𝑥2 𝑑𝑚 = ∫ ∫ 𝑥3 𝑦𝑑𝑥𝑑𝑦 = ∫ ∫ 𝑦𝑑𝑦𝑥3 𝑑𝑥 2 𝑦=0 3 𝑥=0 = ∫ ( 1 2 𝑦2 ) 0 2 𝑥3 𝑑𝑥 3 𝑥=0 = ∫ 2𝑥3 𝑑𝑥 3 𝑥=0 = ( 1 2 𝑥4 ) 0 3 = 81 2 = 9 2 𝑚 d) 𝑰 𝒁 = 𝑰 𝒙 + 𝑰 𝒚 = 18 + 81 2 = 117 2 = 13 18 𝑚 3.21. Kurva y = √ 𝒙 diantara x=0 dan x=2 Tentukan : a) the centroids of the arc b) the volume c) the surface area jawab: a) ∫ 𝑥̅ 𝑑𝑚 = ∫ 𝑥 𝑑𝑚
• 7.
  = ∫ ∫𝑥 𝑑𝑦𝑑𝑥 √ 𝑥 𝑦=0 2 𝑥=0 = ∫ 𝑥 ]0 √ 𝑥 𝑑𝑥 2 𝑥=0 =∫ √ 𝑥 2 𝑥=0 𝑑𝑥 = 2 3 𝑥 3 2]0 2 = 4 3 √2 ∫ 𝑦̅ 𝑑𝑚 = ∫ 𝑦 𝑑𝑚 =∫ ∫ 𝑦 𝑑𝑦𝑑𝑥 √ 𝑥 𝑦=0 2 𝑥=0 =∫ 1 2 𝑦2]0 √ 𝑥 𝑑𝑥 2 𝑥=0 =∫ 1 2 𝑥 𝑑𝑥 2 𝑥=0 = 1 4 𝑥2]0 2 = 1 b) ?????????????????????????????????? c) dA = 2𝜋𝑦 𝑑𝑠 A = ∫ 2𝜋𝑥𝑦 𝑑𝑠 2 𝑥=0 A = ∫ 2𝜋√1 + 4𝑥22 𝑥=0 𝑑𝑥 A = 2𝜋 (1 + 𝑋2)]0 2 A = 10𝜋 3.28 For the curve 𝒚 = √ 𝒙. 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒙 = 𝟎 𝒂𝒏𝒅 𝒙 = 𝟐, 𝒇𝒊𝒏𝒅 𝒕he mass of a wire bent in the shape of the arc if its density (mass per unit length) is √ 𝒙.  Penyelesaian: dm = σ dA M =∫ 𝑑𝑚 σ = √ 𝑥 M = ∫ 𝜎 𝑑𝐴 = ∬ 𝜎 𝑑𝑥 𝑑𝑦 = ∫ ∫ 𝜎 𝑑𝑥 𝑑𝑦 √ 𝑥 𝑦=0 2 𝑥=0 = ∫ 𝑦 ]0 √2 𝜎 𝑑𝑥 2 𝑥=0 =∫ √ 𝑥 2 𝑥=0 √ 𝑥 𝑑𝑥 = ∫ 𝑥 𝑑𝑥 2 𝑥=0 = 1 2 𝑥2]0 2 = 1 2 (2 − 0)2 = 1 2 x 4 = 2
• 8.
  4.1 a. The areaof the disk Dengandi misalkanlingkaranbiruadalahA1dan lingkarandalamdimisalkanA2.Maka luasannya= A1-A2  𝑨 𝟏 = ∫ ∫ 𝒓 𝒅𝒓 𝒅𝜽 𝒂 𝒓=𝟎 𝟐 𝜽=𝟎 𝐴1 = ∫ 1 2 𝑟2]0 𝑎 𝑑𝜃 2 𝜃=0 𝐴1 = ∫ 1 2 𝑎2 − 0 𝑑𝜃 2 𝜃=0 𝐴1 = ∫ 1 2 𝑎2 𝑑𝜃 2 𝜃=0 𝐴1 = 1 2 𝑎2 𝜃] 0 2𝜋 𝐴1 = 1 2 𝑎22𝜋 − 0 𝐴1 = 𝑎2 𝜋  𝑨 𝟐 = ∫ ∫ 𝒓 𝒅𝒓 𝒅𝜽 𝒓 𝒓=𝟎 𝟐 𝜽=𝟎 𝐴2 = ∫ 1 2 𝑟2]0 𝑟 𝑑𝜃 2 𝜃=0 𝐴2 = ∫ 1 2 𝑟2 − 0 𝑑𝜃 2 𝜃=0 𝐴2 = ∫ 1 2 𝑟2 𝑑𝜃 2 𝜃=0 𝐴2 = 1 2 𝑟2 𝜃] 0 2𝜋 𝐴2 = 1 2 𝑟22𝜋− 0 𝐴2 = 𝑟2 𝜋 𝑨 𝟏 − 𝑨 𝟐 = 𝒂 𝟐 𝝅− 𝒓 𝟐 𝝅=𝝅(𝒂 𝟐 − 𝒓 𝟐)
• 9.
