Question

选择适当的坐标计算下列二重积分sin(x^2+y^2),其中D是由圆周x^2+y^2=π^2/4所围成的闭区域

Answer 1

楼上最后错了
显然用极坐标，因为区域是圆
所以
令
x=rcosθ
y=rsinθ
所以原二重积分
=∫∫sin(r^2)rdrdθ
其区域在极坐标下表示为
0<r<π/2 (因为r^2=x^2+y^2<=π^2/4)
0<θ<2π

=∫[0,2π]dθ*∫[0,π/2] sin(r^2)rdr
=2π*(1/2)∫[0,π/2] sin(r^2) d(r^2)
=π(-cos(r^2))|[0,π/2]
=π(-cos(π^2/4)+1)

Answer 2

{D}∫∫sin(x²+y²) dxdy={D}∫∫(sinρ²)ρdρdθ……θ=0~2π，ρ=0~π/2；
=∫(sinρ²)ρdρ∫dθ=-π*(cosρ²)|{0,π/2}=π-cos(π²/4)；