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

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In summary, the conversation discusses the process of rotating a region bounded by y=9-x^2, y=0, x=0 about the x-axis and y=9. The volume obtained for the first rotation is 407 and for the second rotation is 153. There is also a discussion about setting up the correct equations and using the TI-83 calculator. Finally, the person confirms that their initial answers were correct.
  • #1
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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  • #2
Set up the equations you used! :smile:
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
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.
 
  • #7
[tex] \int_{0}^{3} pi*9^2-(pi*(9-9+x^2)^ dx [/tex] is what the second integral maybe?
 
  • #8
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 :),
 

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

Q: What is the formula for calculating volume using integrals?

The formula for calculating volume using integrals is: V = ∫ab A(x) dx. Here, A(x) represents the cross-sectional area of the object at a given point along the x-axis, and the integral is taken over the interval from a to b.

Q: How do I determine the limits of integration for volume using integrals?

The limits of integration for volume using integrals can be determined by identifying the boundaries of the object in the given coordinate system. For example, if the object is bounded by the x-axis and a curve, the limits of integration would be the x-values where the curve intersects the x-axis.

Q: Can volume be calculated using integrals for irregularly shaped objects?

Yes, volume can be calculated using integrals for irregularly shaped objects. In fact, this is one of the main advantages of using integrals for volume calculations, as it allows for more accurate results for non-geometric shapes.

Q: What are some common units of measurement for volume using integrals?

The units of measurement for volume using integrals will depend on the units used for the cross-sectional area and the limits of integration. Some common units include cubic meters, cubic feet, and liters.

Q: Can integrals be used to calculate the volume of a 3D object with varying density?

Yes, integrals can be used to calculate the volume of a 3D object with varying density. In this case, the integrand would include the density function, and the integral would need to be evaluated using the appropriate techniques, such as the method of substitution.

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