= Solution
Use the <C2 root system> realization in an orthonormal basis $\varepsilon_1,\varepsilon_2$ with <simple roots> $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=2\varepsilon_2$. This numbers the <short root> first, as required. The <fundamental weights> and <half-sum of positive roots> are
$$
\omega_1=\varepsilon_1,\qquad\omega_2=\varepsilon_1+\varepsilon_2,\qquad
\rho=2\varepsilon_1+\varepsilon_2,
$$
so $\lambda+\rho=(a+b+2)\varepsilon_1+(b+1)\varepsilon_2$. For the four <positive roots>, the factors of the <Weyl dimension formula> are
$$
\begin{array}{c|c|c}
\alpha&\alpha^\vee&
\langle\lambda+\rho,\alpha^\vee\rangle/\langle\rho,\alpha^\vee\rangle\\\hline
\varepsilon_1-\varepsilon_2&\varepsilon_1-\varepsilon_2&a+1\\
2\varepsilon_2&\varepsilon_2&b+1\\
\varepsilon_1+\varepsilon_2&\varepsilon_1+\varepsilon_2&(a+2b+3)/3\\
2\varepsilon_1&\varepsilon_1&(a+b+2)/2
\end{array}.
$$
Multiplying yields the <Weyl dimension formula for C2>:
$$
\boxed{\dim V(a\omega_1+b\omega_2)=\frac{(a+1)(b+1)(a+b+2)(a+2b+3)}6.}
$$
As checks, $V(\omega_1)$ has dimension $4$, $V(\omega_2)$ dimension $5$, and $V(2\omega_1)$ dimension $10$.
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