Question

设D为区域x^2+y^2≤2x+4y，求二重积分∫∫(x^2+y^2)dxdy

如图所示

Answer 1

答：π (e - 1)

极坐标化简
x = rcosθ
y = rsinθ
x²+y²=r²,0≤r≤1,0≤θ≤2π
∫∫_(D) e^(x²+y²) dxdy
= ∫(0,2π) dθ ∫(0,1) e^r² * r dr
= (2π)∫(0,1) e^r² d(r²)/2
= π * [e^r²](0,1)
= π * (e^1 - e^0)
= π (e - 1)