Centroid of a Hemisphere in Spherical Coordinates

In summary, to find the centroid of the hemisphere z=sqrt(a-x^2-y^2), you need to use the cross product of the partial derivatives of a parametrization of the function as a parametric surface, rather than the gradient. This will give you the correct normal vector with a magnitude of 2a. Using the gradient may result in the wrong answer.
  • #1
whereisnomar
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Homework Statement


Find the centroid of the hemisphere, z=sqrt(a-x^2-y^2).


Homework Equations


z(bar)=(1/m)* Surface integral(z dS)
dS= magnitude of the magnitude of normal vector * dA


The Attempt at a Solution


I use the gradient to the hemisphere to get the magnitude of the normal vector... which comes out to be 2a. But the book says to use the cross product of the partial derivatives of a parametrization of the function as a parametric surface. And I get the wrong answer when I use the gradient.

Why can't you use the gradient to get the normal needed for a surface integral?
 
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  • #2
If you say f(x,y)=sqrt(a^2-x^2-y^2) then the position vector on the surface is r=[x,y,f(x,y)]. And if I take the function to find the gradient of as z-f(x,y), then I get that the gradient vector happens to be the same as the cross product r_x x r_y. But neither one of them has norm 2a. Can you show what you did? In general taking a gradient will give you a normal, but it won't necessarily give you the same normal in terms of magnitude as r_x x r_y. The latter is the correct one to use.
 
  • #3
I would be inclined to write the hemisphere in spherical coordinates with [itex]\rho= a[/itex].
 

1. What is a surface integral problem?

A surface integral problem is a mathematical calculation that involves finding the total value of a function over a given surface. It is a type of double integral, where the integration is performed over a two-dimensional surface instead of a one-dimensional line.

2. How is a surface integral problem different from a regular integral?

Unlike a regular integral, which is performed over a one-dimensional interval, a surface integral is performed over a two-dimensional surface. This means that the limits of integration and the integrand (function being integrated) are in terms of two variables instead of one.

3. What is the purpose of solving a surface integral problem?

The main purpose of solving a surface integral problem is to find the total value of a function over a given surface. This can be useful in a variety of scientific and engineering applications, such as calculating the flux of a vector field through a surface or finding the mass or charge distribution on a surface.

4. What types of surfaces can be used in a surface integral problem?

A surface integral problem can be applied to any two-dimensional surface, including flat surfaces, curved surfaces, and even irregular surfaces. Some common types of surfaces used in surface integral problems include planes, spheres, cylinders, and cones.

5. What are some common methods for solving surface integral problems?

There are several methods for solving surface integral problems, including using parametric equations, converting to polar coordinates, and using surface area elements. Other techniques such as the divergence theorem and Stokes' theorem can also be used to simplify and solve certain types of surface integral problems.

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