Dokumen Tentang MATEMATIKA

(1)

BAB II PEMBAHASAN

A. Integral Lipat Dua dalam Koordinat Polar

1. ₀^𝜋/2 ₀^{cos 𝜃}

𝑟

²

sin 𝜃 𝑑𝑟 𝑑𝜃 Penyelesaian:

𝑟

²

sin 𝜃 𝑑𝑟 𝑑𝜃 = 1

3 𝑟

³

sin 𝜃

𝑟=0 cos 𝜃

𝑑𝜃

𝜋/2 0 cos 𝜃

0 𝜋/2 0

= 1

3 𝑐𝑜𝑠

³

𝜃 sin 𝜃 𝑑𝜃 − 0

𝜋/2 0

= 1

3

^𝜋/2

𝑐𝑜𝑠

³

𝜃 sin 𝜃 𝑑𝜃

0

= − 𝑐𝑜𝑠

⁴

𝜃 12

_𝑟=0

𝜋/2

= − 𝑐𝑜𝑠

⁴

𝜋 2

12 − − 𝑐𝑜𝑠

⁴

(0) 12

= 1 12 ∎

2.

₀^𝜋 ₀^{1− cos 𝜃}

𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 Penyelesaian:

𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃

1− cos 𝜃 0 𝜋 0

= 1

2 𝑟

²

sin 𝜃

𝑟= 0 1− cos 𝜃

𝑑𝜃

𝜋 0

= (1 − cos 𝜃)

²

sin 𝜃

2 − 0

²

sin 𝜃 2 𝑑𝜃

𝜋 0

= sin 𝜃 − sin 2𝜃 + cos 𝜃

²

sin 𝜃

2 𝑑𝜃

𝜋 0

= 1

2 sin 𝜃 − sin 2𝜃 + cos 𝜃

²

sin 𝜃 𝑑𝜃

𝜋 0

= 1

2 sin 𝜃 𝑑𝜃 − sin 2𝜃 𝑑𝜃 + cos 𝜃

²

sin 𝜃 𝑑𝜃

= 1

2 − cos 𝜃 + cos 2𝜃

2 − cos 𝜃

³

2

(2)

= − cos 𝜃

2 + cos 2𝜃

4 − cos 𝜃

³

6

₀

𝜋

= −cos 𝜋

2 +cos 2𝜋

4 −cos 𝜋³

6 − −cos 0

2 +cos 2(0)

4 −cos(0)³ 6

= 1 2 + 1

4 − 1

6 − − 1 2 + 1

4 − 1 6

= 6 + 3 + 2 − (−6 + 3 − 2

12 = 16 12

= 3 4 ∎

Carilah luas daerah yg diberikan dengan 𝑟 𝑑𝑟 𝑑𝐴.

3.

S adalah daerah di dalam lingkarang r = 4 cos 𝜃 dan di luar lingkaran r = 2 Penyelesaian:

r = 2

r = 4 cos 𝜃𝑟

²

= 4𝑟 cos 𝜃

𝑥

²

+ 𝑦

²

= 4𝑥 𝑥

²

− 4𝑥 + 𝑦

²

= 0 𝑥

²

− 4𝑥 + 4 + 𝑦

²

= 4

𝑥 − 2

²

+ 𝑦

²

= 4

2 𝑟 𝑑𝑟 𝑑𝜃 = 2 1

2 𝑟

²

2 4 cos 𝜃

𝑑𝜃

𝜋/3 0 4 cos 𝜃

2 𝜋/3 0

= 2 1

2 (4 cos 𝜃)

²

− 1

2 2

²

𝑑𝜃

𝜋/3 0

= 2 1

2 (4 cos 𝜃)

²

− 2 𝑑𝜃

𝜋/3 0

= 2 8

𝜋3

0

cos(𝜃)

²

𝑑𝜃 − 2𝑑𝜃

𝜋3

0

= 2 8 cos

𝜋/3 0

(𝜃)

²

𝑑𝜃 − 2𝑑𝜃

𝜋/3 0

= 2 8 1 + cos 2𝜃

2 𝑑𝜃 − 2𝑑𝜃

𝜋/3 0 𝜋/3

0

(3)

= 2 8 . 1

2 1 + cos 2𝜃 𝑑𝜃 − 2𝑑𝜃

𝜋/3 0 𝜋/3

0

= 2 4 𝜃 + 2 sin 2𝜃 − 2 𝜃

₀^𝜋/3

= 2 2 𝜃 + 2 sin 2𝜃

₀^𝜋/3

= 2 (2 𝜋

3 + 2 sin 2𝜋

3 ) − (2 sin 0)

= 2 ( 2𝜋

3 + 2 3 2 )

= 2 2𝜋

3 + 3 ∎

4.

S adalah daerah di dalam kardiodida r = 6 – 6 sin 𝜃.

Penyelesaian:

𝑟 𝑑𝑟 𝑑𝜃

6−6 sin 𝜃 0

2𝜋 0

= 1 2 𝑟

²

0 6−sin 𝜃

𝑑𝜃

2𝜋 0

= 1

2 6 − sin 𝜃

²

𝑑𝜃

2𝜋 0

= 1

2 36 𝑑𝜃

2𝜋 0

− 72 sin 𝜃 𝑑𝜃 + 36 sin 𝜃

²

𝑑𝜃

2𝜋 0 2𝜋

0

= 1

2 36 𝜃 + 72 cos 𝜃 + 18 𝜃 − 9 sin 2𝜃

0 2𝜋

= 1

2 54 𝜃 + 72 cos 𝜃 − 9 sin 2𝜃

0 2𝜋

= 27 𝜃 + 36 cos 𝜃 − 9 sin 2𝜃 2

0 2𝜋

= 27 (2𝜋) + 36 (1) −9(0) − (36 (1))

= 54𝜋 ∎

5.

𝑟 𝑑𝑟 𝑑𝜃

₀^𝜋⁴ ₀²

Penyelesaian:

𝑟 𝑑𝑟 𝑑𝜃

²

0 𝜋4

0

= 1 2 𝑟

²

𝜋4

0

𝑑𝜃

= 1 2 (2)

²

𝜋4

0

𝑑𝜃

(4)

= 2

𝜋4

0

𝑑𝜃

= 2𝜃

₀^𝜋/4

= 2𝜋 4 − 0

= 𝜋 2 ∎

6. ₀^𝜋/2 ₀^{cos 𝜃}

𝑟 𝑑𝑟 𝑑𝜃 Penyelesaian:

𝑟 𝑑𝑟 𝑑𝜃

cos 𝜃 0 𝜋/2 0

= 1

2 𝑟

²

0 cos 𝜃

𝑑𝜃

𝜋/2 0

= 1

2 𝑐𝑜𝑠

²

𝜃 𝑑𝜃

𝜋2

0

= 1 2 ( 1

2 + 1

2 cos 2𝜃)

𝜋/2 0

𝑑𝜃

= 1 2

1 2 + 1

2 cos 2𝜃 𝑑𝜃

𝜋/2 0

= 1 2

1 2 + cos 2𝜃 2

𝜋2

0

𝑑𝜃

= 1 2 1

2 𝑑𝜃 + cos 2𝜃 2 𝑑𝜃

= 1 2

1 2 𝜃 + sin 2𝜃 4

= 1

4 𝜃 + sin⁡(2𝜃)

8

₀

𝜋/2

= 1 4 × 𝜋

2 + 𝑠𝑖𝑛 2 × 𝜋 2 8 − 1

4 × 0 + 𝑠𝑖𝑛 2 × 0

8 = 𝜋

8 ∎

(5)

7.

