Evaluating $\iint_D 6x-2x^2y^2+6y\,dx$$dy$ 
Evaluate $\iint_D 6x-2x^2y^2+6y\,dx dy$ where $D$ is the rectangle given by $-2 \leq x \leq 3$ and $-2 \leq y \leq 1$.

I've done this problem two ways. The first time I got $-630$ and then the second time I got $\frac{-1030}{9}$. Lon capa said that they were both wrong . This makes sense because volume cannot be negative.
 A: Firstly lets be clear that 
$$
-2 \leq x \leq 3\\
-2 \leq x \leq 1
$$
Therefore the integral is
$$
\int_{-2}^{3} dx \int_{-2}^{1}6x - 2x^{2}y^{2} + 6y dy
$$
doing the integral over y first we find
$$
\int_{-2}^{3} \left[6xy - \frac{2}{3}x^{2}y^{3} + 3y^{2}\right]_{-2}^{1} = \int_{-2}^{3}18x -6x^{2} - 9 dx
$$
You can finish the integral right?
The result will be exactly 16. 
$\mathrm{\textbf{Edit}}$
I actually get -70 when i recheck the book keeping! 
A: $$ \int_{-2}^1 \int_{-2}^3 6x-2x^2y^2+6y\ dxdy= \int_{-2}^1 \left(6\int_{-2}^3 x\ dx-2y^2\int_{-2}^3 x^2\ dx+6y\int_{-2}^3 dx\right)dy $$
$$ = \int_{-2}^1 \left(6\frac{x^2}{2}\Bigg|_{-2}^3 -2y^2\frac{x^3}{3}\Bigg|_{-2}^3 +6yx\Bigg|_{-2}^3 \right)dy = \int_{-2}^1 \left(3(9-4) -\frac{2}{3}y^2(27+8) +6y(3+2) \right)dy $$
$$ =\int_{-2}^1 \left(15 -\frac{70}{3}y^2 +30y \right)dy= 15\int_{-2}^1 dy -\frac{70}{3} \int_{-2}^1 y^2\ dy + 30\int_{-2}^1 y\ dy $$
$$ = 15y\Bigg|_{-2}^1 -\frac{70}{9}y^3 \Bigg|_{-2}^1 + 15y^2\Bigg|_{-2}^1 =15(1+2)-\frac{70}{9}(1+8)+15(1-4) $$
$$ = 15(3)-70-15(3)=-70 $$
