【高数题,求大佬详解,不清楚如下图】答案有,我看不懂,求详解
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第1个回答 2020-07-26
化成x=rcosθ,y=rsinθ
x²+y²∈[1,4],则r²∈[1,4]
r∈[1,2]
θ∈[0,2π]
∫∫D ln(x²+y²)dσ
=∫(0,2π) dθ∫(1,2) rlnr²dr
=∫(0,2π) dθ (r²lnr-1/2 r²)|(1,2)
=∫(0,2π) (4ln2-1/2×2²-1·ln1+1/2 ×1²)dθ
=∫(0,2π) (4ln2-3/2)dθ
=2π×(4ln2-3/2)
=(8ln2-3)π
∫rlnr²dr=1/2 ∫lnr²dr²=1/2 ∫lntdt (t=r²)
=1/2 (tlnt-∫tdlnt)
=1/2 (tlnt-∫dt)
=1/2 tlnt-1/2 t+C
=1/2 r²ln(r²)-1/2 r²+C
=r²lnr-1/2 r²+C本回答被提问者采纳
x²+y²∈[1,4],则r²∈[1,4]
r∈[1,2]
θ∈[0,2π]
∫∫D ln(x²+y²)dσ
=∫(0,2π) dθ∫(1,2) rlnr²dr
=∫(0,2π) dθ (r²lnr-1/2 r²)|(1,2)
=∫(0,2π) (4ln2-1/2×2²-1·ln1+1/2 ×1²)dθ
=∫(0,2π) (4ln2-3/2)dθ
=2π×(4ln2-3/2)
=(8ln2-3)π
∫rlnr²dr=1/2 ∫lnr²dr²=1/2 ∫lntdt (t=r²)
=1/2 (tlnt-∫tdlnt)
=1/2 (tlnt-∫dt)
=1/2 tlnt-1/2 t+C
=1/2 r²ln(r²)-1/2 r²+C
=r²lnr-1/2 r²+C本回答被提问者采纳