Volume of Solid Revolving Around x & y Axis

In summary, Pappus's 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.
  • #1
Korupt
5
0
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:
Physics news on Phys.org
  • #2
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.
 
  • #3
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:
  • #4
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.
 
  • #5
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.
 

What is the formula for calculating the volume of a solid revolving around the x-axis?

The formula for calculating the volume of a solid revolving around the x-axis is V = π∫ (upper limit: a, lower limit: b) [f(x)]^2 dx, where f(x) is the function representing the cross-sectional area at any given point on the x-axis.

What is the difference between revolving a solid around the x-axis and the y-axis?

The main difference is the orientation of the cross-sectional area. When revolving around the x-axis, the cross-sectional area is perpendicular to the x-axis, while when revolving around the y-axis, the cross-sectional area is perpendicular to the y-axis.

How do you find the limits of integration when calculating the volume of a solid revolving around the x-axis?

The limits of integration can be found by identifying the x-values of the points of intersection between the function and the x-axis. These points will serve as the upper and lower limits of integration.

What is the significance of using integration to calculate the volume of a solid revolving around an axis?

Using integration allows for a more accurate calculation of the volume, as it takes into account the varying cross-sectional areas along the axis of rotation. It also allows for the calculation of more complex shapes, such as those with varying densities.

Are there any real-world applications of calculating the volume of a solid revolving around an axis?

Yes, there are many real-world applications of this concept, such as calculating the volume of rotation of a tire, or the volume of a water tank with a cylindrical shape. It is also used in engineering and architecture to design structures with specific volumes and dimensions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
344
  • Calculus and Beyond Homework Help
Replies
3
Views
273
  • Calculus and Beyond Homework Help
Replies
5
Views
696
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
543
  • Calculus and Beyond Homework Help
Replies
8
Views
876
  • Calculus and Beyond Homework Help
Replies
6
Views
548
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
789
  • Calculus and Beyond Homework Help
Replies
3
Views
958
Back
Top