Question

高等数学 二重积分极坐标如何求解

Answer 1

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Answer 2

过程与结果如图所示

追答

望采纳…

Answer 3

本题用极坐标并不简单。
y = √(1-x^2) , 即上半单位圆 x^2+y^2 = 1, 化为极坐标是 r = 1,
y = x, 化为极坐标是 tant = 1, 则 t = π/4，
x = 0, 化为极坐标是 cost = 0, 则 t = π/2，
积分域 D 是单位圆中 π/4 ≤ t ≤ π/2 的扇形。
I = ∫<π/4, π/2>dt∫<0, 1>√[1-(rsint)^2] rdr
= (1/2)∫<π/4, π/2>dt∫<0, 1>[-1/(sint)^2]√[1-r^2(sint)^2] d[1-r^2(sint)^2]
= (1/2)∫<π/4, π/2>[-1/(sint)^2]dt[(2/3){1-r^2(sint)^2}^(3/2)]<0, 1>
= (1/2)∫<π/4, π/2>[-1/(sint)^2][(cost)^3-1]dt
= (1/2){∫<π/4, π/2>[1/(sint)^2]dt - ∫<π/4, π/2>[1-(sint)^2]dsint/(sint)^2]}
= (1/2){[-cott]<π/4, π/2> + [1/sint + sint]<π/4, π/2>}
= (1/2)[1+(1+1-√2-√2/2)] = (3/4)(2-√2)