Volume using intergals

[flash9286]

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]

Set up the equations you used!

 

[HallsofIvy]

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?

 

[flash9286]

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]

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.

 

[flash9286]

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.

 

[flash9286]

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

 

[flash9286]

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 :),