This is a simple exercise on multiple integration. I am given with $$R=\left \{ (x,y) \in \mathrm{R}^2 \mid y^3 \leq x \leq y, 0 \leq y \leq \sqrt{\frac{\pi}{3}} \right \}$$ and asked to evaluate $$ \iint_R 2\sin y^2 \; dA. $$ I just wanna know if my solution is correct.
\begin{align*} \iint_R 2\sin y^2 \; dA &= \int_0^{\sqrt{\pi /3}} \int_{y^3}^y 2\sin y^2 \; dx \; dy \\ &= \int_0^{\sqrt{\pi /3}} 2y \sin y^2 dy - \int_0^{\sqrt{\pi /3}} 2y^3 \sin y^2 dy \end{align*}
For the first integral, let $u= y^2$ so that $$ \int_0^{\sqrt{\pi /3}} 2y \sin y^2 dy = -\cos y^2 \biggr \rvert_0^{\sqrt{\pi /3}}=\frac{1}{2}. $$ Applying Integration By Parts to the second integral, $$ \int_0^{\sqrt{\pi /3}} 2y^3 \sin y^2 dy = \left( \sin y^2 -y^2 \cos y^2 \right) \biggr \rvert_0^{\sqrt{\pi /3}}=\frac{\sqrt{3}}{2}-\frac{\pi}{6}. $$ Therefore, $$ \iint_R 2\sin y^2 \; dA = \frac{1-\sqrt{3}}{2}+\frac{\pi}{6}. $$
