设半径为a,圆心在原点的圆的面积为s,∫√(a^2-x^2)dx=?,下限0,上线a

Question

设半径为a,圆心在原点的圆的面积为s,∫√(a^2-x^2)dx=?,下限0,上线a

Answer 1

令x=asint,0=∫√(a^2-x^2)dx=a^2∫(cost)^2 dt=a^2∫(cos2t+1)/2 dt
前面的积分为0,∫√(a^2-x^2)dx=a^2*pi/4