∫∫[e^-(x²+y²-π)]sin(x²+y²)dxdy D:x²+y_数学解答

∫∫[e^-(x²+y²-π)]sin(x²+y²)dxdy D:x²+y²≤π

人气：267 ℃　时间：2020-09-10 03:32:07

解答

∫∫ [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...