𝑒

^𝑥²^+𝑦²

𝑑𝐴, dengan S adalah daerah yang dikurung oleh 𝑥

²

+ 𝑦

²

= 4.

Penyelesaian:

B. Penerapan Integral Lipat Dua

Dalam soal-soal 1-10, carilah massa m dan pusat massa (𝑥 , 𝑦 ) dari lamina yang dibatasi oleh kurva-kurva yang diberikan dan dengan kerapatan yang ditunjukkan.

1. 𝑥 = 0, 𝑥 = 4, 𝑦 = 0, 𝑦 = 3; 𝛿 𝑥, 𝑦 = 𝑦 + 1

Penyelesaian:

Massa m

𝒎 = (𝒚 + 𝟏)

𝟒 𝟎

𝒅𝒙𝒅𝒚

𝟑 𝟎

𝑚 = 𝑦 + 1

4

0 𝑑𝑥 𝑑𝑦

3 0

Penjabaran:

𝑦 + 1

4

0 𝑑𝑥 = 𝑥 𝑦 + 1 ₀⁴= 4 𝑦 + 1 − 0 = 4(𝑦 + 1) Jadi,

𝑚 = 𝑦 + 1 ⁴

0 𝑑𝑥 𝑑𝑦

3 0

= 4 𝑦 + 1 𝑑𝑦

3

0

= 4 𝑦 + 1 𝑑𝑦

3

0

= 4 𝑦𝑑𝑦

3 0

+ 1𝑑𝑦

3 0

(6)

= 4 𝑦² 2 ₀

3

+ 1. 𝑦 ₀³

= 4 3²

2 + 1.3

= 4 9

2 + 3

= 4 15 2

= 𝟑𝟎

pusat massa 𝑥 , 𝑦

𝑥 , 𝑦 = 𝑚_𝑦 𝑚 ,𝑚_𝑥

𝑚 𝒎_𝒚 = 𝒙(𝒚 + 𝟏)^𝟒

𝟎

𝒅𝒙𝒅𝒚

𝟑 𝟎

𝑚_𝑦 = 𝑥 𝑦 + 1 ⁴

0 𝑑𝑥 𝑑𝑦

3 0

= (𝑦 + 1) 𝑥𝑑𝑥

4 0

𝑑𝑦

3 0

Penjabaran:

(𝑦 + 1) 𝑥𝑑𝑥

4 0

= 𝑦 + 1 𝑥² 2 ₀

4

= 𝑦 + 1 (8) Jadi,

𝑚_𝑦 = 𝑥 𝑦 + 1 ⁴

0 𝑑𝑥 𝑑𝑦

3 0

= 𝑦 + 1 8𝑑𝑦

3 0

= 8 1𝑑𝑦³

0

+ 𝑦𝑑𝑦³

0

= 8 1. 𝑦 ₀³+ 𝑦² 2 ₀

3

= 8 1.3 + 3² 2

= 8 3 + 9 2

= 8 15 2

= 𝟔𝟎

(7)

𝒎_𝒙= 𝒚(𝒚 + 𝟏)^𝟒

𝟎

𝒅𝒙𝒅𝒚

𝟑 𝟎

𝑚_𝑥 = 𝑦 𝑦 + 1

4

0 𝑑𝑥 𝑑𝑦

3 0

= 𝑥𝑦 𝑦 + 1 ₀⁴𝑑𝑦

3 0

Penjabaran:

𝑥𝑦 𝑦 + 1 ₀⁴ = 4𝑦 𝑦 + 1 Jadi,

𝑚_𝑥 = 𝑦 𝑦 + 1 ⁴

0 𝑑𝑥 𝑑𝑦

3 0

= 4𝑦 𝑦 + 1 𝑑𝑦

3 0

= 4 𝑦(𝑦 + 1)𝑑𝑦³

0

= 4 𝑦²𝑑𝑦

3 0

+ 𝑦𝑑𝑦

3 0

= 4 𝑦³ 3 ₀

3

+ 𝑦² 2 0

3

= 4 3³ 3 + 3²

2

= 4 9 + 9 2

= 4 27 2

= 𝟓𝟒 pusat massa 𝒙 , 𝒚

𝑥 , 𝑦 = 60 30,54

30 𝑥 , 𝑦 = 2, 1.8

(8)

2. 𝒚 =^𝟏_𝒙, 𝒚 = 𝒙, 𝒚 = 𝟎, 𝒙 = 𝟐; 𝜹 𝒙, 𝒚 = 𝒙

Massa m:

𝒎 = 𝒙 𝒅𝒚 𝒅𝒙 + 𝒙^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏 𝒙

𝟎 𝟏 𝟎

 𝑥 𝑑𝑦 = [𝑥𝑦]₀^𝑥 ₀^𝑥 = 𝑥²− 0 =𝑥² 𝒎 = 𝒙^𝟐𝒅𝒙

𝟏 𝟎

+ 𝒙^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 𝑥₀¹ ²𝑑𝑥= ^𝑥₃³

0

1= ¹₃− 0 =¹₃

𝒎 =𝟏

𝟑+ 𝒙^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 ₀^{1 𝑥}𝑥 𝑑𝑦 = [𝑥𝑦]₀^{1 𝑥} = 𝑥¹_𝑥− 0 = 1 − 0 = 1 𝒎 =𝟏

𝟑+ 𝟏 𝒅𝒙^𝟐

𝟏

 1 𝑑𝑥₁² = 1 . 𝑥 ₁²= 2 − 1 = 1

𝒎 =𝟏

𝟑+ 𝟏 =𝟒 𝟑 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒎_𝒚 = 𝒙^𝟐 𝒅𝒚 𝒅𝒙 + 𝒙^{𝟏 𝒙} ^𝟐

