find volume of the solid found by revolving the curve y=√(2x+1), 0 ≤ x ≤ 3 about the x-axis

I do not know if this is correct, but I will try    y= sqrt (2x+1)    Gives the RADIUS of the object that is to be rotated       pi r^2 = area  so:

 

 

Integral (0-3)  pi (sqrt(2x+1)) ^2  =  pi  integral (0-3)  2x+1       =     pi  (0-3)  1/2x^2  + x  + K     K= Constant (which I THINK we can make '0') 

From 0 to 3

pi ( 4.5 + 3 -0) = pi(7.5) = 23.56    ????????????

 

I will NOT guarantee this result.      Can anyone verify this?.     We had a question like this sometime back .... (and I got it WRONG then!)