设L为 ^2+(y)^2=(a)^2 在第一象限内的部份,则 int (e)^sqrt ({x^2+{y)^2}}ds=. ()-|||-A dfrac (pi )(2)a(e)^a-|||-B πa^2e^a-|||-C dfrac (pi )(3)a(e)^a-|||-D )π/4a^2e^a