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**craig** Let $\displaystyle \omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy$.

Find potential function $\displaystyle f$ for $\displaystyle \omega$.

We know$\displaystyle \frac{\partial f}{\partial x} = 3y^2 + 2y$, so $\displaystyle f = 3xy^2 + 2xy + g(x)$

Differentiating wrt $\displaystyle y$, $\displaystyle \frac{\partial f}{\partial y} = 6xy + 2x + g'(x) = 6xy + 2x + y^2$.

So we have $\displaystyle g(x) = \frac{2}{3}y^3$ + constant.

The answers then simply state that the constant is just zero, not quite sure how they get to this?

Thanks in advance