can anyone please help me in finding the definite integral of y (6y-y^2)^1/2 from 0 to 6?
$\displaystyle \int_0^6 y(6y- y^2)^{1/2} dy$
The first thing I would do is "complete the square" inside the square root:
$\displaystyle \int_0^6 y(9- (y- 3)^2)^{1/2}dy$
Let u= y- 3. Of course, y= u+ 3 and when y= 0, u=-3, when y= 6, y= 3. This is now $\displaystyle \int_{-3}^3 (u+3)(9- u^2)^{1/2}du= u\int_{-3}^3 u(9- u^2)^{1/2}du+ 3\int_{-3}^3 (9- u^2)^{1/2}du$
To integrate $\displaystyle \int_{-3}^3 u(9- u^2)^{1/2} du$ let $\displaystyle v= 9- u^2$.
To integrate $\displaystyle 3\int_{-3}^3 (9- u^2)^{1/2}du$ let $\displaystyle u= 3 sin(\theta)$.