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{ y = √x
{ y = x²
==>交点为(0,0),(1,1)
∫∫_D x√y dσ
= ∫(0→1) x ∫(x²→√x) √y dy
= ∫(0→1) x · (2/3)y^(3/2):(x²→√x) dx
= ∫(0→1) (2/3)x · [(√x)^(3/2) - (x²)^(3/2)] dx
= ∫(0→1) [(2/3)x^(7/4) - (2/3)x⁴] dx
= [(2/3)(4/11)x^(11/4) - (2/3)(1/5)x⁵]:(0→1)
= (8/33) - (2/15)
= 6/55
做个简单题目而已也要发邮箱,这麼怕别人抄袭你答案呢。
{ y = x²
==>交点为(0,0),(1,1)
∫∫_D x√y dσ
= ∫(0→1) x ∫(x²→√x) √y dy
= ∫(0→1) x · (2/3)y^(3/2):(x²→√x) dx
= ∫(0→1) (2/3)x · [(√x)^(3/2) - (x²)^(3/2)] dx
= ∫(0→1) [(2/3)x^(7/4) - (2/3)x⁴] dx
= [(2/3)(4/11)x^(11/4) - (2/3)(1/5)x⁵]:(0→1)
= (8/33) - (2/15)
= 6/55
做个简单题目而已也要发邮箱,这麼怕别人抄袭你答案呢。
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