求解一道高数题,题目看图 谢谢! ps 答案:3.34
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∫∫<R>xdA = ∫<0, π/4>dθ∫<2cosθ, 2> rcosθ rdr
= ∫<0, π/4>cosθdθ∫<2cosθ, 2> r^2dr
= (1/3)∫<0, π/4>cosθdθ[r^3]<2cosθ, 2>
= (1/3)∫<0, π/4>cosθ[8-8(cosθ)^3]dθ
= (1/3)∫<0, π/4>[8cosθ-8(cosθ)^4]dθ
= (8/3)[sinθ]<0, π/4> - (1/3)∫<0, π/4>2(1+cos2θ)^2dθ
= 4√2/3 - (1/3)∫<0, π/4>[2+4cos2θ+2(cos2θ)^2]dθ
= 4√2/3 - (1/3)∫<0, π/4>(3+4cos2θ+cos4θ)dθ
= 4√2/3 - (1/3)[3θ+2sin2θ+(1/4)sin4θ]<0, π/4>
= (4√2-2)/3 - π/4
= ∫<0, π/4>cosθdθ∫<2cosθ, 2> r^2dr
= (1/3)∫<0, π/4>cosθdθ[r^3]<2cosθ, 2>
= (1/3)∫<0, π/4>cosθ[8-8(cosθ)^3]dθ
= (1/3)∫<0, π/4>[8cosθ-8(cosθ)^4]dθ
= (8/3)[sinθ]<0, π/4> - (1/3)∫<0, π/4>2(1+cos2θ)^2dθ
= 4√2/3 - (1/3)∫<0, π/4>[2+4cos2θ+2(cos2θ)^2]dθ
= 4√2/3 - (1/3)∫<0, π/4>(3+4cos2θ+cos4θ)dθ
= 4√2/3 - (1/3)[3θ+2sin2θ+(1/4)sin4θ]<0, π/4>
= (4√2-2)/3 - π/4
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