Divergence Theorem: Gauss & Cross-Product Integration

[Apashanka]

From gauss divergence theorem it is known that ##\int_v(\nabla • u)dv=\int_s(u•ds)## but what will be then ##\int_v(\nabla ×u)dv##
Any hint??
The result is given as ##\int_s (ds×u)##

 

[Charles Link]

I had to look this one up in the appendix of an E&M (electricity and magnetism) textbook. I have never used it in any application.
With ## \int \nabla p \, dv =\int p \, \hat{n} ds ##, where the force per unit volume in a fluid is ## f_v=-\nabla p ## and must balance the force of gravity per unit volume ## f_g=-\rho g \hat{z} ##, I have done another proof of Archimedes principle.
Yes, you have it correct, and if you take ## u \times ds ##, it gets a minus sign.

 

[Apashanka]

I had to look this one up in the appendix of an E&M (electricity and magnetism) textbook. I have never used it in any application.
With ## \int \nabla p \, dv =\int p \, \hat{n} ds ##, where the force per unit volume in a fluid is ## f_v=-\nabla p ## and must balance the force of gravity per unit volume ## f_g=-\rho g \hat{z} ##, I have done another proof of Archimedes principle.
Yes, you have it correct, and if you take ## u \times ds ##, it gets a minus sign.

Sir actually I came across this formula ##\int_v(\nabla×u)dv=\int_s(ds×u) ##,but sir I want to just prove this formula once by hand ,but didn't get any idea or hint of how to start with... that's why sir I am asking this...

 

[pasmith]

Let [itex]\mathbf{c}[/itex] be an arbitrary constant vector field and consider [tex]
\mathbf{c} \cdot \int_V \nabla \times \mathbf{u}\,dV = \int_V \mathbf{c} \cdot (\nabla \times \mathbf{u})\,dV
= \int_V \nabla \cdot (\mathbf{u} \times \mathbf{c})\,dV.[/tex]

 

[Apashanka]

Ok sir then ##\int_v\nabla•(u×c)dv=\int_s(u×c)•ds=\int_s(u_jc_k-u_kc_j)ds_i+(u_kc_i-u_ic_k)ds_j+(u_ic_j-u_jc_i)ds_k##rearranging terms having ##c_i,c_j## and ##c_k## coefficients it becomes ##-\int_s(u×ds)•c## and for c being a constt vector Rhs becomes ##\int_v(c•(\nabla×u))##
Thanks @pasmith and @Charles Link