∫∫_D ln(1 + x² + y²) dxdy
= ∫(0→π/2) dθ ∫(0→1) ln(1 + r²) · rdr
= [ln(2) - 1/2] · π/2
= (π/4)(2ln(2) - 1)
= ∫(0→π/2) dθ ∫(0→1) ln(1 + r²) · rdr
= [ln(2) - 1/2] · π/2
= (π/4)(2ln(2) - 1)
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第1个回答 2012-05-29
极坐标
∫∫（D）ln(1+x²+y²)dxdy
=∫∫（D）rln(1+r²)drdθ
=∫[0→2π]dθ∫[0→1] rln(1+r²)dr
=2π∫[0→1] rln(1+r²)dr
=π∫[0→1] ln(1+r²)d(r²)
=πr²ln(1+r²)-2π∫[0→1] r³/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] (r³+r-r)/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] rdr+2π∫[0→1] r/(1+r²)dr
=πr²ln(1+r²)-πr²+π∫[0→1] 1/(1+r²)d(r²)
=πr²ln(1+r²)-πr²+πln(1+r²) |[0→1]
=πln2-π+πln2
=π(2ln2-1)
做错了，当作整圆做的了。 结果再除以4本回答被提问者采纳
∫∫（D）ln(1+x²+y²)dxdy
=∫∫（D）rln(1+r²)drdθ
=∫[0→2π]dθ∫[0→1] rln(1+r²)dr
=2π∫[0→1] rln(1+r²)dr
=π∫[0→1] ln(1+r²)d(r²)
=πr²ln(1+r²)-2π∫[0→1] r³/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] (r³+r-r)/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] rdr+2π∫[0→1] r/(1+r²)dr
=πr²ln(1+r²)-πr²+π∫[0→1] 1/(1+r²)d(r²)
=πr²ln(1+r²)-πr²+πln(1+r²) |[0→1]
=πln2-π+πln2
=π(2ln2-1)
做错了，当作整圆做的了。 结果再除以4本回答被提问者采纳
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