How Do You Apply the First Theorem of Pappus-Guldinus to Calculate Surface Area?

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In summary, the problem involves finding the surface area generated when a curve x = ky^2 is rotated about the x-axis. The first theorem of Pappus is used, along with the equations a = kb^2 and a + 15 = k(12.5)^2, to determine the value of k. The final equation used is A = 2(pi)(yL) = integral(2(pi)(yL), L, 0, b) to find the surface area.
  • #1
O011235813
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Homework Statement


hey, I'm having issues with a problem, and my book doesn't seem to show me how to do it.
so,
i have a curve x = ky^2 and I'm to rotate it about the x-axis. I need to find the surface area generated.

How do i use the first theorem to do this? Thanks for the help

sorry i don't have a pic of the problem. but if you want to see here's a link to my book:
http://books.google.com/books?id=l-...&hl=en&sa=X&oi=book_result&resnum=1&ct=result

problem number 5.60
thanks once again

Homework Equations





The Attempt at a Solution

 
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  • #2
O011235813 said:

Homework Statement


hey, I'm having issues with a problem, and my book doesn't seem to show me how to do it.
so,
i have a curve x = ky^2 and I'm to rotate it about the x-axis. I need to find the surface area generated.

How do i use the first theorem to do this? Thanks for the help

sorry i don't have a pic of the problem. but if you want to see here's a link to my book:
http://books.google.com/books?id=l-...&hl=en&sa=X&oi=book_result&resnum=1&ct=result

problem number 5.60
thanks once again

Homework Equations





The Attempt at a Solution

Welcome to Physics Forums.

Before we can help you, you need to show some effort in solving the problem yourself, either providing an attempted solution or at least detailing your thought.

Stating the theorem of Pappus and then seeing what you can do from there be a good start.
 
  • #3
hey,

thanks for the welcome. I figured it out. It was just a bunch of math that I haven't done in the longest time. i had to figure out what k was which is where i messed up at the beginning. I used:
x = ky^2 and i just put b and a so...
a = kb^2 and found k to be b...which was wrong.

it was a system of equations cause there was a length to the object:

a + 15 = k(12.5)^2
a = k(7.5)^2
solving that i got k to be .15
so x = .15y^2

then applying the theorem:

A = 2(pi)(yL) = integral(2(pi)(yL), L, 0, b)

and solve that to get the answer.
Thanks for willing to help.
 

Related to How Do You Apply the First Theorem of Pappus-Guldinus to Calculate Surface Area?

1. What is the concept of "Statics - Pappus Guldinus"?

The concept of "Statics - Pappus Guldinus" is a principle in statics that explains the relationship between the centroid of a two-dimensional shape and its moments of inertia. It states that the centroid of a shape can be used to calculate its moment of inertia about any axis that passes through the centroid.

2. Who is Pappus Guldinus?

Pappus Guldinus was a Greek mathematician and engineer who lived in the 3rd century AD. He is known for his contributions to mathematics, particularly in the field of geometry and statics. The principle of "Statics - Pappus Guldinus" is named after him.

3. How is the principle of "Statics - Pappus Guldinus" used in engineering?

The principle of "Statics - Pappus Guldinus" is used in engineering to calculate the moments of inertia of two-dimensional shapes. This information is important in determining the stability and strength of structures, such as buildings and bridges.

4. Can the principle of "Statics - Pappus Guldinus" be applied to three-dimensional shapes?

No, the principle of "Statics - Pappus Guldinus" only applies to two-dimensional shapes. In three-dimensional cases, the parallel axis theorem and perpendicular axis theorem are used to calculate moments of inertia.

5. How does the principle of "Statics - Pappus Guldinus" relate to the centroid of a shape?

The principle of "Statics - Pappus Guldinus" states that the moment of inertia of a shape about any axis passing through its centroid is equal to the product of the shape's area and the square of the distance between the centroid and the axis. This means that the centroid is a key factor in determining the moment of inertia of a shape.

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