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:
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  • #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
 
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  • #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.
 

Related to Volume of Solid Revolving Around x & y Axis

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.

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