求曲线积分∫（x+y）ds,其中L为双纽线r^2=a^2cos2t右面一瓣

Question

求曲线积分∫（x+y）ds,其中L为双纽线r^2=a^2cos2t右面一瓣

Answer 1

第一类曲线积分的极坐标形式.
r^2=a^2cos2t>0
取右面一半,那么-π/4x=r(t)cost=acost√cos2t
y=r(t)sint=asint√cos2t
ds=√[r^2+(r')^2]dt=√[a^2cos2t+(-asin2t/√cos2t)^2]dt=adt/√cos2t
所以
∫（x+y）ds=∫（acost√cos2t+asint√cos2t）adt/√cos2t
=a^2∫(-π/4到π/4) (sint+cost)dt
=√2a^2