计算Ⱒ_D鈥(1-x^2-y^2 )dσ,其中D是由y=x,y=0,x^2+y^2=1在第一象限内所围成的区域

2024-12-14 02:39:33
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回答1:

转化为极坐标
∫∫(1-x^2-y^2 )dσ
=∫[0->π/4] dθ∫[0->1] (1-r^2)rdr
=-π/8∫[0->1] (1-r^2)d(1-r^2)
=(-π/8)*(1/2)*(1-r^2)^2 | [0->1]
=π/16