for an inviscid flow the force on a body b is due entirely to pressure

Question

For an inviscid flow, the force on a body B is due entirely to pressure /10 \underline{\mathbf{F}}=-\int_{\partial \mathcal{B}} p_{m} \hat{\mathbf{n}} \mathrm{d} S a) For a steady incompressible inviscid irrotational flow, show that, by integrating the Euler equaions over a volume of fluid exterior to the body, we may alternatively write the force as a surface integral over any closed surface S that encloses the body (see Figure 3) with formula \underline{\mathbf{F}}=-\int_{\mathcal{S}} \rho \hat{\mathbf{n}} \cdot\left[\underline{\mathbf{u}} \underline{\mathbf{u}}-\frac{1}{2}(\underline{\mathbf{u}} \cdot \underline{\mathbf{u}}) \underline{\mathbf{I}}\right] \mathrm{d} S b) Show, using the above formula, that the 'drag' force is zero on any body held fixed in an other-wise uniform inviscid flow U, in other words E.U = 0. Tips: Take the surface integral S very far from B (say |x – xol + x where x, is an arbitrary point on the body and x is on S), and assume the disturbance of the flow due to the body u' = u - U decays fast enough that |u'|² d$ → 0 in this limit.