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏 𝒙

𝟎 𝟏 𝟎

 𝑥₀^𝑥 ²𝑑𝑦 = [𝑥²𝑦]₀^𝑥 = 𝑥³− 0 = 𝑥³ 𝒎_𝒚= 𝒙^𝟏 ^𝟑𝒅𝒙

𝟎

+ 𝒙^{𝟏 𝒙} ^𝟐

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 𝑥₀¹ ³𝑑𝑥 = ^𝑥₄⁴

0

1= ¹₄− 0 =¹₄

(9)

𝒎_𝒚 =𝟏

𝟒+ 𝒙^{𝟏 𝒙} ^𝟐

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 ₀^{1 𝑥}𝑥² 𝑑𝑦 = [𝑥²𝑦]₀^{1 𝑥} = 𝑥^{2 1}_𝑥 − 0 = 𝑥 𝒎_𝒚=𝟏

𝟒+ 𝒙 𝒅𝒙^𝟐

𝟏

 𝑥 𝑑𝑥₁² = ^𝑥₂²

1

2= ⁴₂−¹₂ =³₂

𝒎_𝒚 =𝟏 𝟒+𝟑

𝟐=𝟕 𝟒

𝒎_𝒙 = 𝒙𝒚 𝒅𝒚 𝒅𝒙 + 𝒙𝒚^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏 𝒙

𝟎 𝟏 𝟎

 𝑥𝑦₀^𝑥 𝑑𝑦 = 𝑥 𝑦₀^𝑥 𝑑𝑦 = 𝑥 ^𝑦₂²

0

𝑥 = 𝑥 ^𝑥₂²− 0 =^𝑥₂³

𝒎_𝒙 = 𝒙^𝟑 𝟐

𝟏 𝟎

𝒅𝒙 + 𝒙𝒚^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 ^𝑥³

2 1

0 𝑑𝑥 =¹

2 𝑥₀¹ ³𝑑𝑥 =¹

2 𝑥⁴

4 0 1 =¹

2 1⁴

4 − 0 =¹₂ .¹

4=¹

8

𝒎_𝒙=𝟏

𝟖+ 𝒙𝒚^{𝟏 𝒙}

𝟎

𝒅𝒚 𝒅𝒙

𝟐 𝟏

 ₀^{1 𝑥}𝑥𝑦𝑑𝑦 = 𝑥 ₀^{1 𝑥}𝑦𝑑𝑦 = 𝑥 ^𝑦₂²

0

1 𝑥 = 𝑥 ^{1 𝑥}

2

2 − 0 = 𝑥

1 𝑥2

2 = 𝑥 ¹

2𝑥²= ¹

2𝑥

𝒎_𝒙 =𝟏

𝟖+ 𝟏 𝟐𝒙 𝒅𝒙

𝟐 𝟏

 ¹

2𝑥 𝑑𝑥

2

1 =¹₂ ₁²_𝑥¹ 𝑑𝑥=¹₂[𝑙𝑛 𝑥 ]₁²=¹₂ 𝑙𝑛 2 − 0 =¹₂𝑙𝑛 2 𝒎_𝒙=𝟏

𝟖+𝟏 𝟐𝒍𝒏 𝟐 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒙 , 𝒚 =

𝟕𝟒 𝟒𝟑 ,

𝟏𝟖 +𝟏 𝟐 𝒍𝒏 𝟐

𝟒𝟑

(10)

3. 𝒓 =

𝟐 𝒔𝒊𝒏 𝜽

; 𝜹 𝒓, 𝜽 =

𝒓

Penyelesaian:

𝒎 = 𝒓 𝒓 𝒅𝒓 𝒅𝜽

𝟐 𝐬𝐢𝐧 𝜽 𝟎 𝟐 𝟎

 ₀^{2 sin 𝜃}𝑟 𝑟 𝑑𝑟 = ₀^{2 sin 𝜃}𝑟² 𝑑𝑟 = ^𝑟³

3 0 2 sin 𝜃

= ^{2 sin 𝜃}₃ ³− 0 =^{8𝑠𝑖𝑛}₃³^(𝜃) 𝒎 = 𝟖𝒔𝒊𝒏^𝟑(𝜽)

𝟑

𝟐

𝟎 𝒅𝜽

 ^{8𝑠𝑖𝑛}³^(𝜃)

3 2

0 𝑑𝜃 =⁸

3 𝑠𝑖𝑛₀² ³ 𝜃 𝑑𝜃 =⁸₃ 𝑠𝑖𝑛₀² ²(𝜃)𝑠𝑖𝑛 𝜃 𝑑𝜃

 ⁸

3 1 − 𝑐𝑜𝑠₀² ²(𝜃)𝑠𝑖𝑛 𝜃 𝑑𝜃 =⁸

3 −1 + 𝑢²𝑑𝑢 =⁸

3(− _{𝑐𝑜𝑠 (2)}¹ −1 + 𝑢²𝑑𝑢

𝑐𝑜𝑠 (2) 1

 ⁸

3(−(− _{co s 2}¹ 1𝑑𝑢 + _{co s 2}¹ 𝑢²𝑑𝑢) )

 1𝑑𝑢 = 1 − cos⁡(2)] + [ _{co s 2}¹ 𝑢²𝑑𝑢=^{1−𝑐𝑜𝑠}³²

3 1

co s 2

 ⁸

3 − − 1 − cos(2) +^{1−𝑐𝑜𝑠}₃³² = −^{8(−𝑐𝑜𝑠}³ 2 +3𝑐𝑜𝑠 2 −2

9

𝒎 = −𝟖(−𝒄𝒐𝒔^𝟑 𝟐 + 𝟑𝒄𝒐𝒔 𝟐 − 𝟐

𝟗 =𝟑𝟐

𝟗 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎

𝒎_𝒙 = ^{𝟐 𝐬𝐢𝐧 𝜽} 𝒓 𝒔𝒊𝒏 𝜽 𝒓 𝒓 𝒅𝒓 𝒅𝜽

𝟎 𝟐 𝟎

 ₀^{2 sin 𝜃} 𝑟 𝑠𝑖𝑛 𝜃 𝑟 𝑟 𝑑𝑟= 𝑠𝑖𝑛 𝜃 ₀^{2 sin 𝜃}𝑟³ 𝑑𝑟 = 𝑠𝑖𝑛 𝜃 ^𝑟₄⁴

0 2 sin 𝜃

= 𝑠𝑖𝑛 𝜃 4𝑠𝑖𝑛⁴ 𝜃 = 4𝑠𝑖𝑛⁵(𝜃)

