Let $(x^2 + y^3) \subset k[x, y]$ be an ideal. Let $A = k[x,y]/(x^2 + y^3)$. The goal is to compute $\dim A$.
Here's my thought process: we can identify $A$ with a subring of $k[t]$, so $A$ is a domain, making the ideal $(x^2 + y^3)$ prime. A has fraction field $k(t)$, since $t^3/t^2 = t$. Thus the transcendence degree of $\mathrm{Frac}~A$ over $k$ is 1. But then $\dim A = \dim_k \mathrm{spec}~A = \mathrm{tr.deg}_k k(t) = 1$, since $A$ is irreducible (since $(x^2 + y^3)$ is prime).