设f(x,y)为连续函数,则∫04πdθ∫01f(rcosθ,rsinθ)rdr等于( )。
∫022dx∫x1−x2f(x,y)dy
∫022dx∫01−x2f(x,y)dy
∫022dy∫y1−y2f(x,y)dx
∫022dy∫01−y2f(x,y)dx
- A∫022dx∫x1−x2f(x,y)dy
- B∫022dx∫01−x2f(x,y)dy
- C∫022dy∫y1−y2f(x,y)dx正确答案
- D∫022dy∫01−y2f(x,y)dx