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