Question

高等数学如图第三小题

Answer 1

积分域 D 是 以 O(0, 0), A(2, 0), B(1, √3) 为顶点的正三角形。
直线 AB 方程 y/(x-2) = -√3/(2-1), 即 y = -√3(x-2).
化为极坐标，rsint = -√3(rcost-2),
得 r = 2√3/(sint+√3cost) = √3/sin(t+π/3), 0 ≤ t ≤ π/3.
原积分 I = ∫<0, π/3>dt ∫<0, √3/sin(t+π/3)>rdr/(1+r^2)^2
= (1/2)∫<0, π/3>dt ∫<0, √3/sin(t+π/3)> d(1+r^2)/(1+r^2)^2
= (1/2)∫<0, π/3>dt[-1/(1+r^2)]<0, √3/sin(t+π/3)>
= (1/2)∫<0, π/3>3dt/{3+[sin(t+π/3)]^2}
= (3/2)∫<0, π/3>d(2t)/[6+1-cos(2t+2π/3)] (令u = 2t+2π/3)
= (3/2)∫<2π/3, 4π/3>du/(7-cosu) [令 v = tan(u/2)]
= (3/2)∫<√3, -√3>dv/(3+4v^2) = (1/2)(√3/2)∫<√3, -√3>d(2v/√3)/(1+4v^2/3)
= (√3/4)[arctan(2v/√3)]<√3, -√3> = -(√3/2)arctan2

Answer 2

表示我不会