Line integral

Using $y$ as a parameter, we have $x = 4 - y^2$ , $dx = -2y\ dy$ in the domain $-3 \leq y \leq 2$ . Substituting this in, we have

$\begin{aligned} N &= \int_C (y^2\ dx + x\ dy) \\ &= \int_{-3}^2 (y^2(-2y\ dy) + (4-y^2)\ dy) \\ &= \int_{-3}^2 (-2y^3 -y^2 + 4)\ dy \\ &= \left[-\frac{y^4}{2}-\frac{y^3}{3} + 4y \right]_{-3}^2 \\ &= \left(-\frac{16}{2} -\frac{8}{3} + 8 \right) -\left(-\frac{81}{2} + \frac{27}{3} - 12 \right) \\ &= \frac{245}{6} \\ \end{aligned}$