Volume of Solid Revolving Around x & y Axis

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Homework Help Overview

The discussion revolves around finding the volume of a solid formed by revolving the area bounded by the equations y=-x+3 and y=x^2-3x around the x and y axes. Participants are exploring the application of Pappus's theorem and other methods to solve the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants discuss using Pappus's theorem to find the volume, while others mention using disks or washers as a more familiar method. There is a focus on the complexity introduced by the bounded area crossing the axes of rotation.

Discussion Status

Participants are actively questioning the appropriateness of their approaches and the implications of Pappus's theorem. Some have noted the tricky aspects of the problem, particularly regarding the regions being revolved and the need to consider only external areas when applying the theorem.

Contextual Notes

There is an acknowledgment that the area being revolved crosses the axes, which complicates the calculations. Participants are also considering the implications of the centroid's location in relation to the axes of rotation.

Korupt
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I have to find the volume of the solid whose area is bound by the equations [tex]y=-x+3 \ and \ y=x^2-3x[/tex] as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:

[tex]M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}[/tex]
[tex]M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}[/tex]
[tex]A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}[/tex]
So the center of mass coordinates are:
[tex]\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1[/tex]
[tex]\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}[/tex]
Revolving around x:
[tex]d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}[/tex]
[tex]V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083[/tex]

Revolving around y:
[tex]d=(2\pi)(1) = 2\pi[/tex]
[tex]V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206[/tex]

However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.
 
Last edited:
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Korupt said:
I have to find the volume of the solid whose area is bound by the equations [tex]y=-x+3 \ and \ y=x^2-3x[/tex] as it revolves around the x and y axis. I approached it by finding the center of mass and then using Pappus's theorem:
[tex]M_{x}=\int^{3}_{-1} \frac{1}{2}((-x+3)^2-(x^2-3x)^2)\,dx = \frac{64}{15}[/tex]
[tex]M_{y}=\int^{3}_{-1} x(-x+3-x^2+3x)\,dx = \frac{32}{3}[/tex]
[tex]A=\int^{3}_{-1} -x+3-x^2+3x \, \ dx = \frac{32}{3}[/tex]
So the center of mass coordinates are:
[tex]\overline{x} = \frac{M_{y}}{A} = \frac{\frac{32}{3}}{\frac{32}{3}} = 1[/tex]
[tex]\overline{y} = \frac{M_{x}}{A} = \frac{\frac{64}{15}}{\frac{32}{3}} = \frac{2}{5}[/tex]
Revolving around x:
[tex]d=(2\pi)(\frac{2}{5}) = \frac{4\pi}{5}[/tex]
[tex]V=dA=(\frac{4\pi}{5})(\frac{32}{3}) = \frac{128\pi}{15} \approx 26.8083[/tex]
Revolving around y:
[tex]d=(2\pi)(1) = 2\pi[/tex]
[tex]V=dA=(2\pi)(\frac{32}{3}) = \frac{64\pi}{3} \approx 67.0206[/tex]
However the answers in the book are 63.9848 and 72.9478 respectively. Can anyone tell me where I went wrong or if I am even approaching the problem correctly? Thank you.
First of all: Are you instructed to use Pappus's Theorem ?

What precisely does Pappus's Theorem state?

No matter what method you use, this problem includes some tricky details.
 
I have not been instructed to use any particular method, I just need to find a solution to the problem. For these types of problems I generally use disks or washers to find the volume, but I guess the "tricky" part of this problem is that the bound area crosses the the axis on which it's supposed to rotate, as u can see:
http://www4c.wolframalpha.com/Calculate/MSP/MSP3271a035bi2ii1400fa000012ie2ia8673b6ch5?MSPStoreType=image/gif&s=50&w=381&h=306&cdf=Coordinates&cdf=Tooltips

Pappus's second theorem states that the volume of a solid of revolution generated by the revolution of a region about an external axis is equal to the product of the area of the region and the distance traveled by the region's geometric centroid.

http://mathworld.wolfram.com/PappussCentroidTheorem.html
 
Last edited by a moderator:
Korupt said:
I have not been instructed to use any particular method, I just need to find a solution to the problem. For these types of problems I generally use disks or washers to find the volume, but I guess the "tricky" part of this problem is that the bound area crosses the the axis on which it's supposed to rotate, as u can see:
Yes, this is the tricky part … but it's easily handled. The trickier part is to revolve this around the y-axis. Only part of the area on the left is overlapped by the area on the right.
Pappus's second theorem states that the volume of a solid of revolution generated by the revolution of a region about an external axis is equal to the product of the area of the region and the distance traveled by the region's geometric centroid.
http://mathworld.wolfram.com/PappussCentroidTheorem.html

The part of Pappus's Theorem you seem to be ignoring is: That the axis must be an external axis to the region being rotated. So, when you calculate the centroid, only include the portion of the region above the x-axis . For the other part only the portion to the right of the y-axis , then do the little portion (that's not overlapped) left of the y-axis.
 
SammyS said:
Yes, this is the tricky part … but it's easily handled. The trickier part is to revolve this around the y-axis. Only part of the area on the left is overlapped by the area on the right.

The part of Pappus's Theorem you seem to be ignoring is: That the axis must be an external axis to the region being rotated. So, when you calculate the centroid, only include the portion of the region above the x-axis . For the other part only the portion to the right of the y-axis , then do the little portion (that's not overlapped) left of the y-axis.

Yeah you are right, it is only external. I figured it out by cutting it into smaller parts like you said and then using disks and washers. Thank you for your help.
 

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