Surface Integral of Outward Normal Vector over a Spherical Surface

Click For Summary
The discussion revolves around evaluating the surface integral of the outward normal vector over a spherical surface. The integral of the unit outward normal vector, denoted as n, is found to be zero, but there is confusion regarding the relationship of nn to this result. It is clarified that while the area of the sphere is a scalar quantity, the surface element vector dA, which incorporates the normal vector, leads to a non-zero integral when considering the dot product of n with dA. The key takeaway is that the integral of n dS is zero, but the integral of n dot dA is not. Understanding these distinctions is crucial for correctly evaluating the surface integral.
Xian Xi

Homework Statement



Let n be the unit outward normal of a spherical surface of Radius R, let the surface of the sphere be denoted by S.
Evalute Surface integral of nndS

Homework Equations

The Attempt at a Solution


I have evaluated the surface integral of ndS and found it to be 0. but am not sure how nn relates to it.
 
Physics news on Phys.org
If ##\bf nn## is an inner product, it is positive definite and of magnitude 1. What remains is the integral of ##d\bf S## ?
Can you show how you find the integral of ##{\bf n}d{\bf S } \ ## is zero ?
 
Xian Xi said:

Homework Statement



Let n be the unit outward normal of a spherical surface of Radius R, let the surface of the sphere be denoted by S.
Evalute Surface integral of nndS

Homework Equations

The Attempt at a Solution


I have evaluated the surface integral of ndS and found it to be 0. but am not sure how nn relates to it.
The area of a sphere is a scalar quantity. If bold letters mean vectors, do not use bold for scalars.
Presumably dS is also scalar, the area of a surface element. The outward normal of that surface element is n. So dA=ndS is the surface element vector. Yes, its integral for the whole sphere is zero. But ##\int {\vec n \cdot \vec {dA}} ## is not zero, as @BvU pointed out.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K