计算∫∫D √(X^2 + Y^2) dxdy,其中D是上半圆盘区域:X^2 +Y^2≤4,Y≥0

2024-11-08 05:59:15
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回答(1):

化为极坐标
x=pcosa,y=psina
p∈[0,2]
a∈[0,π]
∫∫ D √(x^2+y^2) dxdy
=∫[0,π]∫ [0,2] p*pdpda
=∫[0,π]da∫ [0,2] p*pdp
=8π/3