Understanding Vector Integrals: A Closer Look at Integral Identity 1

  • Thread starter Thread starter coverband
  • Start date Start date
  • Tags Tags
    Identity Integral
coverband
Messages
170
Reaction score
1
1. By considering each component of the vector A show that \iint A(u.n)ds = \iiint{(u.nambla)A+A(nambla.u)}dV (A,u and n are vectors)



Homework Equations





3. Let A = ai + bj + ck. L.H.S: \iint ai (u.n)ds + \iint bj (u.n)ds +\iint ck (u.n)ds
R.H.S. = \iiint(u.nambla)ai dV+ \iiint(u.nambla)bj dV+ \iiint(u.nambla)ck dV + \iiinta(nambla.u)i dV ...


Sorry my latex is all over the place maybe just ignore this
 
Physics news on Phys.org
That gradient symbol is called "nabla." "nambla" is something else entirely.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top