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

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Discussion Overview

The discussion revolves around calculating the volume of a region bounded by the equations y=9-x^2, y=0, and x=0 when rotated about the x-axis and the line y=9. Participants are exploring the setup of integrals for these calculations and questioning the accuracy of their results.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant claims to have calculated the volume for rotation about the x-axis as 407 and for rotation about y=9 as 153, but expresses uncertainty about these results.
  • Another participant requests the setup of the equations used for the volume calculations.
  • A participant notes that a "\pi" factor is typically involved in such volume calculations and asks for the integration details.
  • One participant proposes the first integral as pi*(9-x^2)^2 from 0 to 3, but is unsure about the second integral for rotation about y=9.
  • Another participant challenges the initial volume calculation for rotation about the x-axis, suggesting that the setup is correct but the result of 407 is likely incorrect.
  • A participant mentions using a TI-83 calculator to arrive at their results and expresses difficulty visualizing the shape formed when rotating around y=9.
  • One participant attempts to provide a potential form for the second integral but does not express confidence in its correctness.
  • A later reply indicates that the participant has verified their initial answers using Maple, expressing a desire to stop second-guessing themselves.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the volume calculations, with multiple competing views on the setup and results of the integrals presented.

Contextual Notes

There are unresolved questions regarding the correct setup of integrals and the interpretation of the rotation about different axes, as well as potential miscalculations in the volume results.

flash9286
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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 "[itex]\pi[/itex]" 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.
 
[tex]\int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx[/tex] 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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