1. **Problem statement:** We need to evaluate the triple integral $$\iiint_Q \frac{\sqrt[3]{x^2 + y^2 + z^2}}{2} \, dV$$ using spherical coordinates. 2. **Recall spherical coordinates:** - $$x = \rho \sin\phi \cos\theta$$ - $$y = \rho \sin\phi \sin\theta$$ - $$z = \rho \cos\phi$$ - Volume element: $$dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$$ 3. **Rewrite the integrand:** Since $$x^2 + y^2 + z^2 = \rho^2$$, we have $$\sqrt[3]{x^2 + y^2 + z^2} = \sqrt[3]{\rho^2} = \rho^{2/3}$$. 4. **Substitute into the integral:** $$\iiint_Q \frac{\rho^{2/3}}{2} \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \iiint_Q \frac{\rho^{2/3 + 2}}{2} \sin\phi \, d\rho \, d\phi \, d\theta = \iiint_Q \frac{\rho^{8/3}}{2} \sin\phi \, d\rho \, d\phi \, d\theta$$. 5. **Set the limits of integration:** Assuming $$Q$$ is the ball of radius $$R$$ centered at the origin, - $$\rho$$ from 0 to $$R$$ - $$\phi$$ from 0 to $$\pi$$ - $$\theta$$ from 0 to $$2\pi$$ 6. **Evaluate the integral:** $$\int_0^{2\pi} d\theta = 2\pi$$ $$\int_0^{\pi} \sin\phi \, d\phi = 2$$ $$\int_0^R \frac{\rho^{8/3}}{2} \, d\rho = \frac{1}{2} \cdot \frac{\rho^{11/3}}{11/3} \Big|_0^R = \frac{3}{22} R^{11/3}$$ 7. **Combine all parts:** $$2\pi \times 2 \times \frac{3}{22} R^{11/3} = \frac{12\pi}{22} R^{11/3} = \frac{6\pi}{11} R^{11/3}$$ **Final answer:** $$\boxed{\frac{6\pi}{11} R^{11/3}}$$