求∮√x^2+y^2ds（其中L为圆周x^2+y^2=ax）的积分值

Question

求∮√x^2+y^2ds（其中L为圆周x^2+y^2=ax）的积分值

Answer 1

求∮√(x²+y²)ds（其中L为圆周x²+y²=ax）的积分值
L：x²-ax+y²=(x-a/2)²+y²-a²/4=0,故得(x-a/2)²+y²=a²/4,这是一个圆心在(a/2,0),半径r=a/2的圆；故写成参数形式就是：x=(a/2)(1+cos2t),y=(a/2)sin2t,t∈[-π/2,π/2].
ds=√[(dx/dt)²+(dydt)²]dt=√[(-asin2t)²+(acos2t)²]dt=adt
故∮√(x²+y²)ds=[-π/2,π/2]a∫√[(a²/4)(1+cos2t)²+(a²/4)sin²2t]dt
=[-π/2,π/2](a²/2)∫√(2+2cos2t)dt=[-π/2,π/2][(√2)a²/2]∫√(1+cos2t)dt
=[-π/2,π/2][(√2)a²/2]∫√(2cos²t)dt=[-π/2,π/2](a²)∫costdt
=(a²)sint︱[-π/2,π/2]=a²[sin(π/2)-sin(-π/2)]=2a².