# Proving a Connected Surface is Contained in a Sphere

• bruno321
In summary, If all normals to a connected surface pass through the origin, the surface is contained in a sphere. This is because if all points on the surface can be connected to the origin through normal lines, then the surface must be contained within a sphere. This can be shown using the fact that a connected surface can be locally represented as the graph of a differentiable function, and that the differential of a constant function is always zero. Therefore, the surface must be contained within a sphere.

## Homework Statement

Show that if all normals to a connected surface pass through the origin, the surface is contained in a sphere.

## The Attempt at a Solution

I know a surface is locally the graph of a differentiable function, so in a neighbourhood of a point p, the points satisfy the equation F(x,y,z) = 0. Then a normal vector would be grad(F)(p), and the parameterized normal line would be X(t) = p + t*grad(F)(p).

I know that line passes through the origin, so for some t, X(t)=(0,0,0). But then I am lost. I don't really know how to handle the problem.

I also thought of taking a parameterization of a neighbourhood of p, then a basis for the tangent plane is given by the partial derivatives of the parameterization, and a normal vector is the vector product of those derivatives.

Thanks for any help :)

Hint: If v is a vector function on R3, and v(x0) passes through the origin, then we can write v(x0) = kx0 for some scalar k.

If the surface is connected then any two points on the surface can connected with a curve c(t). c'(t) is perpendicular to the normal. VKint's point is that c(t) is parallel to the normal. What is the derivative of c(t) dotted with itself?

Well, I managed to do it differently. I didn't realize the exercise preceding this one was going to help me :P

This other exercise said that if S is a connected surface, f: S->R a differentiable function, and the differential of f is always 0, then f is a constant function. This is easily proved using the mean value theorem for one variable and the chain rule.

On the exercise I posted above, then, all I need is to take the norm function squared: f(x)=||x||^2 , which is a differentiable function, and S is a connected surface.

If . is the inner product, then the differential df_p(v) = grad(f)(p).v = 2 p.v = 0 because v is in TpS and p obviously lies on the normal line through p.

Then using the result posted above, f is constant => the norm squared is constant => the norm is constant => the surface is contained on a sphere.

## 1. What is a connected surface?

A connected surface is a surface that can be traced continuously without any breaks or holes. In other words, all points on the surface are reachable by a continuous path.

## 2. How do you prove that a connected surface is contained in a sphere?

To prove that a connected surface is contained in a sphere, we must show that all points on the surface are equidistant from the center of the sphere. This can be done by constructing a line segment from the center of the sphere to any point on the surface and showing that the length of this segment is the same for all points on the surface.

## 3. What is the significance of proving that a connected surface is contained in a sphere?

Proving that a connected surface is contained in a sphere is significant because it demonstrates the topological property of being "closed." This means that the surface has no boundaries or edges and is completely contained within itself.

## 4. Are there any other ways to prove that a connected surface is contained in a sphere?

Yes, there are other ways to prove that a connected surface is contained in a sphere. For example, we can use the concept of curvature to show that the surface is curved in a way that is consistent with being contained in a sphere. Another method is to use the Gauss-Bonnet theorem, which relates the curvature of a surface to its topology.

## 5. Can a connected surface be contained in a sphere of any size?

No, a connected surface can only be contained in a sphere of a specific size. This is because the surface must have a finite area and be able to fit within the surface of the sphere without any overlap or gaps. The size of the sphere will depend on the size and shape of the surface.