Could someone please check my work?

$\displaystyle \int \int_D (x-y) dA$

where D is bounded by $\displaystyle y=\sqrt{x}$ and $\displaystyle y=x^2$

I evaluated the double integral as

$\displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} (x-y) dy dx$

Solving the definite integral, the answer got was $\displaystyle \frac{1}{15}$.

Is this correct?