Question

请教一道高数二重积分的问题

∫∫D(-2-2y)dxdy,其中D是星形线

Answer 1

星形线x = a(cost)^3，y = a(sint)^3
D应该是星形线构成的闭域：x = r(cost)^3，y = r(sint)^3
(其中域D‘：0 <= r <= a，0 <= t <= 2pai)
Jacobi行列式J= [ (cost)^3 -3rsint(cost)^2 ]
[ (sint)^3 3rcost(sint)^2 ]
因此 | det(J) | = 3r(sint)^2 (cost)^2
∫∫D (-2-2y) dxdy = ∫∫D‘ (-2-2r(sint)^3)* 3r(sint)^2 (cost)^2 drdt
= ∫∫D‘ (-3/2)(sin2t)^2 drdt + ∫∫D‘ 6(r^2)*((sint)^4)*((cost)^2))*(-sint) drdt
= -(3/4)*pai*a^2

方法二(改进)：
∫∫D (-2-2y) dxdy = ∫∫D (-2) dxdy + ∫∫D (-2y) dxdy
= (-2)∫∫D dxdy (第二项为零原因：-2y关于x轴是奇函数；D对称)
= (-2)S (S = ∫∫D dxdy 是闭域D的面积，可由对称性只计算第I象限的部分)
= 4*(-2)∫∫D‘’ 3r(sint)^2 (cost)^2 drdt (其中域D‘：0 <= r <= a，0 <= t <= pai/2)
= (-8) * (3/32)*pai*a^2
= -(3/4)*pai*a^2

Answer 2

先利用对称性简化，再利用极坐标