A(a,0), O(2a,0), B(2a,a) 其中AB為1/4圓,O為圓心
圓弧AB及Y=0及X=2a 圍成的封閉面積
繞Y軸旋轉1/4圈後,圓弧AB所掃過的表面積=?
更新1:
小弟另有一解法:想法直接,積分較繁 (x-2a)^2+y^2=a^2, y=√a^2-(x-2a)^2 -> dy/dx= -(x-2a)/√a^2-(x-2a)^2 circular arc segment = ds rotational segment= x dθ [x: radius of rotation) Area=∫ds *∫x dθ [rectangular]
更新2:
ds=√dx^2+dy^2 = √1+(dy/dx)^2 * dx Area=∫√1+[-(x-2a)/√a^2-(x-2a)^2]^2 dx * x [π/2 – 0] =π/2 ∫x * √1+[-(x-2a)/√a^2-(x-2a)^2]^2 dx Let t=x-2a, after integration, Area=πa^2(π-1)/2 Where ∫tdt/√a^2-t^2 = √t^2-a^2, ∫dt/√a^2-t^2 = arcsin (t/a)