  b. The centroidofone quardiant of the disk  𝑥̅ = ∫ ∫ 𝑥 𝑑𝑥 𝑑𝑦 𝑎 𝑟 𝑎 0 𝑥̅ = ∫ 1 2 𝑥2] 𝑟 𝑎 𝑑𝑦 𝑎 0 𝑥̅ = ∫ 1 2 𝑎2 − 1 2 𝑟2 𝑑𝑦 𝑎 0 𝑥̅ = [ 1 2 𝑎2 𝑦 − 1 2 𝑟2 𝑦] 0 𝑎 𝑥̅ = 1 2 𝑎2(𝑎) − 1 2 𝑟2(𝑎) 𝑥̅ = 1 2 𝑎3 − 1 2 𝑎𝑟2  𝑦̅ = ∫ ∫ 𝑥 𝑑𝑥 𝑑𝑦 𝑎 𝑟 𝑎 0 𝑦̅ = ∫ 1 2 𝑥2] 𝑟 𝑎 𝑑𝑦 𝑎 0 𝑦̅ = ∫ 1 2 𝑎2 − 1 2 𝑟2 𝑑𝑦 𝑎 0 𝑦̅ = [ 1 2 𝑎2 𝑦 − 1 2 𝑟2 𝑦] 0 𝑎 𝑦̅ = 1 2 𝑎2(𝑎) − 1 2 𝑟2(𝑎) 𝑦̅ = 1 2 𝑎3 − 1 2 𝑎𝑟2 c. The momentof inertiaof the diskabout diameter 𝐼 𝑥 = ∫ 𝑦2 𝑑𝑚 𝐼 𝑥 = ∫∫ 𝑦2 𝜎 𝑑𝐴 𝐼 𝑥 = ∫∫ 𝑦2 𝜎 𝑟 𝑑𝑟 𝑑 𝜃 𝐼 𝑥 = ∫ ∫ 𝑦2 𝜎 𝑟 𝑑𝑟 𝑑𝜃 𝑎 𝑟=𝑟 𝜋 2 𝜃=0 𝐼 𝑥 = ∫ ∫ (r sin 𝜃)2 𝜎 𝑟 𝑑𝑟 𝑑𝜃 𝑎 𝑟=𝑟 𝜋 2 𝜃=0 𝐼 𝑥 = ∫ ∫ 𝑟2 𝑠𝑖𝑛2 𝜃 𝜎 𝑟 𝑑𝑟 𝑑𝜃 𝑎 𝑟=𝑟 𝜋 2 𝜃=0 𝐼 𝑥 = 𝜎 ∫ ∫ 𝑟3 𝑠𝑖𝑛2 𝜃 𝑑𝑟 𝑑𝜃 𝑎 𝑟=𝑟 𝜋 2 𝜃=0 𝐼 𝑥 = 𝜎 ∫ 1 4 𝑟4] 𝑟 𝑎 𝑠𝑖𝑛2 𝜃 𝑑𝜃 𝜋 2 𝜃=0 𝐼 𝑥 = 𝜎 ∫ ( 1 4 𝑎4 − 1 4 𝑟4) 𝑠𝑖𝑛2 𝜃 𝑑𝜃 𝜋 2 𝜃=0
• 10.
