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 \mathbf{c} be an arbitrary constant vector field and consider <br /> \mathbf{c} \cdot \int_V \nabla \times \mathbf{u}\,dV = \int_V \mathbf{c} \cdot (\nabla \times \mathbf{u})\,dV<br /> = \int_V \nabla \cdot (\mathbf{u} \times \mathbf{c})\,dV.

 

[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