1. The document discusses double integrals and their application to moments of inertia. It explains how to calculate the volume under a surface using double integrals as the limit of Riemann sums as the number of partitions goes to infinity. 2. It also discusses how double integrals can be evaluated by changing the order of integration, and some properties of double integrals like linearity. 3. Examples are given of calculating double integrals to find volumes and moments of inertia, like finding the moment of inertia of a rod about its center of mass.

1. PENJELASAN INTEGRAL LIPAT
DUA & PENERAPAN PADA
MOMEN INERSIA
By
Mochammad Purinda Bisma Samudra
(4216100076)

2. PENDAHULUAN

3. 𝑎 ≤ 𝑥 ≤ 𝑏
[𝑥𝑖−1 𝑥1]
∆ 𝑥 = (𝑏 − 𝑎)/𝑛
𝑖=1
𝑛
𝑓 𝑥1 ∆𝑥

4.  𝑛 → ∞
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 =
𝑖=1
𝑛
𝑓(𝑥1)∆𝑥

𝑎
𝑏
𝑓 𝑥 𝑑𝑥


5. INTEGRAL LIPAT DUA

6. 
𝑅2

 𝑥𝑜𝑦

7. 𝑹 = [𝒂, 𝒃] 𝒙[𝒄, 𝒅] = {(𝒙, 𝒚) ∈ 𝑹 𝟐|𝒂 ≤ 𝒙 ≤ 𝒃, 𝒄 ≤ 𝒚 ≤ 𝒅}
𝑓 (𝑥, 𝑦) ≥ 0
𝒛 = 𝒇 ( 𝒙, 𝒚)
(𝑺 = {( 𝒙, 𝒚, 𝒛) ∈ 𝑹 𝟑|0 ≤ 𝒛 ≤ 𝒇 ( 𝒙, 𝒚), ( 𝒙, 𝒚) ∈ 𝑹}
[𝑥𝑖−1 𝑥1] ∆𝒙 =
𝒃−𝒂
𝒎
[𝑦𝑖−1 𝑦1] ∆ 𝒚 =
𝒅−𝒄
𝒏

8. 𝑹 = 𝑥𝑖−1 𝑥1 𝒙 𝑦𝑖−1 𝑦1 = {( 𝒙, 𝒚)|𝑥𝑖−1 ≤ 𝒙 ≤ 𝑥𝑖 𝑦𝑖−1 ≤ 𝒚𝒊}

9. 𝒙𝒊𝒋 𝒚𝒊𝒋
𝑹𝒊𝒋
𝑹𝒊𝒋
𝑹𝒊𝒋 𝒙𝒊𝒋 𝒚𝒊𝒋
(𝒇 𝒙𝒊𝒋 𝒚𝒊𝒋 𝒙 ∆ 𝑨 = 𝒇 𝒙𝒊𝒋 𝒚𝒊𝒋 𝒙 ∆𝒙∆ 𝒚
𝑣 ≈
𝑖=𝑥
𝑚
𝑗=1
𝑛
𝒇 𝒙𝒊𝒋 𝒚𝒊𝒋 ∆ 𝑨

10. 𝑣 = lim
𝑚,𝑛→∞
𝑖=1
𝑚
𝑗=1
𝑛
𝒇 𝒙𝒊𝒋 𝒚𝒊𝒋 ∆ 𝑨
𝒇 ( 𝒙, 𝒚) 𝒅𝑨 ≥ 0
𝒛 = 𝒇 ( 𝒙, 𝒚)
𝑅
𝒇 𝒙𝒊𝒋 𝒚𝒊𝒋 𝑑𝐴

11. 1.
2. 𝑅
𝑐𝑓 𝒙, 𝒚 𝑑𝐴 = 𝑐 𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 , 𝑐 𝑎𝑑𝑎𝑙𝑎ℎ 𝑘𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑎
3. Jika 𝒇 𝒙, 𝒚 ≥ 𝒈 𝒙, 𝒚 𝑢𝑛𝑡𝑢𝑘 ∀ 𝒙, 𝒚 ∈ 𝑅, 𝑚𝑎𝑘𝑎
𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 ≥
𝑅
𝑔 𝒙, 𝒚 𝑑𝐴

