设函数f(x,y)连续,则二次积分∫2ππdx∫sinx1f(x,y)dy等于( )。
∫01dy∫π+arcsinyπf(x,y)dx
∫01dy∫π−arcsinyπf(x,y)dx
∫01dy∫2ππ+arcsinyf(x,y)dx
∫01dy∫2ππ−arcsinyf(x,y)dx
- A∫01dy∫π+arcsinyπf(x,y)dx
- B∫01dy∫π−arcsinyπf(x,y)dx正确答案
- C∫01dy∫2ππ+arcsinyf(x,y)dx
- D∫01dy∫2ππ−arcsinyf(x,y)dx