  𝐼 𝑥 =𝜎 [( 1 4 𝑎4 − 1 4 𝑟4)( 1 2 𝜃 − 1 4 sin 2𝜃)] 𝜃=0 𝜋 2 𝐼 𝑥 = 𝜎 ( 1 4 𝑎4 − 1 4 𝑟4)( 1 2 𝜋 2 − 1 4 sin 2 𝜋 2 − 1 2 0 − 1 4 sin 2𝑜) 𝐼 𝑥 = 𝜎 ( 1 4 𝑎4 − 1 4 𝑟4)( 1 2 𝜋 2 − 1 4 sin 2 𝜋 2 − 0 − 0) 𝐼 𝑥 = 𝜎 ( 1 4 𝑎4 − 1 4 𝑟4)( 1 2 𝜋 2 − 1 4 sin 2 𝜋 2 ) 𝐼 𝑥 = 𝜎 ( 1 4 𝑎4 − 1 4 𝑟4)( 𝜋 4 − sin 𝜋 4 ) 𝐼 𝑥 = 𝜎 ( 1 4 𝑎4 − 1 4 𝑟4)( 𝜋 4 − 0,5 4 ) 𝐼 𝑥 = 𝜎 1 4 ( 𝑎4 − 𝑟4) 1 4 ( 𝜋 − 0,5) 𝐼 𝑥 = 𝜎 1 16 ( 𝑎4 − 𝑟4)( 𝜋 − 0,5) d. The circumference of the circle r=a 𝑆 = ∫ 𝑟 𝑑𝜃 2𝜋 𝜃=0 𝑆 = ∫ 𝑎 𝑑𝜃 2𝜋 𝜃=0 𝑆 = 𝑎𝜃] 𝜃=0 2𝜋 𝑆 = 𝑎2𝜋 − 0 𝑆 = 2𝜋𝑎 e. The centoroidof a quarter circle  𝑥̅ = ∫ ∫ 𝑥 𝑑𝑥 𝑑𝑦 𝑎 𝑟 𝑎 0 𝑥̅ = ∫ 1 2 𝑥2] 𝑟 𝑎 𝑑𝑦 𝑎 0 𝑥̅ = ∫ 1 2 𝑎2 − 1 2 𝑟2 𝑑𝑦 𝑎 0 𝑥̅ = [ 1 2 𝑎2 𝑦 − 1 2 𝑟2 𝑦] 0 𝑎 𝑥̅ = 1 2 𝑎2(𝑎) − 1 2 𝑟2(𝑎) 𝑥̅ = 1 2 𝑎3 − 1 2 𝑎𝑟2
• 11.
   𝑦̅ =∫ ∫ 𝑥 𝑑𝑥 𝑑𝑦 𝑎 𝑟 𝑎 0 𝑦̅ = ∫ 1 2 𝑥2] 𝑟 𝑎 𝑑𝑦 𝑎 0 𝑦̅ = ∫ 1 2 𝑎2 − 1 2 𝑟2 𝑑𝑦 𝑎 0 𝑦̅ = [ 1 2 𝑎2 𝑦 − 1 2 𝑟2 𝑦] 0 𝑎 𝑦̅ = 1 2 𝑎2(𝑎) − 1 2 𝑟2(𝑎) 𝑦̅ = 1 2 𝑎3 − 1 2 𝑎𝑟2 Section5 5.5 𝑧2 = 3( 𝑥2 + 𝑦2) 𝑧2 3 = 𝑥2 + 𝑦2 𝑧2 3 + 𝑧2 = 16 𝑧2 + 3𝑧2 = 48 4𝑧2 = 48 𝑧2 = 12 𝑧 = 2√3 𝑟⃗ = 𝑧 √3 cos∅ 𝑖̂ + 𝑧 √3 sin ∅ 𝑗̂ + 𝑧𝑘 ( 𝑧, ∅) ∈ (0,2√3) × (0,2𝜋) 𝜕𝑟⃗ 𝜕𝑧 = [ cos∅ √3 sin∅ √3 1] ; 𝜕𝑟⃗ 𝜕∅ = [ −zsin∅ √3 zcos∅ √3 0] | 𝜕𝑟⃗ 𝜕𝑧 × 𝜕𝑟⃗ 𝜕∅ | = √3 3 𝑧 ∫ ∫ √3 3 𝑧𝑑∅𝑑𝑧 = ∫ √3 3 𝑧2𝜋 = [ √3 6 𝑧22𝜋]2√3 0 = √3 6 . 12.2𝜋 2√3 0 = 4√3𝜋 2𝜋 0 2√3 0
• 12.
  -2 -1 0 1 2 0 100 200300 400 sin sudut sin sudut 6.10 y = sinx mencari titikpuncak 𝑦′ = 0 Cos x = 0 X = 900 Dimasukkanke persamaany= sinx maa didapatkany = sin(900 ) = 1 Maka koordinattitikpuncakadalah( 1,𝜋/2) a. b. Untuk Volume yangdiputarpadasumbux maka ∫ 𝑦21 −1 = 1 3 𝑦3 −1 1 ⌋ = 1 3 (1)3 - 1 3 (−1)3 1 3 - (- 1 3 ) = 2 3 c. Ix = ∫ 𝑦21 0 dm ∫ 𝑦21 0 σ dx dy 2 3 ∫ ∫ 𝑦2 𝜋 2 0 1 0 dx dy 2 3 ∫ 𝑦2 𝑑𝑦 1 0 ∫ 𝑑𝑥 𝜋 2 0 2 3 1 3 𝑦3 0 1 ⌋ 𝑥0 𝜋 2 ⌋ 2 3 1 3 𝜋 2 2𝜋 18