急求星形线x^(2/3)+y^(2/3)=a^(2/3)与x=a及y=a围成的图形的面积?谢谢

化成参数方程,
x=a(cost)^3,
y=a(sint)^3,
图形星形线第一象限和正方形之间所围成图形,
S=∫ [0,a] ydx=4∫ [π/2,0] a(sint)^3d[a (cost)^3]
=a^2∫ [π/2,0] (sint)^3 *[3(cost)^2*(-sint)]dt
=(-3a^2)∫ [π/2,0](sint)^4(cost)^2dt
=(-3/8)∫ [π/2,0] [1-2cos2t+(cos2t)^2](1+cos2t)dt
=(-3/8)a^2 [π/2,0][t-t/2-sin4t/8+(1/2)sin2t-(sin2t)^3/3]
=(-3a^2/8)[0-(π/4-0+0-0]
=3πa^2/32,
第一象限星形线外是正方形,面积为a^2,
∴二者所围成面积为:a^2-3πa^2/32.

极坐标和参数都可以做，多看下书，画图，画出图形知道求哪块就简单了''

设a>0，x=a(cosθ)^3，y=a(sinθ)^3

θ=0时，x=a，y=0；θ=π/2时，x=0，y=a

星形线在第一象限的面积

=∫(θ:π/2→0)ydx

=-∫(θ:0→π/2)ydx

=-∫(0→π/2)a(sinθ)^3*3*a(cosθ)^2*(-sinθ)dθ

=3a^2∫(0→π/2)(sinθ)^4*(cosθ)^2*dθ

=3a^2∫(0→π/2)(sinθ)^4*[1-(sinθ)^2]dθ

=3a^2*[∫(0→π/2)(sinθ)^4*dθ-∫(0→π/2)(sinθ)^6*dθ]

=3a^2*(3/4*1/2*π/2-5/6*3/4*1/2*π/2)

=3πa^2/32

所求图形的面积=a^2-3πa^2/32=(1-3π/32)a^2