1. **State the problem:** We need to evaluate the double integral $$\int_{-2}^{2} \int_{x^2}^{2} \cos\left(\sqrt{y^3}\right) \, dy \, dx.$$ This integral is over the region where $x$ ranges from $-2$ to $2$ and $y$ ranges from $x^2$ to $2$. 2. **Understand the integral:** The inner integral is with respect to $y$, and the outer integral is with respect to $x$. The function to integrate is $\cos\left(\sqrt{y^3}\right)$. 3. **Consider changing the order of integration:** The region of integration is bounded by $y = x^2$ and $y = 2$, with $x$ between $-2$ and $2$. Since $y$ goes from $x^2$ to $2$, and $x$ from $-2$ to $2$, we can describe the region in terms of $y$ first: - $y$ ranges from $0$ (since $x^2 \geq 0$) to $2$. - For each fixed $y$, $x$ ranges between $-\sqrt{y}$ and $\sqrt{y}$. 4. **Rewrite the integral with reversed order:** $$\int_{y=0}^{2} \int_{x=-\sqrt{y}}^{\sqrt{y}} \cos\left(\sqrt{y^3}\right) \, dx \, dy.$$ 5. **Evaluate the inner integral with respect to $x$:** Since the integrand does not depend on $x$, the inner integral is: $$\int_{-\sqrt{y}}^{\sqrt{y}} \cos\left(\sqrt{y^3}\right) \, dx = \cos\left(\sqrt{y^3}\right) \times \left(\sqrt{y} - (-\sqrt{y})\right) = 2\sqrt{y} \cos\left(\sqrt{y^3}\right).$$ 6. **Substitute back into the outer integral:** $$\int_0^2 2\sqrt{y} \cos\left(\sqrt{y^3}\right) \, dy.$$ 7. **Make a substitution to solve the integral:** Let $$t = \sqrt{y^3} = y^{3/2}.$$ Then, $$t = y^{3/2} \implies dt = \frac{3}{2} y^{1/2} dy = \frac{3}{2} \sqrt{y} dy.$$ Rearranged: $$\sqrt{y} dy = \frac{2}{3} dt.$$ 8. **Change the limits for $t$:** When $y=0$, $t=0^{3/2}=0$. When $y=2$, $t=2^{3/2} = 2^{1.5} = 2 \times \sqrt{2} = 2\sqrt{2}.$ 9. **Rewrite the integral in terms of $t$:** $$\int_0^{2} 2 \sqrt{y} \cos(t) \, dy = 2 \int_0^{2} \sqrt{y} \cos(t) \, dy = 2 \int_0^{2\sqrt{2}} \cos(t) \times \frac{2}{3} dt = \frac{4}{3} \int_0^{2\sqrt{2}} \cos(t) \, dt.$$ 10. **Integrate with respect to $t$:** $$\int_0^{2\sqrt{2}} \cos(t) \, dt = \sin(t) \Big|_0^{2\sqrt{2}} = \sin(2\sqrt{2}) - \sin(0) = \sin(2\sqrt{2}).$$ 11. **Final answer:** $$\frac{4}{3} \sin(2\sqrt{2}).$$