𝒎_𝒙= 𝟒𝒔𝒊𝒏^𝟐 ^𝟓(𝜽)𝒅

𝟎

𝜽

 4 𝑠𝑖𝑛₀² ⁵ 𝜃 𝑑𝜃 = 4 𝑠𝑖𝑛₀² ⁴ 𝜃 si n 𝜃 𝑑𝜃 = 4 1 − 𝑐𝑜𝑠₀² ² 𝜃 ²(sin⁡(𝜃)𝑑𝜃

 1 − 𝑐𝑜𝑠₀² ² 𝜃 ²(sin⁡(𝜃)𝑑𝜃 = 1 − 𝑢² ²𝑠𝑖𝑛 𝜃 − ¹

si n 𝜃 𝑑𝑢

 −1 + 2𝑢²− 𝑢⁴𝑑𝑢 = 4 ₁^{𝑐𝑜𝑠 (2)}−1 + 2𝑢²− 𝑢⁴𝑑𝑢= 4 _{𝑐𝑜𝑠 (2)}¹ −1 + 2𝑢²− 𝑢⁴𝑑𝑢

(11)

 4 − − _{co s 2}¹ −1𝑑𝑢 + _{co s 2}¹ 2𝑢²𝑑𝑢− _{co s 2}¹ 𝑢⁴𝑑𝑢

 4 − − 1 − 𝑐𝑜𝑠 2 +^{2 1−𝑐𝑜𝑠}₃ ³²−^{1−𝑐𝑜𝑠}⁵²

5

 −4 −1 + 𝑐𝑜𝑠 2 +^{3𝑐𝑜𝑠}⁵^{2 −10𝑐𝑜𝑠}³^{2 +7}

15

𝒎_𝒙 = −𝟒 −𝟏 + 𝒄𝒐𝒔 𝟐 +𝟑𝒄𝒐𝒔^𝟓 𝟐 − 𝟏𝟎𝒄𝒐𝒔^𝟑 𝟐 + 𝟕

𝟏𝟓 =𝟔𝟒

𝟏𝟓 𝒎_𝒚= 𝟎

pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒙 , 𝒚 = 𝟎

𝟑𝟐𝟗 ,

𝟔𝟒𝟏𝟓 𝟑𝟐𝟗

= (𝟎, 𝟏. 𝟐)

4. 𝒓 = 𝟐 + 𝟐 𝐜𝐨𝐬 𝜽; 𝜹 𝒓, 𝜽 = 𝒓

Penyelesaian:

𝒎 = ^{𝟐+𝟐 𝐜𝐨𝐬 𝜽}𝒓𝒓𝒅𝒓𝒅𝜽

𝟎 𝟐𝒙 𝟎

 ₀^{2+2 cos 𝜃}𝑟𝑟𝑑𝑟= ^𝑟₃³

0 2+2 𝑐𝑜𝑠𝜃

=¹

3 8 + 24𝑐𝑜𝑠 𝜃 + 24𝑐𝑜𝑠²𝜃 + 8𝑐𝑜𝑠³𝜃 𝒎 = 𝟏

𝟑 𝟖 + 𝟐𝟒𝒄𝒐𝒔 𝜽 + 𝟐𝟒𝒄𝒐𝒔^𝟐𝜽 + 𝟖𝒄𝒐𝒔^𝟑𝜽 𝒅𝜽

𝟐𝒙 𝟎

 ¹

3 0 + 24𝜋 + 16𝜋 =^40𝜋

3

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎

𝒎_𝒚= 𝒓 𝒓𝒄𝒐𝒔𝜽 𝒓𝒅𝒓𝒅𝜽

𝟐+𝟐 𝐜𝐨𝐬 𝜽 𝟎

𝟐𝒙 𝟎

 _𝟎^{𝟐+𝟐 𝐜𝐨𝐬 𝜽}𝒓 𝒓𝒄𝒐𝒔𝜽 𝒓𝒅𝒓 = ^𝟏

𝟒𝒓^𝟒𝒄𝒐𝒔𝜽

𝟎 𝟐+𝟐𝒄𝒐𝒔𝜽

= ^𝟏_𝟒(𝟐 + 𝟐𝒄𝒐𝒔𝜽)^𝟒𝒄𝒐𝒔𝜽 𝒎_𝒚= 𝟏

𝟒(𝟐 + 𝟐𝒄𝒐𝒔𝜽)^𝟒𝒄𝒐𝒔𝜽 𝒅𝜽

𝟐𝒙 𝟎

= 𝟐𝟖𝝅

(12)

𝒎_𝒙= 𝟎 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝟐𝟖𝝅 40𝜋3

, 𝟎 40𝜋3

= 𝟐𝟏 𝟏𝟎, 𝟎

5. Persegi dengan titik sudut 𝟎, 𝟎 , 𝟎, 𝒂 , 𝒂, 𝒂 , 𝒂, 𝟎 ; 𝜹 𝒙, 𝒚 = 𝒙 + 𝒚

Penyelesaian:

𝐼_𝑥 = (𝑥 + 𝑦)𝑦^𝑎 ²

0 𝑎 0

𝑑𝑥𝑑𝑦

 𝑥 + 𝑦 𝑦₀^𝑎 ²𝑑𝑥 = 𝑦² 𝑦 + 𝑥₀^𝑎 𝑑𝑥 = 𝑦²

 𝑦𝑑𝑥 + 𝑥₀^𝑎 ₀^𝑎 𝑑𝑥 =𝑦² 𝑎𝑦 +^𝑎²

2

𝐼_𝑥 = 𝑦² 𝑎𝑦 +𝑎² 2

𝑎 0

𝑑𝑦

 𝑎𝑦₀^𝑎 ³ 𝑎𝑦 +^𝑎²₂^𝑦² 𝑑𝑦 = 𝑎𝑦₀^𝑎 ³+ ₀^𝑎^𝑎²₂^𝑦²𝑑𝑦 = ^𝑎₄⁵+^𝑎₆⁵ = ₁₂⁵ 𝑎⁵ 𝐼_𝑦 = (𝑥 + 𝑦)𝑦^𝑎 ²

0 𝑎 0

𝑑𝑥𝑑𝑦

 𝑥 + 𝑦 𝑦₀^𝑎 ²𝑑𝑥 = 𝑦² 𝑦 + 𝑥₀^𝑎 𝑑𝑥 = 𝑦²

 𝑦𝑑𝑥 + 𝑥₀^𝑎 ₀^𝑎 𝑑𝑥 =𝑦² 𝑎𝑦 +^𝑎²

2

𝐼_𝑦 = 𝑦² 𝑎𝑦 +𝑎² 2

𝑎 0

𝑑𝑦

 𝑎𝑦³ 𝑎𝑦 +^𝑎²^𝑦²

2 𝑎

0 𝑑𝑦 = 𝑎𝑦₀^𝑎 ³+ ^𝑎²^𝑦²

2 𝑎

0 𝑑𝑦 = ^𝑎₄⁵+^𝑎⁵

6 = ⁵

12 𝑎⁵ 𝐼_𝑧 = (𝑥 + 𝑦)𝑦²

𝑎 0 𝑎 0

𝑑𝑥𝑑𝑦

(13)

Dalam soal-soal 15-20 diberikan integral berulang dalam koordinat persegi panjang atau polar.

