Surface Integral of Outward Normal Vector over a Spherical Surface

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SUMMARY

The discussion focuses on evaluating the surface integral of the outward normal vector over a spherical surface of radius R, denoted as S. The integral of the unit outward normal vector, n, multiplied by the surface area element dS, is confirmed to be zero. However, the integral of the inner product nn is clarified as being related to the scalar area of the sphere, which is positive definite. The integral of n with respect to the surface area element dA is not zero, highlighting the distinction between vector and scalar quantities in this context.

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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.
 
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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.
 

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