Question

一道关于对弧长的曲线积分的问题计算∮(ax+by)²ds积分区域是x²+y²=1

Answer 1

令x=cost,y=sint

ds=（x‘)^2+(y')^2=1*dt=dt t∈[0,2π]

∫[0,2π](acost+bsint)^2dt

=∫[0,2π](a^2*cost^2+b^2sint^2+2absintcost)dt

=∫[0,2π)(1/2*a^2(1+cos2t)+1/2*b^2*(1-cos2t)+2absintcost)dt

=1/2*a^2*(2π）+1/2*b^2*(2π)+0

=π（a^2+b^2)