How Do I Apply the Theorem of Pappus to Find the Volume of a Revolved Region?

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In summary, Pappus' theorem states that the volume of a figure created by rotating a region of area A around a line is equal to 2πRA, where R is the distance from the centroid of the region to the line. To find the volume, calculate the distance from the centroid to the line, find the area of the region, and use the formula V= 2πRA. In the given example, the area of the region is πa^2/2 and the distance from the centroid to the line y=x-a is |0-(4a/3π)-a|/√2. Therefore, the volume is (2πa^2/2)*((4a/3π)+a
  • #1
blumfeld0
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The theorem of pappus seems too simple that i do not get it.
ok
suppose I have the 1. (x cm., y c.m) meaning i have the x and y center of mass (the centroids). the center of mass is of some function
2. y(x) how do i find the volume when i revolve this region about the line 3. y(x) = mx+b

thank you
 
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  • #2
No, the center of mass is not "some function". The center of mass is a single point: of some 2 dimensional region you have not mentioned.

And since you haven't mentioned the region, you haven't mentioned the area of the region which is the the cross-sectional area of the volume swept out. Pappus' theorem says that if you have a region of area A, rotated about a line, the volume of the figure created is 2[itex]\pi[/itex]RA where R is the distance from the centroid of the region.

Calculate the distance, R, from the centroid of the given region to the line y= mx+ b, calculate the area of the region, and then use [itex]V= 2\pi RA[/itex].
 
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  • #3
Ok here is a sample of what I mean.
Given the centroid of the region enclosed by the x-axis and
y = Sqrt[a^2-x^2] (a semicircle) is located at (0,4a/3Pi).
find the volume of the solid generated by revolving this region about the y= x-a.

So I know Volume = 2Pi*area*distance from centroid to line y=x-a

Area = Pi * a^2 right?
distance to from centroid to line y = x-a is what??

thank you
 
  • #4
The distance from a point, [itex](x_0, y_0)[/itex], to a line Ax+ By+C= 0 is given by
[tex]\frac{|Ax_0+ By_0+ C|}{\sqrt{A^2+ B^2}}[/itex]

The area of a circle is [itex]\pi r^2[/itex]. The area of a semi- circle is half that.
 
  • #5
Hi Thank you very much. quick question what does
|Ax +By+c| mean?
I do not understand what the tabs means?

can you please give me a quick example.
In my case

y= x-a
so |1x-1y-a|
does this equal x^2-y^2-a^2"
thank you
 
  • #6
blumfeld0 said:
Hi Thank you very much. quick question what does
|Ax +By+c| mean?
I do not understand what the tabs means?

can you please give me a quick example.
In my case

y= x-a
so |1x-1y-a|
does this equal x^2-y^2-a^2"
thank you
| | is absolute value. But let's back up a little. WHY are you doing this problem? You don't know Pappus' theorem, you got the wrong area for a semi- circle and you have the wrong value for the centroid! If you are taking a multi-variable Calculus course, you should know all those things before you attempt a problem like this. I recommend you go back, re-read the problem, take a deep breath, and start all over again.
 
  • #7
We skipped this stuff when I was in calc II and I really want to get it so I am doing a couple of problems for fun. I know the area of a semicircle.
The problem says
"the centroid of the region enclosed by the x-axis and the semicircle
y= Sqrt[a^2-x^2] lies at the point (0,4a/3Pi). Find the volume of the solid generated by revolving this region about the line y = x-a"

Volume = 2Pi A *b(distance)

A= Pi*a^2/2
b = (1x0-1y0-a)/Sqrt[(1+1)] = |0-(4a/3Pi)-a|/Sqrt[2] =

(2*Pi*Pi*a^2/2) * ((4a/3Pi)+a) /Sqrt[2]

is that right?
thank you!
 

Related to How Do I Apply the Theorem of Pappus to Find the Volume of a Revolved Region?

What is the Theorem of Pappus?

The Theorem of Pappus, also known as the Pappus-Guldin Theorem, is a mathematical principle that relates the volume of a solid to the area of its generating curve. It is named after the ancient Greek mathematician Pappus of Alexandria and was later expanded by Paul Guldin.

What is the formula for the Theorem of Pappus?

The formula for the Theorem of Pappus is V = A * d, where V is the volume of the solid, A is the area of the generating curve, and d is the distance traveled by the centroid of the generating curve when rotated about an axis.

How is the Theorem of Pappus used in real life?

The Theorem of Pappus has many real-life applications, particularly in engineering and architecture. It is used to calculate the volume of three-dimensional objects such as gears, pipes, and barrels. It is also used in the design and construction of structures such as bridges and buildings.

What are the conditions for the Theorem of Pappus to be applied?

The Theorem of Pappus can only be applied if the generating curve is a closed curve and the axis of rotation is perpendicular to the plane of the curve. Additionally, the centroid of the curve must lie on the axis of rotation.

How does the Theorem of Pappus relate to other mathematical principles?

The Theorem of Pappus is closely related to other mathematical concepts such as the centroid and the parallel axis theorem. It can also be extended to calculate the surface area of a solid generated by a curved line rotating about an axis.

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