Integral lipat dua itu memberi massa suatu lamina R. sketsakan lamina R dan tentukan kerapatan. Kemudian carilah massa dan pusat massa.

6. 𝒌𝒚 𝒅𝒚 𝒅𝒙_𝟎^𝟏 _𝒙^𝟐

Penyelesaian:

𝒎 = 𝒌𝒚^𝟐

𝒙 𝟏 𝟎

𝒅𝒚 𝒅𝒙

 [₀¹ ^𝑘₂𝑦²]_𝑥²𝑑𝑥 = ₀¹^𝑘₂ 4 − 𝑥² 𝑑𝑥 = 2𝑘 −₀¹ ^𝑘𝑥₂²𝑑𝑥 𝒎 = 𝟐𝒌 −𝒌𝒙^𝟐

𝟐

𝟏

𝟎 𝒅𝒙

 2𝑘𝑥 −^𝑘𝑥³

6 0

1 = 2𝑘 1 −^{𝑘 1}³

6 = 2𝑘 −^𝑘

6

𝒎 = 𝟐𝒌 −𝒌

𝟔=𝟏𝟐𝒌 − 𝒌

𝟔 =𝟏𝟏

𝟔 𝒌 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒎_𝒙 = 𝒌𝒚^𝟐

𝒙 𝟏 𝟐 𝟎

𝒅𝒚 𝒅𝒙

 ^𝑘₃𝑦³

𝑥 1 2

0 𝑑𝑥 = ₀¹^𝑘₃(8 − 𝑥³)𝑑𝑥 = ^8𝑘₃ −^𝑘𝑥³

3 1

0 𝑑𝑥

𝒎_𝒙= 𝟖𝒌 𝟑 −𝒌𝒙^𝟑

𝟑

𝟏 𝟎

𝒅𝒙

 ^8𝑘𝑥

3 −^𝑘𝑥⁴

12 0

1 =^{8𝑘 1}

3 −^{𝑘 1}⁴

12 =^8𝑘

3 − ^𝑘

12

𝒎_𝒙=𝟖𝒌 𝟑 − 𝒌

𝟏𝟐=𝟑𝟐𝒌 − 𝒌

𝟏𝟐 =𝟑𝟏

𝟏𝟐𝒌 𝒎_𝒚= 𝒌𝒙𝒚

𝟐 𝒙 𝟏 𝟎

𝒅𝒚 𝒅𝒙

 [₀¹ ^𝑘𝑥₂ 𝑦²]_𝑥²𝑑𝑥 = ₀¹^𝑘𝑥₂ 4 − 𝑥² 𝑑𝑥 = ₀¹^4𝑘𝑥₂ −^𝑘𝑥₂³𝑑𝑥

(14)

𝒎_𝒚= 𝟒𝒌𝒙 𝟐 −𝒌𝒙^𝟑

𝟐

𝟏 𝟎

𝒅𝒙 = 𝟐𝒌𝒙 −𝒌𝒙^𝟑 𝟐

𝟏 𝟎

𝒅𝒙

 𝑘𝑥²−^𝑘𝑥₈⁴

0

1 = 𝑘(1)²−^𝑘(1)₈ ⁴ = 𝑘 −^𝑘₈

𝒎_𝒚= 𝒌 −𝒌

𝟖=𝟖𝒌 − 𝒌

𝟖 =𝟕

𝟖𝒌 pusat massa 𝒙 , 𝒚

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒙 , 𝒚 =

𝟕𝟖 𝒌 𝟏𝟏𝟔 𝒌

, 𝟑𝟏𝟏𝟐 𝒌 𝟏𝟏𝟔 𝒌

= 𝟐𝟏 𝟒𝟒,𝟗𝟑

𝟒𝟒

7. 𝒌𝒓_𝟎^𝝅 _𝟏^𝟑 ^𝟐 𝒅𝒓 𝒅𝜽

Penyelesaian:

𝑚 = 𝑘𝑟³ ²

1 𝜋 0

𝑑𝑟𝑑𝜃

 ^𝑘𝑟₃³

1 𝜋 3

0 𝑑𝜃 = (₀^𝜋 ^{𝑘 3}₃ ³−^𝑘(1)₃ ³)𝑑𝜃 = (₀^𝜋 ^27𝑘₃ −^𝑘₃)𝑑𝜃 = ₀^𝜋^26𝑘₃ 𝑑𝜃 𝑚 = 26𝑘

3

𝜋 0

𝑑𝜃

 ^26𝑘

3 0

𝜋 = ^26𝑘𝜋

3

𝑚_𝑥 = 𝑘𝑟³ ²

1 𝜋 0

𝑟 sin 𝜃 𝑑𝑟𝑑𝜃

 [₀^𝜋 ^𝑘𝑟₄⁴sin 𝜃] ₁³𝑑𝜃 = ₀^𝜋 ^{𝑘 3}₄⁴sin 𝜃 −^{𝑘 1}₄ ⁴sin 𝜃 𝑑𝜃

 ₀^𝜋 ^81𝑘₄ sin 𝜃 −^𝑘₄sin 𝜃 𝑑𝜃 = 20𝑘 sin 𝜃 ₀^𝜋 𝑑𝜃 𝑚_𝑥 = 20𝑘 sin 𝜃

𝜋 0

𝑑𝜃

 20𝑘 sin 𝜃₀^𝜋 𝑑𝜃 = 20[− cos 𝜃]₀^𝜋 = 20(− cos 𝜋 − − cos 0 𝑚_𝑥 = 20 − −1 − −1 = 20𝑘 2 = 40𝑘

𝑚_𝑦 = 0

(15)

𝑚_𝑦 = 𝑘𝑟²

3 1 𝜋 0

𝑟 cos 𝜃 𝑑𝑟𝑑𝜃

 [₀^𝜋 ^𝑘𝑟₄⁴cos 𝜃] ₁³𝑑𝜃 = ^{𝑘 3}₄ ⁴cos 𝜃 −^{𝑘 1}⁴

4 cos 𝜃

𝜋

0 𝑑𝜃

 ₀^𝜋 ^81𝑘₄ cos 𝜃 −^𝑘₄cos 𝜃 𝑑𝜃 = 20𝑘 cos 𝜃 ₀^𝜋 𝑑𝜃 𝑚_𝑦 = 20𝑘 cos 𝜃 ^𝜋

0

𝑑𝜃

 20𝑘 cos 𝜃₀^𝜋 𝑑𝜃 = 20[sin 𝜃]₀^𝜋 = 20(sin 𝜋 − sin 0 𝑚_𝑦 = 20 0 = 0

𝒙 , 𝒚 = 𝒎_𝒚 𝒎 ,𝒎_𝒙

𝒎 𝒙 , 𝒚 = 𝟎

26𝑘𝜋3 , 𝟒𝟎𝒌

26𝑘𝜋3

= (𝟎, 𝟔𝟎 𝟏𝟑𝝅)

C. Integral Lipat Tiga dalam Koordinat Cartesius

1.

