𝐶â𝑢 1
𝜕𝑓′(𝑀)
𝜕𝑢⃗ = (𝑓
′
𝑥 , 𝑓′
𝑦 , 𝑓′
𝑧) 𝑢⃗
|𝑢|= (−4𝑥 ,2𝑦𝑧
2 , 2𝑧𝑦2+ 1) (0, 1
√2,
1
√2)
𝑡ℎ𝑎𝑦 𝑡ọ𝑎 độ 𝑀(1,1, −2) 𝑣à𝑜 𝑡𝑎 đượ𝑐 ∶ 𝜕𝑓′(𝑀)
𝜕𝑢⃗ = 4√2 +
−3
√2=
5
√2
𝐶â𝑢 2
𝑋é𝑡 𝐼 = ∫(𝑥𝑦 − 𝑦2)𝑑𝑥 + (2𝑥𝑦 + 𝑥2)𝑑𝑦
𝐷𝑜 𝑚𝑖ề𝑛 𝑐ủ𝑎 đề 𝑏à𝑖 𝑙à 𝑚𝑖ề𝑛 𝑘í𝑛 𝑛ê𝑛 𝑠ử 𝑑ụ𝑛𝑔 𝐺𝑟𝑒𝑒𝑛 𝑣à đ𝑖 𝑡ℎ𝑒𝑜 𝑐ℎ𝑖ề𝑢 𝑘𝑖𝑚 đồ𝑛𝑔 ℎồ 𝑛ê𝑛
𝐼 = − ∬(2𝑦 + 2𝑥 − 𝑥 + 2𝑦)𝑑𝑥𝑑𝑦 = − ∫ 𝑑𝜑
𝜋
3𝜋 4
∫[4𝑟𝑠𝑖𝑛𝜑 + (2 + 𝑟𝑐𝑜𝑠𝜑)
2
0
]𝑟𝑑𝑟 =20√2
3 − 𝜋 +
−32 3
𝐶â𝑢 3
𝐶â𝑢 4
𝐼 = ∬(𝑥3− 3𝑦𝑧)𝑑𝑦𝑑𝑧 − (𝑦2+ 2𝑥𝑦)𝑑𝑧𝑑𝑥 + (𝑧 − 𝑥)𝑑𝑥𝑑𝑦
𝐼1= ∬(𝑥3− 𝑦𝑧)𝑑𝑦𝑑𝑧 = 0(𝑑𝑜 𝑚ặ𝑡 𝑡𝑟ụ 𝑧 = 4 − 𝑦2 𝑠𝑜𝑛𝑔 𝑠𝑜𝑛𝑔 𝑣ớ𝑖 𝑂𝑥 )
Trang 2𝐼2= ∬ −(𝑦2+ 2𝑥𝑦)𝑑𝑧𝑑𝑥 = ∬ −𝑦2𝑑𝑧𝑑𝑥 + ∬ −2𝑥𝑦𝑑𝑧𝑑𝑥
𝑇𝑎 có ∬– 𝑦2𝑑𝑧𝑑𝑥 = 0 𝑣ì 𝑡í𝑛ℎ đố𝑖 𝑥ứ𝑛𝑔 𝑣à ℎà𝑚 𝑐ℎẳ𝑛 𝑡ℎ𝑒𝑜 𝑦
𝑆𝑢𝑦 𝑟𝑎 𝐼2 = − ∬ 2𝑥𝑦𝑑𝑧𝑑𝑥
𝑆1∪𝑆2
= − [− ∬ 2𝑥√4 − 𝑧
𝐷𝑥𝑧
𝑑𝑥𝑑𝑧 + ∬ 2𝑥(−√4 − 𝑧)𝑑𝑥𝑑𝑧
𝐷𝑥𝑧
]
= 2 ∬ 2𝑥√4 − 𝑧
𝐷𝑥𝑧
𝑑𝑥𝑑𝑧 = 2 ∫ 𝑑𝑥
2
0
∫ 2𝑥√4 − 𝑧𝑑𝑧 =−1184
21
4−2𝑥
0
𝐼3= ∬(𝑧 − 𝑥)𝑑𝑥𝑑𝑦 = − ∬(4 − 𝑦2− 𝑥)𝑑𝑥𝑑𝑦 = − ∫ 𝑑𝑦
2
−2
∫ (4 − 𝑦2− 𝑥)𝑑𝑥 = −8
3
𝑦2 2
0
𝐷𝑥𝑦
𝑉ậ𝑦 𝐼 =−1184
8 3
𝐶â𝑢 5
𝐶â𝑢 6
Trang 3𝐶â𝑢 7