AK903
幼苗
共回答了17个问题采纳率:94.1% 举报
∫∫_D x²y dxdy
= _D₁ x²y dxdy + ∫_D₂ x²y dxdy
= ∫(0→1) dy ∫(0→-√(1 + y²)) x²y dx + ∫(0→1) dy ∫(0→√(1 + y²) x²y dx
= ∫(0→1) [y · x³/3 |(0→-√(1 + y²)) + y · x³/3 |(0→√(1 + y²))] dy
= (1/3)∫(0→1) [- y(1 + y²)^(3/2) + y(1 + y²)^(3/2)] dy
= 0
1年前
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