求二重积分∫∫√（x²+y²）dxdy，积分区域D由直线y＝x及曲线y＝x²围成。
想用极坐标做（一般方法好像不怎么好做，也不会。。。），但是ρ的范围好像不怎么好确定，答案错误。。。请大佬们帮忙，谢谢了！
第1个回答 2019-08-10
y = x^2, ρsint = ρ^2(cost)^2, 得 ρ = sint/(cost)^2,
I = ∫∫<D>√(x²+y²)dxdy
= ∫<0, π/4> dt ∫<0, sint/(cost)^2> ρ·ρdρ
= (1/3)∫<0, π/4> dt [ρ^3]<0, sint/(cost)^2>
= (1/3)∫<0, π/4> [(sint)^3/(cost)^6]dt
= (-1/3)∫<0, π/4> [1-(cost)^2]dcost/(cost)^6
= (-1/3)[(-1/5)/(cost)^5-(-1/3)/(cost)^3]<0, π/4>
= (1/45)[3/(cost)^5-5/(cost)^3]<0, π/4>
= (1/45)[3·4√2-5·2√2] = (2/45)√2本回答被网友采纳
I = ∫∫<D>√(x²+y²)dxdy
= ∫<0, π/4> dt ∫<0, sint/(cost)^2> ρ·ρdρ
= (1/3)∫<0, π/4> dt [ρ^3]<0, sint/(cost)^2>
= (1/3)∫<0, π/4> [(sint)^3/(cost)^6]dt
= (-1/3)∫<0, π/4> [1-(cost)^2]dcost/(cost)^6
= (-1/3)[(-1/5)/(cost)^5-(-1/3)/(cost)^3]<0, π/4>
= (1/45)[3/(cost)^5-5/(cost)^3]<0, π/4>
= (1/45)[3·4√2-5·2√2] = (2/45)√2本回答被网友采纳
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