₋₃⁷ ₀^2𝑥 _𝑦^𝑥−1

𝑑𝑧 𝑑𝑦 𝑑𝑥

Penyelesaian :

𝑑𝑧 𝑑𝑦 𝑑𝑥 = (𝑥 − 1 − 𝑦))

2𝑥 0

𝑑𝑦 𝑑𝑥

7

−3 𝑥−1

𝑦 2𝑥 0 7

−3

= 𝑥𝑦 − 𝑦 − 𝑦

²

2

₀

2𝑥

𝑑𝑥

7

−3

= 𝑥 2𝑥 − 2𝑥 − 2𝑥

²

2 − 0 𝑑𝑥

7

−3

= 2𝑥

²

− 2𝑥 − 4𝑥

²

2 𝑑𝑥

7

−3

= 2𝑥

⁷ ²

− 2𝑥 − 2𝑥

²

𝑑𝑥

−3

= −2𝑥 𝑑𝑥

7

−3

= 2 𝑥 𝑑𝑥

⁷

−3

= −2 𝑥

²

2

7 −3

(16)

= −2 7

²

2 − −3

²

2 = −2 49 2 − 9

2 = −40

2. 6𝑥𝑦

₀⁵ ₋₂⁴ ₁² ²

𝑧

³

𝑑𝑥 𝑑𝑦 𝑑𝑧 Penyelesaian :

6𝑥𝑦

²

𝑧

³

𝑑𝑥 𝑑𝑦 𝑑𝑧

2 1 4

−2 5 0

= 6 𝑧

³

𝑑𝑧 𝑦

²

𝑑𝑦 𝑥 𝑑𝑥

2 1 4

−2 5

0

= 6 𝑧

⁴

4 5 0

𝑦

³

3

4 −2 𝑥

²

2 2 1

=6

⁵₄⁴

⁴₃³

− −

²₃³

²₂²

−

¹₂²

= 6 625 4 64

3 + 8 3 4

2 − 1 2

= 33750

3.

₀² ₀^𝑧 ₀^𝑥/𝑧

2𝑥𝑦𝑧 𝑑𝑦 𝑑𝑥 𝑑𝑧 Penyelesaian :

2𝑥𝑦𝑧 𝑑𝑦 𝑑𝑥 𝑑𝑧 = 2𝑥 𝑦

²

2 𝑧 𝑥/𝑧 0

𝑧 0 2 0

𝑑𝑥 𝑑𝑧

𝑥/𝑧 0 𝑧 0 2 0

= [𝑥𝑦

^𝑧 ²

𝑧

0

]

₀^𝑥/𝑧

2 0

𝑑𝑥 𝑑𝑧

= 𝑥 𝑥 𝑧

𝑧 2 0

𝑧

2 0

𝑑𝑥 𝑑𝑧

= 𝑥

^𝑧 ²

𝑑𝑥 𝑑𝑧

0 2 0

= 𝑥

³

3 𝑧 0 𝑑𝑧

2 0

= 1

3 𝑧

² ³

𝑑𝑧

0

(17)

= 1 3

𝑧

⁴

4 2 0

= 1 3

2

⁴

4 = 1 3

16 4

= 4 3

4.

₀^𝜋² _{sin 2𝑥}⁰ ₀^2𝑦𝑥

sin

^𝑥_𝑦

𝑑𝑥 𝑑𝑦 𝑑𝑧 Penyelesaian :

sin 𝑥

𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑧

2𝑦𝑥 0 0 sin 2𝑥 𝜋2

0

= −𝑦 𝑐𝑜𝑠 𝑥 𝑦

0 sin 2𝑥 𝜋2

0

2𝑦𝑥 0

𝑑𝑦 𝑑𝑧

= −𝑦 𝑐𝑜𝑠 2𝑦𝑥 𝑦

0 sin 2𝑥 𝜋2 0

− −𝑦 𝑐𝑜𝑠 0

𝑦 𝑑𝑦 𝑑𝑧

= −𝑦 𝑐𝑜𝑠 2𝑥

0 sin 2𝑥 𝜋2

0

− −𝑦 𝑑𝑦 𝑑𝑧

=

⁰

−𝑦 𝑐𝑜𝑠 2𝑥

sin 2𝑥 𝜋2

0

+ 𝑦 𝑑𝑦 𝑑𝑧

=

⁰

−𝑦 1 − 2 sin

²

𝑥 + 𝑦

sin 2𝑥

𝑑𝑦 𝑑𝑧

𝜋2

0

= −𝑦 + 2𝑦 sin

²

𝑥 + 𝑦

0 sin 2𝑥

𝑑𝑦 𝑑𝑧

𝜋2

0

= 2𝑦 sin

²

𝑥

0 sin 2𝑥

𝑑𝑦 𝑑𝑧

𝜋2

0

= 2 sin

²

𝑥 𝑦

0 sin 2𝑥

𝑑𝑦 𝑑𝑧

𝜋2

0

= 2 sin

²

𝑥 𝑦

²

2 0 sin 2𝑥

𝑑𝑧

𝜋2

0

= sin

²

𝑥 𝑦

²

𝑑𝑦 0 sin 2𝑥 𝑑𝑧

𝜋2

0

(18)

= 0 − sin

²

𝑥

𝜋2

0

sin

²

2𝑥 𝑑𝑧

= − sin

²

𝑥 sin

²

2𝑥

𝜋2

0

𝑑𝑧

= [− sin

²

2𝑥 sin

²

𝑥] 𝜋

2 − [− sin

²

2𝑥 sin

²

𝑥]0

= −𝜋 sin

²

𝑥 sin

²

𝑥 2

5. 𝑆 ∶ { 𝑥, 𝑦, 𝑧 ∶ 0 ≤ 𝑥 ≤

¹₂

𝑦 , 0 ≤ 𝑦 ≤ 4, 0 ≤ 𝑧 ≤ 2}

Penyelesaian :

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧

12𝑦

0 4 0 2 0

6. 𝑆 ∶ { 𝑥, 𝑦, 𝑧 ∶ 0 ≤ 𝑥 ≤ 𝑦

²

, 0 ≤ 𝑦 ≤ 𝑧, 0 ≤ 𝑧 ≤ 1}

Penyelesaian :

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑥 𝑑𝑦 𝑑𝑧

^𝑦

2 0 𝑧 0 1 0

(19)

7. S adalahdaerah di oktanpertama yang dibatasi oleh tabung𝑦

²

+ 𝑧

²

= 1 dan bidang-bidang𝑥 = 1 dan 𝑥 = 4

Penyelesaian :

𝑓 𝑥, 𝑦, 𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑥

1−𝑦² 0 1 0 4 1

D. Integral Lipat Tiga dalam Koordinat Tabung dan Bola

Dalam soal-soal 1-6, hitunglah integral yang diberikan dalam koordinat silindris atau sferis, dan uraikan daerah integrasi R.

