## Calculus (3rd Edition)

The iterated integral after changing the order of integration is $\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{y^2}} \sin {y^3}{\rm{d}}x{\rm{d}}y$ Evaluate: $\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{{y^2}} \sin {y^3}{\rm{d}}x{\rm{d}}y \simeq 0.3818$ We have $\mathop \smallint \limits_0^4 \mathop \smallint \limits_{\sqrt x }^2 \sin {y^3}{\rm{d}}y{\rm{d}}x$. From the limits of the integrals we obtain the domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 4,\sqrt x \le y \le 2} \right\}$ Notice that this is a vertically simple region. Using this description we sketch the the domain of integration. Please see the figure attached. We change the order of integration such that the domain becomes a horizontally simple region. In this case, the lower and upper boundaries are $y=0$ and $y=2$, respectively. Whereas, the left boundary is $x=0$. To find the right boundary, we use $y = \sqrt x$ and get $x = {y^2}$. Thus, the new domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 2,0 \le x \le {y^2}} \right\}$ So, the iterated integral after changing the order of integration is $\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{y^2}} \sin {y^3}{\rm{d}}x{\rm{d}}y$ Next, we evaluate this integral: $\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^{{y^2}} \sin {y^3}{\rm{d}}x{\rm{d}}y$ $= \mathop \smallint \limits_{y = 0}^2 \sin {y^3}\left( {x|_0^{{y^2}}} \right){\rm{d}}y$ $= \mathop \smallint \limits_{y = 0}^2 {y^2}\sin {y^3}{\rm{d}}y$ $= - \frac{1}{3}\cos {y^3}|_0^2$ $= - \frac{1}{3}\cos 8 + \frac{1}{3}$ $\simeq 0.3818$