Question

如图，请问这个定积分怎么解，过程咋写？

Answer 1

朋友，你好！学习过程rt所示，希望能帮到你解决问题

Answer 2

2x-x^2 = 1-(x-1)^2, 令 x = 1+sint， 得 dx = costdt， 积分变为
I = 2π∫<-π/2→0>(1-sint) (cost -1-sint)dt
= 2π∫<-π/2→0>[cost -1-2sint-sintcost+(sint)^2]dt
= 2π∫<-π/2→0>[cost -1/2-2sint-sintcost-(1/2)cos2t]dt
= 2π[sint-t/2+2cost-(1/2)(sint)^2-(1/4)sin2t]<-π/2→0>
= 2π[2-(-1+π/4-1/2)] = 2π(7/2-π/4) = π(7-π/2)

Answer 3

虽然我看到这个问题了，但是我不会。

Answer 4

先看根号下的式子：
=1 - (x² - 2x + 1)
=1 - (x-1)²
因为有二次根号，且有 (x-1)² < 1 的条件限制，所以把它转换成正弦函数来做就行了。
设 1 - x = sinθ，则 x = 1 - sinθ，dx = -cosθ * dθ。原积分范围就可以变换为：π/2 → 0
那么，原积分就可以变换成：
V = 2π * ∫(1+sinθ) * [√(1-sin²θ) - 1 + sinθ] * (-cosθ) * dθ
= -2π * ∫(1+sinθ)(cosθ+sinθ-1) * cosθ * dθ
= -2π * ∫(cosθ+sinθcosθ+sinθ+sin²θ -1 -sinθ) * cosθ * dθ
= -2π * ∫(cosθ + sinθcosθ + sin²θ - 1) * cosθ * dθ
= -2π * ∫[cos²θ + sinθ * cos²θ + sin²θ * cosθ - cosθ] * dθ
= -2π * [1/2 * ∫2cos²θdθ + ∫cos²θ * sinθdθ + ∫sin²θ *cosθdθ - ∫cosθdθ]
= -2π * {1/2 * ∫[1+cos(2θ)]dθ - ∫cos²θ *d(cosθ) + ∫sin²θ * d(sinθ) - ∫d(sinθ)}
= -2π * [1/2 * ∫dθ + 1/2 * ∫cos(2θ)dθ - 1/3 * cos³θ + 1/3 * sin³θ - θ]
= -2π * [ 1/2 * θ + 1/4 * ∫cos(2θ) * d(2θ) - 1/3 * cos³θ + 1/3 * sin³θ - θ]
= -2π * [1/4 * sin(2θ) - 1/3 * cos³θ + 1/3 * sin³θ - θ/2]
= -π/2 * sin(2θ)|θ=π/2→0 + 2π/3 * cos³θ|θθ=π/2→0 - 2π/3 * sin³θ|θ=π/2→0 + π * θ|θ=π/2→0
= -π/2 * (sin0 - sinπ) + 2π/3 * [cos³0 - cos³(π/2)] - 2π/3 * [sin³0 - sin³(π/2)] + π * (0 - π/2)
= 0 + 2π/3 + 2π/3 - π²/2
= 4π/3 - π²/2