Question

计算二重积分∫∫（x+y）dxdy，其中D为x^2+y^2≤2x

Answer 1

楼上错的，楼上当作矩形区域算了

首先本题区域关于x轴对称，y关于y是一个奇函数，因此积分为0，所以被积函数中的y可去掉。
∫∫（x+y）dxdy
=∫∫xdxdy
用极坐标，x²+y²=2x的极坐标方程为：r=2cosθ
=∫[-π/2---->π/2] dθ∫[0---->2cosθ] rcosθ*rdr
=∫[-π/2---->π/2] cosθdθ∫[0---->2cosθ] r²dr
=∫[-π/2---->π/2] (cosθ)*(1/3)r³ |[0---->2cosθ] dθ
=(8/3)∫[-π/2---->π/2] cos⁴θ dθ
=(16/3)∫[0---->π/2] cos⁴θ dθ
=(16/3)∫[0---->π/2] [1/2(1+cos2θ)]² dθ
=(4/3)∫[0---->π/2] (1+cos2θ)² dθ
=(4/3)∫[0---->π/2] (1+2cos2θ+cos²2θ) dθ
=(4/3)∫[0---->π/2] (1+2cos2θ+1/2(1+cos4θ)) dθ
=(4/3)∫[0---->π/2] (3/2+2cos2θ+1/2cos4θ) dθ
=(4/3)(3/2θ+sin2θ+1/8sin4θ) |[0---->π/2]
=(4/3)(3/2)*(π/2)
=π

Answer 2

D可化为：（x-1）²+y²≤1，得：0≤x≤1，-1≤y≤1
1 1 1
所以：∫∫（x+y）dxdy=∫ dx ∫(x+y)dy=∫ 2xdx=4
0 -1 0

也可以先对x积分

Answer 3

支持二楼