∫∫[e^-(x²+y²-π)]sin(x²+y²)dxdy D:x²+y²≤π
问题描述：
∫∫[e^-(x²+y²-π)]sin(x²+y²)dxdy D:x²+y²≤π
答
∫∫ [e^-(x²+y²-π)]sin(x²+y²) dxdy=∫∫ [e^-(r²-π)]sin(r²) rdrdθ=e^π∫[0→2π]dθ∫[0→√π] re^(-r²)sin(r²)dr=2πe^π∫[0→√π] re^(-r²)sin(r²)d...