12. 𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 =
𝑅
𝑓 𝒙, 𝒚 𝑑𝑥𝑑𝑦 =
𝑎
𝑏
𝑦=𝑓1(𝑦)
𝑦=𝑓2(𝑦)
𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦
𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 =
𝑅
𝑓 𝒙, 𝒚 𝑑𝑦𝑑𝑥 =
𝑎
𝑏
𝑦=𝑓1(𝑦)
𝑦=𝑓2(𝑦)
𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

13. CONTOH SOAL

14. 𝐴 =
𝑅
𝑑𝐴 =
0
2
2𝑦−4
2−𝑦
𝑑𝑥𝑑𝑦 =
0
2
𝑥
2𝑦−4
2−𝑦
𝑑𝑡
=
0
2
2 − 𝑦 − 2𝑦 − 4 𝑑𝑦 =
0
2
6 − 3𝑦 𝑑𝑦
= 6𝑦 −
3
2
𝑦2
0
2
= (12 − 6) = 6

15. 𝒛 = 𝒙 𝟐
+ 𝒚 𝟐
𝑦 = 𝑥2
𝐷 = 𝑥, 𝑦 0 ≤ 𝑥 ≤ 2, 𝑥2
≤ 𝑦 ≤ 2𝑥
𝒛 = 𝒙 𝟐
+ 𝒚 𝟐
𝑣 =
𝐷
𝑥2
+ 𝑦2
𝑑𝐴 =
0
2
𝑥
2𝑥
(𝑥2
+ 𝑦2
)𝑑𝑥𝑑𝑦
𝑣 = ?
=
0
2
𝑥2
𝑦 +
𝑦3
3 𝑦=𝑥2
𝑦=2𝑥
𝑑𝑥 =
0
2
𝑥2
2𝑥 +
2𝑥3
3
− 𝑥2
𝑥2
(𝑥2
)3
3
𝑑𝑥
−𝑥 6
3
− 𝑥4
+
14𝑥3
3
𝑑𝑥 =
−𝑥7
21
−
𝑥5
5
+
7𝑥4
6
0
2
=
216
35

16. PENERAPAN INTEGRAL LIPAT DUA PADA
MOMEN INERSIA

17. 𝑬𝒌 =
1
2
𝒎𝒗 𝟐

18. 𝑟 𝜔
𝑬𝒌 =
1
2
(𝑟2 𝒎)𝜔2
𝑬𝒌 =
1
2
𝑰𝜔2

19. 𝐼 𝑥 = lim
𝑚,𝑛→∞
𝑖=1
𝑚
𝑗=1
𝑛
𝑦𝑖𝑗
2
𝑝 𝑥𝑖𝑗 𝑦𝑖𝑗 ∆𝐴 =
𝐷
𝑦2
𝐼 𝑥 = lim
𝑚,𝑛→∞
𝑖=1
𝑚
𝑗=1
𝑛
𝑥𝑖𝑗
2 𝑝 𝑥𝑖𝑗 𝑦𝑖𝑗 ∆𝐴 =
𝐷
𝑥2

20. 𝑟 =
𝐼
𝑚

21. 𝑟 =
𝐼
𝑚
𝑬𝒌 =
1
2
(𝑟2
𝒎)𝜔2

22. KESIMPULAN

23. 𝑅2.

24. 𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 =
𝑅
𝑓 𝒙, 𝒚 𝑑𝑥𝑑𝑦 =
𝑎
𝑏
𝑦=𝑓1(𝑦)
𝑦=𝑓2(𝑦)
𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦
𝑅
𝑓 𝒙, 𝒚 𝑑𝐴 =
𝑅
𝑓 𝒙, 𝒚 𝑑𝑦𝑑𝑥 =
𝑎
𝑏
𝑦=𝑓1(𝑦)
𝑦=𝑓2(𝑦)
𝑓(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

25. 𝑚𝑟 2
𝐼 = 𝑚𝑟2

26. 𝐼 𝑥 =
𝑠
𝑦2 𝛿 𝑥, 𝑦 𝑑𝐴
𝐼 𝑦 =
𝑠
𝑥2 𝛿 𝑥, 𝑦 𝑑𝐴
𝐼𝑧 =
𝑠
(𝑥2+𝑦2)𝛿 𝑥, 𝑦 𝑑𝐴 = 𝐼 𝑥 + 𝐼 𝑦