math, line integral
回答 (2)
For path A, the equation is r(t)=ti+tj (0<=t<=1)
r'(t)=i+j
∫F(r(t)).r'(t)dt [from 0 to 1]
=∫(ti+tj)(i+j)dt
=∫2t dt
=1
For path B, the equation is r(t)=ti+t^2j (0<=t<=1)
r'(t)=i+2tj
∫F(r(t)).r'(t)dt [from 0 to 1]
=∫(ti+t^2j)(i+2tj)dt
=∫t+2t^3 dt
=1
For path C, the equation is r1(t)=ti (0<=t<=1) r2(t)=tj
r1'(t)=i r2'(t)=j
∫F(r(t)).r1'(t)+F(r(t)).r2'(t)dt [from 0 to 1]
=∫(ti)(i)+(tj)(j) dt
=1
shows that the integration is path-independent
ok?^_^
收錄日期: 2021-04-21 00:27:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080508000051KK00825
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