∫[0,π/2]d(α)∫[0,cosα]f(rcosα,rsinα)rd(r)
=∫[0,1]dx∫[0,根号（1/4-(x-1/2)^2）]f(x,y)dy
就是把r=cosα-->r^2=rcosα-->x^2+y^2=x
然后画图,就可以看出来了
∫[0,π/2]d(α)∫[0,cosα]f(rcosα,rsinα)rd(α)转化为直角坐标形式为
1个回答
相关问题
-
已知点M(rcosα,rsinα),N(rcosβ,rsinβ),(-π/2
-
已知α∈﹙0,π/2）,且2Sinα-SinαCosα-3Cosα=0.求[Sin（α+π/4)]/(Sin2α+Cos
-
若α∈（0,2π）且|cosα|=cos（π-α）
-
α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin
-
已知f（α）=sin(π−α)cos(2π−α)cos(−α+32π)cos(π2−α)sin(−π−α)
-
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2
-
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2
-
π〈α〈2π且f(α）={√2-2 cosα}/{（1-sinα-cosα）（cosα/2+sinα/2）}请化简f（α
-
已知f（α）=tan(π−α)•cos(2π−α)•sin(π2+α)cos(−α−π)．
-
已知tanα=2,sinα+cosα＜0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π