= Solution
Use the <outgoing Green function for the three-dimensional Helmholtz equation>
$$
G_k(\mathbf r,\mathbf r')=\frac{e^{ik|\mathbf r-\mathbf r'|}}{4\pi|\mathbf r-\mathbf r'|},\qquad (\Delta_{\mathbf r'}+k^2)G_k=-\delta_{\mathbf r}.
$$
Let $\mathbf n'$ point out of the obstacle $D$ into the exterior. Apply <Green second identity> to the field and $G_k$ in a large exterior region. Their <Sommerfeld radiation conditions> make the outer-sphere contribution vanish. The normal on the exterior region's inner boundary is $-\mathbf n'$, so the <Kirchhoff–Helmholtz representation> is
$$
\boxed{\psi(\mathbf r)=\int_VG_k(\mathbf r,\mathbf r')Q(\mathbf r')\,d\mathbf r'+\int_{\partial D}[\psi(\mathbf r')\partial_{n'}G_k-G_k\partial_{n'}\psi(\mathbf r')]\,dS',\quad\mathbf r\notin\overline D.}
$$
The first integral is the incident source field $\psi_i$. Therefore the scattered field is the surface term. Since $\psi_i$ solves the homogeneous <Helmholtz equation> within the obstacle, its surface term is zero at exterior observation points; the same representation can be written using only scattered-field traces:
$$
\boxed{\psi_s(\mathbf r)=\int_{\partial D}[\psi_s\partial_{n'}G_k-G_k\partial_{n'}\psi_s]\,dS'.}
$$
The unknown boundary condition determines these traces, but the representation itself does not require choosing that condition.
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