My question: $$\iint_D xy \,dA\,,$$ where $D$ is the region bounded by $y = x - 1$ and parabola $y^{2} = 2x + 6$
Quick question, what does the inner integrand represent?
Anyway here's how I set up the double integral with respect to y as the inner integrand:
$$\int_{-1}^{5} \int_{-2}^{4} xy \,dy \,dx$$ $$\int_{-1}^{5} \int_{-2}^{4} (y+1)y \,dy \,dx$$ $$\int_{-1}^{5} \int_{-2}^{4} (y^2 + y \,dy \,dx$$ $$\int_{-1}^{5} \left[\left(\frac{y^3}{3} + \frac{y^2}{2}\right)\right]_{-2}^{4} \,dy \,dx$$
Limits $[-1,5] \times[-2,4]$ gives whole rectangle and cannot be true. If you "stay" on axis $OY$ having head up to $OX$ direction, then floor is $y^{2} = 2x + 6$ and roof $y=x-1$.
So you have limits $$\left\{\begin{array}{} -2 \leqslant y \leqslant 4\\ \frac{1}{2}y^2-3 \leqslant x \leqslant y+1 \end{array}\right\}$$ i.e. $$\int\limits_{-2}^{4}\,dy\int\limits_{\frac{1}{2}y^2-3}^{y+1} xy \,dx$$