請教一題複變積分問題 - 考試
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By Charlotte
at 2013-07-13T23:11
at 2013-07-13T23:11
Table of Contents
可以合在一起算比較快 避免算兩個積分式易眼殘
代入路徑參數 y = -x 則積分上下線變成 x=0 --> x=2
2-2i 2-2i
∫ |z| dz = ∫ (x^2 + (y)^2)^1/2 (dx + idy)
0 0
2-2i 2
∫ |z| dz = ∫ (x^2 + (-x)^2)^1/2 (dx - idx)
0 0
2-2i 2
∫ |z| dz = ∫ (x^2 + (-x)^2)^1/2 (1 - i)dx
0 0
2
= (1 - i) 2^1/2 ∫ x dx
0
= 2*2^1/2 * (1 - i)
※ 引述《raefield (Rae)》之銘言:
: [課業] 國考課業相關問題,非歷屆考題的討論,如學理觀念的釐清。
: 寫得有點亂 不好意思
: http://ppt.cc/mMP8
: 題目於圖左上
: 想請教一下這題目這樣的做法是否正確?
: 算是算出來了...但自己都搞不清楚自己到底怎麼算的|||
: 煩請各位大大指教|||OTL
--
代入路徑參數 y = -x 則積分上下線變成 x=0 --> x=2
2-2i 2-2i
∫ |z| dz = ∫ (x^2 + (y)^2)^1/2 (dx + idy)
0 0
2-2i 2
∫ |z| dz = ∫ (x^2 + (-x)^2)^1/2 (dx - idx)
0 0
2-2i 2
∫ |z| dz = ∫ (x^2 + (-x)^2)^1/2 (1 - i)dx
0 0
2
= (1 - i) 2^1/2 ∫ x dx
0
= 2*2^1/2 * (1 - i)
※ 引述《raefield (Rae)》之銘言:
: [課業] 國考課業相關問題,非歷屆考題的討論,如學理觀念的釐清。
: 寫得有點亂 不好意思
: http://ppt.cc/mMP8
: 題目於圖左上
: 想請教一下這題目這樣的做法是否正確?
: 算是算出來了...但自己都搞不清楚自己到底怎麼算的|||
: 煩請各位大大指教|||OTL
--
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考試
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By Damian
at 2013-07-17T05:37
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at 2013-07-18T01:44
at 2013-07-18T01:44
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