1. **State the problem:** We have parametric equations $x(t) = 2 - 4 \sin t$ and $y(t) = 4 \cos t$ for $t$ in $[0, \frac{\pi}{2}]$. We want to find the volume generated when the curve is rotated around the y-axis. 2. **Formula used:** The volume $V$ of a solid of revolution around the y-axis for parametric curves is given by $$V = 2\pi \int_a^b x(t) y'(t) \, dt$$ where $y'(t) = \frac{dy}{dt}$. 3. **Calculate $y'(t)$:** $$y'(t) = \frac{d}{dt} (4 \cos t) = -4 \sin t$$ 4. **Set up the integral:** $$V = 2\pi \int_0^{\frac{\pi}{2}} (2 - 4 \sin t)(-4 \sin t) \, dt$$ 5. **Expand the integrand:** $$= 2\pi \int_0^{\frac{\pi}{2}} (-8 \sin t + 16 \sin^2 t) \, dt$$ 6. **Split the integral:** $$= 2\pi \left( \int_0^{\frac{\pi}{2}} -8 \sin t \, dt + \int_0^{\frac{\pi}{2}} 16 \sin^2 t \, dt \right)$$ 7. **Integrate each term:** - For $\int_0^{\frac{\pi}{2}} -8 \sin t \, dt$: $$= -8 [-\cos t]_0^{\frac{\pi}{2}} = -8 (-\cos \frac{\pi}{2} + \cos 0) = -8 (0 + 1) = -8$$ - For $\int_0^{\frac{\pi}{2}} 16 \sin^2 t \, dt$ use the identity $\sin^2 t = \frac{1 - \cos 2t}{2}$: $$= 16 \int_0^{\frac{\pi}{2}} \frac{1 - \cos 2t}{2} \, dt = 8 \int_0^{\frac{\pi}{2}} (1 - \cos 2t) \, dt$$ $$= 8 \left[ t - \frac{\sin 2t}{2} \right]_0^{\frac{\pi}{2}} = 8 \left( \frac{\pi}{2} - 0 \right) = 4\pi$$ 8. **Sum the integrals:** $$= 2\pi (-8 + 4\pi) = 2\pi (4\pi - 8) = 8\pi^2 - 16\pi$$ 9. **Final answer:** $$\boxed{8\pi^2 - 16\pi}$$ This is the volume of the solid formed by rotating the curve around the y-axis.