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

  • Thread starter Thread starter Murr14
  • Start date Start date
  • Tags Tags
    Integral
AI Thread Summary
The discussion focuses on solving an integral related to a sphere of radius R, centered at a vector Ro, with a constant vector c inside the sphere. Ro represents the center of the sphere, while c is a fixed point within it, and r is a variable point on the sphere's surface. The integral involves the infinitesimal outward area vector da at point r and the distance |c - r| in the denominator. To solve the integral, one must understand the relationship between these vectors and apply the definition of a double integral, utilizing the dot product for integration. Clarifying these components is essential for successfully completing the exam question.
Murr14
Messages
3
Reaction score
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


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!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top