Finding the Integral of a Sphere: Understanding Ro, c, and r | Physics Forums

  • Thread starter Murr14
  • Start date
  • Tags
    Integral
In summary, the question involves a sphere with a center at Ro and a constant vector c that specifies a point on the surface. The integral we need to solve is over the surface of the sphere and involves the distance between c and a point r on the surface. To solve it, we will need to use the concept of surface area and the unit normal vector. The final answer will be in terms of Ro and c.
  • #1
Murr14
3
0
hey everyone, this is my first post at this physicsforums.com dealy. i have a take home exam and I'm havin a rough time with the way the prof worded this question:

Consider a sphere of radius R centered at Ro (Ro is a vector) and a constant vector c which specifies a point in the sphere, where the origin of these vectors is arbitrarily chosen somewhere in space.

A) find the following integral:

(double integral sign) da / |c - r |

where da is an infinitesimal outward area vector at r on the surface S of the sphere. Express your answer in terms of Ro and c .



...so that's it...i'm confused about the components of Ro and c and r...and whether or not i even need them to do the equation...if anyone could get me started (or even finished...hehe) that would be really appreciated! THANKS
 
Last edited:
Physics news on Phys.org
  • #2


Hello and welcome to the forum! I understand that this question may seem confusing at first, but I will do my best to explain it to you.

First, let's break down the components of the question. We have a sphere with a radius R and a center at Ro, which is a vector. We also have a constant vector c that specifies a point on the surface of the sphere. The origin of these vectors is arbitrarily chosen, meaning it can be anywhere in space.

Next, we have the integral that we need to solve. The integral is over the surface S of the sphere and we are integrating with respect to the infinitesimal outward area vector da at a point r on the surface. This means that we are essentially adding up the infinitesimal areas on the surface of the sphere.

Now, let's look at the expression inside the integral, |c - r|. This represents the distance between the constant vector c and a point r on the surface of the sphere. This is important because it will help us determine the magnitude of the infinitesimal area vector da at that point.

To solve this integral, we will need to use the concept of surface area and use the unit normal vector to calculate the infinitesimal area vector da. The unit normal vector is perpendicular to the surface at a given point and has a magnitude of 1.

So, the final answer will be in terms of Ro and c, as they are the given parameters in the question. To get started, you can try to set up the integral using the surface area formula and the unit normal vector. You can also look up resources on surface integrals and unit normal vectors for further guidance.

I hope this helps you get started on solving the problem. Good luck with your exam!
 
  • #3


Hi there! It's great that you're seeking help with your exam question. Let's break down the components of Ro, c, and r so we can better understand the integral.

Ro is a vector representing the center of the sphere. It is a point in space that is arbitrarily chosen as the origin for the vectors in this problem.

c is a constant vector that specifies a point within the sphere. This means that c is a fixed vector that does not change throughout the problem. It is also important to note that c is not necessarily equal to Ro - it can be any point within the sphere.

r is a variable vector that represents a point on the surface of the sphere. This means that r will change as we move along the surface of the sphere. It is important to note that r is not necessarily equal to Ro or c - it can be any point on the surface of the sphere.

Now let's look at the integral. The da represents an infinitesimal outward area vector at r on the surface of the sphere. This means that da is a small vector that is perpendicular to the surface of the sphere at point r. It is important to note that da is not a constant vector - it will change as we move along the surface of the sphere.

The |c - r| in the denominator represents the distance between the constant vector c and the variable vector r. This distance will also change as we move along the surface of the sphere.

To solve the integral, you will need to use the definition of a double integral, which involves taking the limit of a sum as the area of each infinitesimal outward area vector approaches zero. This will involve using the dot product between da and the vector (c - r) to find the component of da that is parallel to (c - r). Then you can integrate over the surface of the sphere to find the total area.

I hope this helps to clarify the components and process for solving the integral. Good luck with your exam!
 

1. What is a confusingly worded integral?

A confusingly worded integral is an integral (a mathematical concept used in calculus) that is written in a way that is difficult to understand or interpret. This can be due to unclear notation, ambiguous phrasing, or other factors.

2. Why do confusingly worded integrals exist?

Confusingly worded integrals can exist for a variety of reasons. They may be a result of a mistake made by the person writing the integral, or they may be intentionally worded in a confusing way as a trick or puzzle. Additionally, different notations and conventions can also contribute to the confusion.

3. How do I solve a confusingly worded integral?

Solving a confusingly worded integral can be challenging, but there are a few strategies that can help. First, try to rewrite the integral in a clearer form by using different notation or breaking it down into smaller parts. You can also use your knowledge of calculus and integration techniques to guide your approach. If all else fails, seeking help from a math tutor or professor can also be beneficial.

4. Can a confusingly worded integral have multiple correct solutions?

Yes, a confusingly worded integral can have multiple correct solutions. This is because there are often different ways to interpret and solve an integral, and the confusing wording may lead to different approaches and solutions. It is important to carefully check your work and make sure your solution satisfies the original integral.

5. How can I avoid creating a confusingly worded integral?

To avoid creating a confusingly worded integral, it is important to use clear and consistent notation, and to communicate your thoughts and intentions clearly when writing mathematical expressions. It can also be helpful to have someone else review your work for clarity and understanding before finalizing it.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
808
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
8K
  • Calculus and Beyond Homework Help
Replies
21
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
11K
  • Other Physics Topics
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Back
Top