1. 𝑟₀^2𝜋 ₀³ ₀¹² 𝑑𝑧 𝑑𝑟 𝑑𝜃 Penyelesaian:

𝑟¹²

0

𝑑𝑧 𝑑𝑟 𝑑𝜃

3 0 2𝜋 0

= 𝑟𝑧 ³ ₀¹² 𝑑𝑟 𝑑𝜃

0 2𝜋 0

= 12𝑟 𝑑𝑟 𝑑𝜃³

0 2𝜋 0

= 6𝑟² ₀³ 𝑑𝜃

3 0 2𝜋 0

(20)

= 54 𝑑𝜃

2𝜋 0

= 54𝜃 ₀^2𝜋 = 108𝜋 ∎

2. ₀^𝜋 ₀^{sin 𝜃} 𝑟₀² 𝑑𝑧 𝑑𝑟 𝑑𝜃 Penyelesaian:

𝑟

2 0

𝑑𝑧 𝑑𝑟 𝑑𝜃

sin 𝜃 0 𝜋 0

= 𝑟𝑧 ₀² 𝑑𝑟 𝑑𝜃

sin 𝜃 0 𝜋 0

= 2𝑟 𝑑𝑟 𝑑𝜃

sin 𝜃 0 𝜋 0

= 𝑟^𝜋 ² ₀^{sin 𝜃} 𝑑𝜃

0

= 𝑠𝑖𝑛² 𝜃 𝑑𝜃

𝜋 0

= 1 2−1

2cos 2𝜃 𝑑𝜃

𝜋 0

= 1 2𝜃 −1

2sin 2𝜃

0 𝜋

=1 2𝜋 ∎

Dalam soal-soal 7-14, gunakan koordinat silindris untuk mencari besaran yang ditunjukkan.

3. Volume benda pejal yang dibatasi oleh paraboloida 𝑧 = 𝑥²+ 𝑦² dan bidang 𝑧 = 4.

Penyelesaian:

Daerah S dalam Koordinat Cartesius :

𝑆 = 𝑥, 𝑦, 𝑧 | − 2 ≤ 𝑥 ≤ 2, − 4 − 𝑥² ≤ 𝑦 ≤ 4 − 𝑥², 𝑥²+ 𝑦²≤ 𝑧 ≤ 4 Dalam Koordinat Tabung :

𝑆 = 𝑟, 𝜃, 𝑧 |0 ≤ 𝑟 ≤ 2,0 ≤ 𝜃 ≤ 2𝜋, 𝑟²≤ 𝑧 ≤ 4

(21)

Sehingga, volume benda pejalnya adalah

𝑉 = 𝑟 𝑑𝑧 𝑑𝜃 𝑑𝑟

4

𝑟² 2𝜋

0 2

0

= 𝑟𝑧 _𝑟42

2𝜋

0 2

0

𝑑𝜃 𝑑𝑟

= 𝑟𝑧 _𝑟42

2𝜋

0 2

0

𝑑𝜃 𝑑𝑟

= (4𝑟 − 𝑟³)

2𝜋

0 2

0

− 0 𝑑𝜃 𝑑𝑟

= 4𝑟𝜃 − 𝑟³𝜃 ₀^2𝜋

2

0

𝑑𝑟

= 4𝑟2𝜋 − 𝑟³2𝜋 − 0

2

0

𝑑𝑟

= 8𝜋𝑟 − 2𝜋𝑟³

2

0

𝑑𝑟

= 4𝜋𝑟²−1 2𝜋𝑟⁴

0 2

= 4𝜋 × 2²−1

2𝜋 × 2⁴ − 0

= 8𝜋 ∎

4. Volume benda pejal yang di atas dibatasi oleh bidang 𝑧 = 𝑦 + 4, di bawah oleh 𝑏𝑖𝑑𝑎𝑛𝑔 − 𝑥𝑦, dan secara menyamping oleh tabung lingkaran tegak yang mempunyai jejari 4 dan yang sumbunya adalah adalah 𝑠𝑢𝑚𝑏𝑢 − 𝑧.

Penyelesaian:

Daerah S dalam Koordinat Tabung :

𝑆 = 𝑟, 𝜃, 𝑧 |0 ≤ 𝑟 ≤ 4,0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝑧 ≤ 4 + 𝑟 sin 𝜃 Sehingga, volume benda pejalnya adalah

𝑉 = 𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃

4+𝑟 sin 𝜃

0 4

0 2𝜋

0

= 𝑟𝑧 ₀^{4+𝑟 sin 𝜃}

4

0 2𝜋

0

𝑑𝑟 𝑑𝜃

(22)

= 4𝑟 + 𝑟²sin 𝜃 − 0

4

0 2𝜋

0

𝑑𝑟 𝑑𝜃

= 2𝑟²+1

3𝑟³sin 𝜃

0 2𝜋 4

0

𝑑𝜃

= 2 × 4²+1

3× 4³sin 𝜃

2𝜋

0

− 0 𝑑𝜃

= 32 +64 3 sin 𝜃

2𝜋

0

𝑑𝜃

= 32𝜃 −64 3 cos 𝜃

0 2𝜋

= (32 × 2𝜋 −64

3 cos 2𝜋) − 0 = 64𝜋 ∎

5. Pusat massa benda pejal homogen yang di atas dibatasi oleh 𝑧 = 12 − 2𝑥²− 2𝑦² dan di bawah oleh 𝑧 = 𝑥²+ 𝑦².

Penyelesaian:

Daerah S dalam Koordinat Tabung :

𝑆 = 𝑟, 𝜃, 𝑧 |0 ≤ 𝑟 ≤ 2,0 ≤ 𝜃 ≤ 2𝜋, 𝑟² ≤ 𝑧 ≤ 12 − 2𝑟² Maka,

 𝑚 = _𝑟^12−2𝑟2 ²𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃

2 0 2𝜋 0

= 𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃

12−2𝑟²

𝑟² 2

0 2𝜋

0

= 𝑟𝑧 12−2𝑟_𝑟2 ²𝑑𝑟 𝑑𝜃

2

0 2𝜋

0

= 𝑟 12 − 2𝑟² − 𝑟 × 𝑟² 𝑑𝑟 𝑑𝜃

2

0 2𝜋

0

= 12𝑟 − 3𝑟³ 𝑑𝑟 𝑑𝜃

2

0 2𝜋

0

= 6𝑟²−3 4𝑟⁴

0 2𝜋 2

0

𝑑𝜃

(23)

