The centroid of a hemispherical surface

In summary, the conversation discusses finding the centroid of a hemispherical surface using Pappus's theorem, which is not applicable. The correct method involves calculating the width of the shell and integrating using simple integrals and rectangular coordinates. The conversation also addresses the difference between the centroid of a 2D semicircle and a 3D hemispherical surface.
  • #1
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I have to find the centroid of a hemispherical surface with radius r.
I want to use Pappus's theorem.S=2*Pi*x*s
The surface S, is 2*Pi*r^2 and the length of the curve s, is Pi*r so the distance from the centroid ,which lies on the y-axis by symmetry, to the x-axis is S/(2*Pi*Pi*r) = r/Pi

This is not the answer by textbook gives for the y coordinate of the centroid. What am I doing wrong? Please help me.
 
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  • #2
x is the centroid of the generating surface which is a semicircle. And the centroid of a semicircle (a 2D object) is different to that of a hemispherical surface (a 3D object). I'm afraid you can't use Pappus.
 
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  • #3
Try this instead - http://mathworld.wolfram.com/Hemisphere.html

I believe the hemispherical is a semi-circle rotated by pi, or a quarter circle rotated by 2pi about an axis corresponding to a radius and corresponding to either end of the arc.
 
  • #4
I understand I can't use Pappus's theorem.
So I try to cut the hemispherical shell into horizontal shells up the y axis. Each shell should have the surface 2*Pi*R where R=sqrt(r^2-y^2)
So dS = 2*Pi*R dy
But there is something wrong here, When I integrate from 0 to r I don't get the right answer for the surface of the hemispherical surface.
Could you help me some more, please.
 
  • #5
I got it right. I had to calculate the width of the shell.And then integrate.
 
  • #6
How can I calculate the centroid of the hemisphere surface using only simple integrals and rectangular coordinates?
 

1. What is the centroid of a hemispherical surface?

The centroid of a hemispherical surface is the point located at the exact center of the hemisphere. It is the point at which the surface can be balanced on a single point without toppling over.

2. How is the centroid of a hemispherical surface calculated?

The centroid of a hemispherical surface can be calculated by finding the average of the x, y, and z coordinates of all the points on the surface. This can be done using an integration formula or by breaking the surface into smaller, simpler shapes and calculating the centroid of each shape.

3. Why is the centroid of a hemispherical surface important?

The centroid of a hemispherical surface is important because it is used in various engineering and physics applications, such as determining the stability of a structure or the distribution of weight. It is also a key concept in the study of fluid mechanics and heat transfer.

4. Can the centroid of a hemispherical surface be outside of the surface itself?

No, the centroid of a hemispherical surface will always be located within the surface itself. This is because the centroid is the point of balance for the surface, and if it were outside of the surface, the surface would not be able to balance and would topple over.

5. How does the centroid of a hemispherical surface differ from the center of mass?

The centroid of a hemispherical surface is a geometric property, while the center of mass is a physical property. The center of mass takes into account the distribution of mass within the surface, while the centroid only considers the shape and dimensions of the surface. In a uniform hemisphere, the centroid and center of mass will coincide, but in irregular shapes, they may be different points.

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