Volume using intergals

If the region bounded by y=9-x^2,y=0,x=0. Is rotated about the x axis the what would the volume be. I'm got 407. For the same region rotate about the y=9, and i got 153. Can someone check these answers i think they are wrong.

arildno
Homework Helper
Gold Member
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Set up the equations you used!

HallsofIvy
Homework Helper
You know, I could have sworn that when you rotate something a "$\pi$" would show up!

How about showing us exactly what you integrated and how you got those answers?

The first integral would be, I think: b=3,a=0, on the integral pi*(9-x^2)^2

The second one, which I'm not sure about, would be same b and a, on the integral pi*(9-(9-x^2))^2

HallsofIvy
$$\int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx$$ is what the second integral maybe?