= 6 × 2²−3 4× 2⁴

2𝜋

0

− 0 𝑑𝜃

= 12

2𝜋

0

𝑑𝜃 = 12𝜃 ₀^2𝜋 = 12 × 2𝜋 − 0 = 24𝜋

 𝑀_𝑥𝑦 = ₀^2𝜋 ₀² _𝑟^12−2𝑟₂ ²𝑧𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃 = 1

2𝑟𝑧²

𝑟² 12−2𝑟² 2

0 2𝜋

0

𝑑𝑟 𝑑𝜃

= 𝑟 12 − 2𝑟² ² 2

2

0 2𝜋

0

−𝑟 𝑟² ² 2 𝑑𝑟 𝑑𝜃

= 72𝑟 − 24𝑟³+3 2

2

0 2𝜋

0

𝑟⁵ 𝑑𝑟 𝑑𝜃

= 36𝑟²− 6𝑟⁴+𝑟⁶ 4 ₀

2𝜋 2

0

𝑑𝜃

= 36 × 2²− 6 × 2⁴+2⁶ 4 − 0

2𝜋

0

𝑑𝜃

= 64

2𝜋

0

𝑑𝜃 = 64𝜃 ₀^2𝜋 = 64 × 2𝜋 − 0 = 128𝜋

 𝑀_𝑦𝑧 = ₀^2𝜋 ₀² _𝑟^12−2𝑟₂ ²𝑥𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃 = 𝑟𝑥𝑧 _𝑟12−2𝑟2 ²

2

0 2𝜋

0

𝑑𝑟 𝑑𝜃

= 𝑟𝑥 12 − 2𝑟²

2

0 2𝜋

0

− 𝑟𝑥 𝑟² 𝑑𝑟 𝑑𝜃

= 12𝑟𝑥 − 3𝑟³𝑥

2

0 2𝜋

0

𝑑𝑟 𝑑𝜃

(24)

= 6𝑟²𝑥 −3𝑟⁴𝑥 4 ₀

2𝜋 2

0

𝑑𝜃

= 6 × 2²𝑥 −3 × 2⁴𝑥 4 − 0

2𝜋

0

𝑑𝜃

= 12𝑥

2𝜋

0

𝑑𝜃 = 12𝑥𝜃 ₀^2𝜋 = 12𝑥 × 2𝜋 − 0 = 24𝑥𝜋

 𝑀_𝑥𝑧 = ₀^2𝜋 ₀² _𝑟^12−2𝑟₂ ²𝑦𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃 = 𝑟𝑦𝑧 _𝑟12−2𝑟2 ²

2

0 2𝜋

0

𝑑𝑟 𝑑𝜃

= 𝑟𝑦 12 − 2𝑟²

2

0 2𝜋

0

− 𝑟𝑦 𝑟² 𝑑𝑟 𝑑𝜃

= 12𝑟𝑦 − 3𝑟³𝑦

2

0 2𝜋

0

𝑑𝑟 𝑑𝜃

= 6𝑟²𝑦 −3𝑟⁴𝑦 4 ₀

2𝜋 2

0

𝑑𝜃

= 6 × 2²𝑦 −3 × 2⁴𝑦 4 − 0

2𝜋

0

𝑑𝜃

= 12𝑦

2𝜋

0

𝑑𝜃 = 12𝑦𝜃 ₀^2𝜋 = 12𝑦 × 2𝜋 − 0 = 24𝑦𝜋

 𝑧 =^𝑀^𝑥𝑦

𝑚 =^128𝜋

24𝜋 =¹⁶

3

 𝑥 =^𝑀_𝑚^𝑦𝑧 =^24𝑥𝜋_24𝜋 = 1

 𝑦 =^𝑀^𝑥𝑧

𝑚 =^24𝑦𝜋

24𝜋 = 1, karena 𝑥 = 𝑦 = 1 maka simetri ∎

(25)

6. Massa benda pejal di dalam bola berjejari 2a dan di luar tabung lingkaran berjejari a, yang sumbunya adalah diameter bola tersebut, jika kerapatan berbanding lurus dengan kuadrat jarak dari pusat bola.

Penyelesaian:

8 𝑘𝜌² 𝜌²

2𝑎

𝑎 csc∅

sin∅ 𝜋 2

0 𝜋 2

𝜋 6

𝑑𝜌 𝑑𝜃 𝑑∅= 8 𝑘𝜌⁵ sin∅

5 _{𝑎 csc}

∅

𝜋 2𝑎 2

0 𝜋 2

𝜋 6

𝑑𝜃 𝑑𝜑

= 8 𝑘 × 2𝑎 ⁵sin∅

5 − 𝑘 × 𝑎 csc∅⁵sin∅

5

𝜋2

0 𝜋2

𝜋6

𝑑𝜃 𝑑∅

= 8 ^32𝑎

5𝑘 sin ∅ − 𝑘 × 𝑎 csc ∅ ⁵sin ∅ 5

𝜋 2

0 𝜋2

𝜋6

𝑑𝜃 𝑑∅

= 8 ^32𝑎

5𝑘𝜃 sin ∅ − 𝑘𝜃 × 𝑎 csc ∅ ⁵sin ∅

5 0

𝜋2 𝜋 2

𝜋 6

𝑑∅

= 8 ^32𝜋𝑎

5𝑘 sin ∅ − 𝜋𝑘 × 𝑎 csc ∅ ⁵sin ∅

10

𝜋 2

𝜋 6

𝑑∅

= 8 7 3 𝜋𝑘𝑎⁵

5

=56 3 𝜋𝑘𝑎⁵ 5 ∎

7. Momen Inersia benda pejal dalam Soal 18 terhadap sumbu simetrinya.

Penyelesaian:

𝐼_𝑧= 𝑥²+ 𝑦² 𝑘 𝑥²+ 𝑦² ¹²

𝑆

𝑑𝑉

= 𝑘 𝜌⁵ 𝑠𝑖𝑛⁴∅

𝑎

0 2𝜋

0 𝜋2

0

𝑑𝜌 𝑑𝜃 𝑑∅

= 𝑘 𝜌⁶𝑠𝑖𝑛⁴∅

6 ₀

2𝜋 𝑎

0 𝜋2

0

𝑑𝜃 𝑑∅

= 𝑘 𝑎⁶𝑠𝑖𝑛⁴∅ 6

2𝜋

0

− 0

𝜋2

0

𝑑𝜃 𝑑∅