Question

利用柱面坐标系求三重积分z=x^2+y^2 z=2y. 求∫∫∫Zdv

求高手 我想了很久了

Answer 1

该立体投影到xoy面为x²+y²=2y，即Dxy：x²+(y-1)²=1，其极坐标方程为：r=2sinθ
∫∫∫zdv
=∫∫ (∫[0--->2y]zrdz) drdθ
=∫∫ (∫[0--->2rsinθ]zrdz) drdθ
=1/2∫∫ z²r |[0--->2rsinθ] drdθ
=2∫∫ r³sin²θ drdθ
=2∫[0--->π]∫[0--->2sinθ] r³sin²θ drdθ
=1/2∫[0--->π] r⁴sin²θ |[0--->2sinθ] dθ
=8∫[0--->π] sin⁶θ dθ
=∫[0--->π] (1-cos2θ)³ dθ
=∫[0--->π] (1-3cos2θ+3cos²2θ-cos³2θ) dθ
=∫[0--->π] (1-3cos2θ+3/2(1+cos4θ)-sin³2θ) dθ
=∫[0--->π] (5/2-3cos2θ+3/2cos4θ-sin³2θ) dθ
=(5/2)θ-(3/2)sin2θ+(3/8)sin4θ-∫[0--->π] sin³2θdθ
=(5/2)θ-(3/2)sin2θ+(3/8)sin4θ+1/2∫[0--->π] sin²2θd(cos2θ)
=(5/2)θ-(3/2)sin2θ+(3/8)sin4θ+1/2∫[0--->π] (1-cos²2θ)d(cos2θ)
=(5/2)θ-(3/2)sin2θ+(3/8)sin4θ+1/2cos2θ-1/6cos³2θ |[0--->π]
=5π/2