Question

How to integrate sqrt(2y-y^2)*y dy?

Answer 1

@ganeshie8 @Luigi0210 @paki @abb0t @Destinymasha

Answer 2

\[\int\limits \sqrt{2y-y^2}y dy\]

Answer 3

@hartnn

Answer 4

I would start by completing the square under the radical

Answer 5

\[\int\limits_{}^{} y*\sqrt{2y-y^2}dy= \int\limits_{}^{} y*\sqrt{1-(y-1)^2}dy\]

Answer 6

then do a u-substitution, u = y-1

Answer 7

How did you get 1-(y-1)^2?

Answer 8

2y - y^2
= -y^2 + 2y
= - ( y^2 - 2y )
complete the square
= - ( y^2 - 2y + 1 - 1 )
= - (y^2 -2y +1) - (-1)
= -(y - 1)^2 + 1

Answer 9

So if u=y-1, then du=dy, then what?

Answer 10

Changing to \(u\), you get
\[\int(u+1)\sqrt{1-u^2}\,du\]
Try a trigo sub, \(u=\sin t\).

Answer 11

Thank you guys~.