What is the volume of a rotated region bounded by y=9-x^2, y=0, and x=0?

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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.
 
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Set up the equations you used! :smile:
 
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
 
The first is set up correctly but that surely is not "407"!

When you rotate around the y-axis, your "disks" will be moving up the y axis. The radius of each disk will be x as a function of y and their "thickness" is dy.
 
Thats what i get when I put it into my ti-83. I'm pretty sure that the first one is right. But I'm having a hard time imaging the shape you get when you rotate the area around y=9.
 
\int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx is what the second integral maybe?
 
Never mind guys, I just checked my work on maple and the answers I had first were right, I should stop second guessing my self :),
 
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