Solids of Revolution - Negative Volume

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Discussion Overview

The discussion revolves around calculating the volume of a solid of revolution formed by rotating the area bounded by the curve \(y=x^2-5x+6\) and the line \(y=0\) about the y-axis. Participants explore different methods for computing this volume, particularly focusing on the cylindrical shell method and addressing issues related to negative volume results.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the volume using the cylindrical shell method and obtains a negative result, suggesting that the height of the shells is incorrectly represented by a function that returns negative values.
  • Another participant challenges the initial approach, indicating that the volume should be calculated using the formula \(V=\int_2^3 \pi (r(x))^2 dx\) but notes that this is not appropriate for rotation about the y-axis.
  • A participant clarifies that the height of each shell should be the difference between the upper and lower graphs, leading to the conclusion that the height is indeed \(-f(x)\).
  • Some participants express the importance of visualizing the solid of revolution and suggest that understanding the geometry can aid in deriving the correct formulas rather than relying solely on memorization.
  • There is a discussion about the usefulness of making drawings and sketches to aid in understanding the volume elements involved in the problem.
  • Several participants mention the use of the Desmos API for visualizing solids of revolution, indicating a desire to explore graphical representations of the problem.
  • Some participants express a preference for re-deriving formulas based on their understanding of the volume elements rather than memorizing them, highlighting a more intuitive approach to solving such problems.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the volume, with some agreeing on the need to adjust the height in the cylindrical shell method while others propose alternative methods. The discussion remains unresolved regarding the best approach to take, as multiple competing views are presented.

Contextual Notes

Participants note that the negative volume result arises from the choice of function for height, and there is uncertainty about the correct representation of the volume element. The discussion reflects various assumptions about the methods and approaches to solving the problem.

Who May Find This Useful

This discussion may be useful for students and educators interested in understanding the complexities of calculating volumes of solids of revolution, particularly those exploring different methods and the importance of visualization in mathematical reasoning.

  • #31
I like Serena said:
Hmm... can you redo the following integral with a couple more steps? (Wondering)
$$\int_{-1}^{1} (x^2+y^2)dx$$

I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1=\frac{2}{3}$$

EDIT: Somehow, I thought the integral of a constant was 0. Let me re-work this.
 
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  • #32
Rido12 said:
I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1$$

Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$
 
  • #33
I like Serena said:
Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$

It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...
 
  • #34
Rido12 said:
It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...

Oh, I know.
The derivative of a constant is 0.
It means that the anti-derivative of 0 is a constant, which is the other way around.
I suspect you were just mixing them up, which is a pretty normal thing to do.
 
  • #35
$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D
 
  • #36
Rido12 said:
$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D

Yep! (Emo)
 
  • #37
Yup, using advice from this thread, I can now solve most types of these questions. In particular, I'm happy to have gotten this one right: Find the volume of the frustum of a cone whose lower base is of radius R, upper base is of radius r, and altitude is h.

I also want to point out, although not another "method", but some questions require the use of this formula (which is kind of obvious): (I guess you can say the disc/washer method uses this too, technically)

$$V= \int A(x) dx$$

"A solid has a circular base of radius 4 units. Find the volume of the solid if every plane perpendicular to the fixed diameter is an isosceles right triangle with the hypotenuse in the plane of the base".
 
Last edited:

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