Question

求助高数大神！！！

Answer 1

联立解 x^2+y^2+2x+2y = 0 与 x^2+y^2+2x-2y = 0 得交点 O(0, 0), P(-2, 0),
D 由 x^2+y^2+2x+2y = 0 与 x^2+y^2+2x-2y = 0 围成的两弓形之和。
D 上半部分，x^2+y^2+2x+2y ≤ 0, r^2 + 2r(cost+sint) ≤ 0,
r = -2(cost+sint), 3π/4 ≤ t ≤ π;
D 下半部分，x^2+y^2+2x-2y ≤ 0, r^2 + 2r(cost-sint) ≤ 0,
r = -2(cost-sint)， π ≤ t ≤ 5π/4.
I = ∫∫<D>√(x^2+y^2)dxdy
= ∫<3π/4, π>dt∫<0, -2(cost+sint)>r^2dr + ∫<π, 5π/4>dt∫<0, -2(cost-sint)>r^2dr
= (-8/3)∫<3π/4, π>(cost+sint)^3dt + (-8/3)∫<π, 5π/4>(cost-sint)^3dt
= (-8/3)∫<3π/4, π>2√2[sin(t+π/4)]^3dt + (-8/3)∫<π, 5π/4>2√2[cos(t+π/4)]^3dt
= (-16√2/3)∫<π, 5π/4>(sinu)^3du + (-16√2/3)∫<5π/4, 3π/2>(cosu)^3du
= (16√2/3)∫<π, 5π/4>[1-(cosu)^2]dcosu - (16√2/3)∫<5π/4, 3π/2>[1-(sinu)^2]dsinu
= (16√2/3)[cosu-(1/3)(cosu)^3]<π, 5π/4> - (16√2/3)[sinu-(1/3)(sinu)^3]<5π/4, 3π/2>
= (16√2/3)[-1/√2+(1/6)/√2+1-1/3] - (16√2/3)[-1+1/3+1/√2-(1/6)/√2]
= (16√2/3)[-(5/3)/√2+4/3] = (16/9)(4√2-5)

Answer 2