PENYELESAIAN FISMAT MARRY L BOAS

1. TUGAS FISIKA UNTUK MATEMATIKA 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 A thin 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 area of 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 centroidof one 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
0 100 200 300 400
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