Past exam of the mathematical methods II course of the University of Oxford / 2017 resit / 10 / c

Integrating the mixed partial derivative successively gives $z=x^3{y}^2/6+A(x)+B(y)$. The first boundary value gives $A(x)=x^2-1-B(0)$. Substitution at $x=1$ then gives $B(y)-B(0)=\sin y-y^2/6$. ThereforeIts mixed derivative is $x^{2}y$, and direct evaluation checks both boundary values. Their common corner is compatible: both prescribe $z(1,0)=0$.