Let $D\subset \mathbb{R}^3$. Let $D$ be a connected subset of $\mathbb{R}^3$. Show that if there is a solution of the system of equations \begin{equation} \Delta u=f \text{ in } D, \frac{du}{dn}=g\text{ on boundary of } D, \end{equation} then $\int_D\ f \ dV=\int_{\text{boundary of} D} g \ ds$.
My partial answer:
Assume that $\int_D\ f \ dV\neq\int_{\text{boundary of} D} g \ dS$ and $u$ is the solution of the system of equation, then
\begin{equation} \int_D \ \Delta u \ dV= \int_D\ f \ dV\neq\int_{\text{boundary of} D} g \ dS= \int_{\text{boundary of} D} \frac{du}{dn} \ dS. \end{equation} This contradicts Green's first identity.
Please let me know that idea of my answer is correct. Is it possible to prove this question without using a contradiction.
The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.
In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.