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(1) The region bounded by the given curves is rotated about y = 9. x=(y-10)^2 , x=1. Find the volume V of the resulting solid by any method. and (2) The region bounded by the given curves is rotated about the y-axis. y=-x^2+15x-54, y=0. Find the volume V of the resulting solid by any method. I got (...

I integrated from 11 to 9 pi*((y-10)^2)^2 dy
I know its not correct but wot did u get and hw did u do it. I did not know wot to do with the x=1 line

I actually got that at some point but min was a negative anwser and why is it (y-9) and not (9-y).

Thank You for showing me how to do it. I usually get confused when you are meant to rotate functions about lines other than the y-axis and x-axis.

How would you evaluate this integral ?

x^2/(36-x^2)^(3/2)

Parts or Substitution ?(I tried using substitution) Please explain.

Integrate sqrt(x)/(x^2+x) from 1/3 to 3

How do you go about the integration

I am studying for an exam and I don't you know how to go about this? Volume of solid under the curve y= 7/(x^2 +5x +6) from x = 0 to x = 1 if rotated about the y-axis. I made x the subject of the formula getting (sqrt(y+28)-5 sqrt(y))/(2 sqrt(y)) and then I found what y is when x= 0 and x=1, which w...

because we are rotating about the y axis

Yes that is what I did right. pi*(f(y))^2 dy

and so are b and a 7/6 and 7/12 ?

Am I correct????????????