Question

请问大家这道题怎么算啊 在线等详细过程 急急急急

Answer 1

用极坐标计算。
I = ∫<0, π/2>dt∫<0, 2cost>(1-r)rdr = ∫<0, π/2>dt∫<0, 2cost>(r-r^2)dr
= ∫<0, π/2>dt[r^2/2-r^3/3]<0, 2cost>
= ∫<0, π/2>[2(cost)^2-(8/3)(cost)^3]dt
= ∫<0, π/2>(1+cos2t)dt - (8/3)∫<0, π/2>[1-(sint)^2]dsint
= [t+(1/2)sin2t]<0, π/2> - (8/3)[sint-(sint)^3/3]<0, π/2>
= π/2 - (8/3)(2/3) = π/2 - 16/9
答案 2/3 不对。

Answer 2

这个二重积分由 xoy 坐标系 转换成极坐标系 得到的积分式是正确的。
那么，继续计算：
=∫dθ∫(r - r²)dr
=∫dθ * (∫rdr - ∫r²dr)
=∫dθ * [1/2 * r² - 1/3 * r³]|r=0→2cosθ
=∫dθ * [1/2 * 4cos²θ - 1/3 * 8cos³θ]
=∫(2cos²θ - 8/3 * cos³θ) * dθ
=∫2cos²θdθ - 8/3 * ∫cos²θ * cosθdθ
=∫[1+cos(2θ)]dθ -8/3 * ∫(1-sin²θ) d(sinθ) 注：d(sinθ) = cosθ * dθ
=[∫dθ + ∫cos(2θ)dθ] - 8/3 * [∫d(sinθ) - ∫sin²θ *d(sinθ)]
=[θ + 1/2 * ∫cos(2θ)d(2θ)] - 8/3 * [sinθ - 1/3 * sin³θ]|θ=0→π/2
=[θ + 1/2 * sin(2θ)]|θ=0→π/2 - 8/3 * [sin(π/2) - sin0 - 1/3 * sin³(π/2) + 1/3 * sin³0]
=[(π/2 - 0) + 1/2 * sin(2*π/2) - 1/2 * sin(2*0)] - 8/3 * [1 - 1/3]
=π/2 + 1/2 - 8/3 * 2/3
=π/2 + 1/2 - 16/9
=π/2 - 23/18

追问

答案是2/3