Question

大一高等数学利用极坐标计算二重积分 第三题的三个小题

Answer 1

(1) I = ∫<0, 2π>dt∫<π,2π> sinr^2 rdr
= π∫<π,2π> sinr^2dr^2 = π[-cosr^2]<π,2π>
= π[cos(π^2)-cos(4π^2)]
(2) I = ∫<-π/2,π/2>dt∫<0,Rcost>√(R^2-r^2) rdr
= -∫<0,π/2>dt∫<0,Rcost>√(R^2-r^2)d(R^2-r^2)
= (2R^3/3)∫<0,π/2>[1-(sint)^3]dt
= (2R^3/3)[t+cost-(1/3)(cost)^3]<0,π/2> = π/3 - 4/9
(3) I = ∫<0,π/2>dt∫<0,2acost>r^2 rdr
= 4a^4∫<0,π/2>(cost)^4dt = a^4∫<0,π/2>(1+cos2t)^2dt
= a^4∫<0,π/2>[1+2cos2t+(cost)^2]dt
= a^4∫<0,π/2>[3/2+2cos2t+(1/2)cos4t]dt
= a^4[3t/2+sin2t+(1/8)sin4t]<0,π/2> = (3/4)πa^4