设D是圆域:x2+y2≤1,则∫∫Dxdσ=()
A.2∫0πdθ∫01r3sinθdθ
B.∫02πdθ∫01r2coxθdr
C.4∫0π/4dθ∫01rcosθdr
D.2∫0π/4dθ∫01r2sinθdr
正确答案是B
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