Question

求学霸指教，望给出详细解答过程

Answer 1

0≤x≤y²
0≤y≤2
原式=∫(0,2)dy∫(0,y²)1/√(1+y²)dx
=∫(0,2)y²/√(1+y²) dy
令y=tant
dy=sec²tdt
原式=∫(0,arctan2)tan²t/sect ·sec²tdt
=∫(0,arctan2)(sint/cost)²·1/costdt
=∫(0,arctan2)sint/cos³tdt
=-∫(0,arctan2)1/cos³t dcost
=1/2 ·1/cos²t|(0,arctan2)
=1/2 sec²t|(0,arctan2)
=1/2 (tan²t+1)|(0,arctan2)
=1/2(4+1-1)
=2

追问

不是1+y²是1+y³

不过还是谢谢你了