Problem
Calculate $$\int_R \int \nabla^2 \big( x^3 - y^3 \big) ~dx~ dy$$ in the region bounded by $0 \leq y \leq x^2$ and $|x|\leq 2$.[1]
My Attempt
$$\nabla \big( x^3 - y^3 \big) = 3x^2 \hat{\bf{i}} - 3y^2 \hat{\bf{j}}$$ $$\nabla^2 \big( x^3 - y^3 \big) = 6x - 6y$$ $$ \int_{-2}^2 \Biggl[\int_0^{x^2} 6x - 6y ~dy \Biggl]dx$$ $$ \int_{-2}^2 \Biggl[ 6xy - 3y^2 \Bigg|^{x^2}_0 \Biggl]dx$$ $$ \int_{-2}^2 6x^3 - 3x^2 ~dx$$ $$\tfrac{3}{2}x^4 - x^3\Bigg|_{-2}^2$$ $$ - (2)^3--(-2)^3$$ $$ - 8-8$$ $$ -16$$
So this looks pretty good to me; I can't find any problem. However, the answer in the back of the book says $-38.4$. Where is my mistake?
[1] From Kreyszig 10th edition problem 10.4.17
You have a minor error in your calculation
$$\dots=\int_{-2}^{2}\left[6xy-3y^{2} \Bigg|_{y=0}^{y=x^{2}}\right]{\rm d}x=\int_{-2}^{2}\left(6x^{3}-3x^{\color{red}{4}}\right){\rm d}x=\frac{3}{2}x^{4}-\color{red}{\frac{3}{5}x^{5}}\Bigg|_{-2}^{2}=\color{red}{38.4}$$