What is the volume of the resulting solid?

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Discussion Overview

The discussion revolves around calculating the volume of solids obtained through various methods of rotation, specifically focusing on problems involving curves and geometric shapes. Participants share their approaches, challenges, and calculations related to these volume problems.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes difficulty in finding the volume of a solid formed by rotating the region bounded by the curves y = x^2 and y = 1 about y = 2, using the Shell method.
  • Another participant discusses the volume of a solid obtained by rotating the curve y = 40x - 8x^2 about the y-axis, suggesting it resembles an upside-down cone and recommending the use of horizontal disks for integration.
  • There is a suggestion to split the volume calculation into inner and outer volumes for a solid with a cylindrical hole, with a proposed integral setup for the calculation.
  • One participant expresses confusion over determining the outer radius in their volume calculation, seeking clarification on the approach taken by another participant.
  • A later reply proposes a method for calculating the volume of a sphere with a cylindrical hole, using vertical washers, but indicates that their result does not match the expected answer from an online program.
  • Another participant suggests a different axis of rotation for one of the problems, leading to further discussion about the correct setup for the integral.
  • Participants express frustration over discrepancies in their answers compared to an online homework program, with some suggesting potential bugs in the program.

Areas of Agreement / Disagreement

Participants express varying opinions on the correct setup for volume calculations, with no consensus reached on the methods or answers for the problems discussed. Multiple competing views remain regarding the correct approach to the problems.

Contextual Notes

Some participants mention specific integral setups and ranges, but there are unresolved questions about the correctness of these setups and the assumptions made in their calculations. The discussion reflects uncertainty about the parameters involved in the volume calculations.

Caldus
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I am having a lot of trouble with these last 3 problems out of 10 that I have done relating to volumes. I have tried just about every method for each of these and I am just not getting the right answer (according to this online math program I am using).

2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:
y = x^2, y = 1; about y = 2.

The resulting solid for me looks like a volcano. So I used the Shell method like so:

A(x) = length * width
A(x) = 2*pi*(x)*x^2 = 2*pi*x^3

V = integral b/w 0 and 1 of A(x) ...

Not getting the right answer here either...

3. A ball of radius 11 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

I tried using vertical washers like so:

A(x) = pi*11^2 - pi*4^2

V = integral b/w 1 and 11 of A(x) = ...

Incorrect here as well. Not sure how else to approach this one.

Thanks for help on any of these. At least I got the other 7 problems done on my own. :P
 
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Caldus said:
1. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = 40x - 8x^2, y = 0; about the y-axis.

Alright so I graph this function on my calculator. When I imagine the solid created after rotating about the y-axis. It looks like an upside down cone correct?
[itex]40x-8x^2[/itex] is a parabola with it's maximum at x=2.5. So the region bounded by this parabola and the line y=0 is like a hill.
When you rotate it around the y-axis you get something like a donut, but with the upper part like a parabola and the bottom part removed, so it's flat on the bottom side.
Using horizontal disks is right, but you have to integrate from y=0 to y=50. (From bottom to top).
Each horizontal disk is like a ring. Try setting up the integral from here.
 
Thanks, I eventually figured it out.

Still stuck on the other two problems...
 
In the second, I would suggest that you split the volume up in the following manner:

Inner Volume: Radius 2-1=1
The volum of this cylinder is subtracted from:

Outer Volume: Radius:[tex]2-x^{2}[/tex]

Hence, your volume should be:
[tex]V=\int_{0}^{1}\pi((2-x^{2})^{2}-1^{2})dx[/tex]
 
I ended up getting 5.864306287 as the answer but the online homework program says it's still incorrect.

I agree that the small radius is 1. How did you come up with 2 - x^2 as the big radius though? I don't get how to determine what the outer radius is in this case. Thanks for any help.

Any ideas on the last problem?
 
OK, got an idea for the last problem but still not getting the right answer.

OK so if I place a sphere of radius 11 in the center of the x-y plane, then the top right quadrant curve of that sphere will equal (r^2 - x^2)^(1/2) correct? So then I can do the volume like so (using vertical washers):

V = 2pi * Integral b/w 0 and 11 of (11^2 - x^2) - (4^2 - x^2) dx

But it isn't right according to this program.
 
Caldus said:
I ended up getting 5.864306287 as the answer but the online homework program says it's still incorrect.

I agree that the small radius is 1. How did you come up with 2 - x^2 as the big radius though? I don't get how to determine what the outer radius is in this case. Thanks for any help.

Any ideas on the last problem?
You are to rotate it around y=2, right?
The distance from y=2 to the parabola is 2-x^2
 
Yeah I suppose so. Grrr this stupid program. It must be a bug or something...
 
For the last problem, you want to rotate the area bounded by [itex]y=\sqrt{11^2-x^2}[/itex] and [itex]y=2[/itex] about the x-axis.

The answer I get is [itex]\frac{4268\pi}{3}\approx 4469.44[/itex]
 
  • #10
Wouldn't it be y = 4 and not y = 2 for the other one?

I tried your answer and it did not work while I tried my answer using y = 4 and of course it says incorrect.

I'm running out of ideas as to what I could have done wrong for both of these problems.
 
  • #11
Anyone? These problems are really frustating me...
 
  • #12
Caldus said:
Wouldn't it be y = 4 and not y = 2 for the other one?
Yes it is. Sorry for that. It should be 4506.84 (if I did it right this time).
Does that confirm?

Also, the integral Arildno gave should run from -1 to 1, so the answer will be doubled.
 
  • #13
Galileo said:
Yes it is. Sorry for that. It should be 4506.84 (if I did it right this time).
Does that confirm?

Also, the integral Arildno gave should run from -1 to 1, so the answer will be doubled.
AARGH!
That's where my mistake was!
Thx, Galileo, sorry Caldus..:redface:
 

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