展开全部
∫(0->2) ρ^3 dρ
=(1/4)[ρ^4]|(0->2)
=4
∫(0->2π) [∫(0->2) ρ^3 dρ] dθ
=∫(0->2π) 4 dθ
=8π
∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ
=[ (2/3)sinθ.ρ^3 - (1/4)ρ^4 ]|(0->2sinθ)
=(16/3)(sinθ)^4 -(1/4)(sinθ)^4
=(61/12)(sinθ)^4
∫(0->π) [∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ] dθ
=∫(0->π) (61/12)(sinθ)^4 dθ
=(61/48)∫(0->π) (1-cos2θ)^2 dθ
=(61/48)∫(0->π) [ 1-2cos2θ + (cos2θ)^2 ] dθ
=(61/96)∫(0->π) ( 3-4cos2θ + cos4θ) dθ
=(61/96) [ 3θ-2sin2θ + (1/4)sin4θ] |(0->π)
=(61/32)π
∫(0->2π) [∫(0->2) ρ^3 dρ] dθ +2∫(0->π) [∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ] dθ
=8π +2[(61/32)π]
=8π +(61/16)π
=(189/16)π
=(1/4)[ρ^4]|(0->2)
=4
∫(0->2π) [∫(0->2) ρ^3 dρ] dθ
=∫(0->2π) 4 dθ
=8π
∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ
=[ (2/3)sinθ.ρ^3 - (1/4)ρ^4 ]|(0->2sinθ)
=(16/3)(sinθ)^4 -(1/4)(sinθ)^4
=(61/12)(sinθ)^4
∫(0->π) [∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ] dθ
=∫(0->π) (61/12)(sinθ)^4 dθ
=(61/48)∫(0->π) (1-cos2θ)^2 dθ
=(61/48)∫(0->π) [ 1-2cos2θ + (cos2θ)^2 ] dθ
=(61/96)∫(0->π) ( 3-4cos2θ + cos4θ) dθ
=(61/96) [ 3θ-2sin2θ + (1/4)sin4θ] |(0->π)
=(61/32)π
∫(0->2π) [∫(0->2) ρ^3 dρ] dθ +2∫(0->π) [∫(0->2sinθ) ρ(2ρsinθ -ρ^2) dρ] dθ
=8π +2[(61/32)π]
=8π +(61/16)π
=(189/16)π
更多追问追答
追问
我算的也是这样,但是答案是9π。
追答
那你查一下题目有没有错!
不然的话,答案也不一定是对的!
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询
广告 您可能关注的内容 |