十九题，高数的，求过程？！！

解：∵(cosθ)^5是偶函数，sinθ(cosθ)^4是奇函数
∴∫<-π/2,π/2>(cosθ)^5dθ=2∫<0,π/2>(cosθ)^5dθ
∫<-π/2,π/2>sinθ(cosθ)^4dθ=0
故 原式=∫<-π/2,π/2>dθ∫<0,2acosθ>(rcosθ+2rsinθ)r*rdr (作极坐标变换)
=4a^4∫<-π/2,π/2>(cosθ+2sinθ)(cosθ)^4dθ
=4a^4[∫<-π/2,π/2>(cosθ)^5dθ+2∫<-π/2,π/2>sinθ(cosθ)^4dθ]
=4a^4[2∫<0,π/2>(cosθ)^5dθ+2*0]
=8a^4∫<0,π/2>(cosθ)^5dθ
=8a^4∫<0,π/2>cosθ(cosθ)^4dθ
=8a^4∫<0,π/2>[1-2(sinθ)^2+(sinθ)^4]d(sinθ)
=8a^4[sinθ-2(sinθ)^3/3+(sinθ)^5/5]│<0,π/2>
=8a^4(1-2/3+1/5)
=64a^4/15。