1. **Problem Statement:** We are given a solid region $Q$ bounded by: - Outer sphere: $x^2 + y^2 + z^2 = 16$ - Inner sphere: $z^2 + y^2 = 4 - x^2$ - Planes: $y \geq x$, $z \geq 0$, and $z \leq 4$ We need to: (i) Derive the integral for $\iiint_Q \frac{\sqrt[3]{(x^2 + y^2 + z^2)^2}}{2} \, dV$ using spherical coordinates. (ii) Determine the volume of the solid $Q$. --- 2. **Understanding the boundaries:** - Outer sphere radius $R=4$ since $x^2 + y^2 + z^2 = 16$. - Inner sphere rewritten: $z^2 + y^2 = 4 - x^2$ implies $x^2 + y^2 + z^2 = 4$, so inner sphere radius $r=2$. - Planes restrict the region further. 3. **Spherical coordinates:** Use: $$ \begin{cases} x = \rho \sin\phi \cos\theta \\ y = \rho \sin\phi \sin\theta \\ z = \rho \cos\phi \end{cases} $$ where $\rho \geq 0$, $0 \leq \phi \leq \pi$, $0 \leq \theta < 2\pi$. Volume element: $$dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$$ 4. **Bounds for $\rho$:** Between inner and outer spheres: $$2 \leq \rho \leq 4$$ 5. **Bounds for $\phi$ and $\theta$ from planes:** - $z \geq 0 \Rightarrow \cos\phi \geq 0 \Rightarrow 0 \leq \phi \leq \frac{\pi}{2}$ (upper hemisphere) - $z \leq 4$ is automatically satisfied since $\rho \leq 4$ and $\cos\phi \leq 1$. - $y \geq x$ means $\sin\phi \sin\theta \geq \sin\phi \cos\theta$; since $\sin\phi \geq 0$ for $0 \leq \phi \leq \pi$, divide both sides by $\sin\phi$: $$\sin\theta \geq \cos\theta \Rightarrow \tan\theta \geq 1$$ This implies: $$\theta \in \left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$$ 6. **Integral for (i):** The integrand: $$\frac{\sqrt[3]{(x^2 + y^2 + z^2)^2}}{2} = \frac{\sqrt[3]{\rho^4}}{2} = \frac{\rho^{4/3}}{2}$$ Integral: $$ \iiint_Q \frac{\rho^{4/3}}{2} \, dV = \int_{\theta=\pi/4}^{5\pi/4} \int_{\phi=0}^{\pi/2} \int_{\rho=2}^4 \frac{\rho^{4/3}}{2} \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$ Simplify integrand: $$\frac{\rho^{4/3}}{2} \cdot \rho^2 = \frac{\rho^{4/3 + 2}}{2} = \frac{\rho^{10/3}}{2}$$ So: $$ \int_{\pi/4}^{5\pi/4} \int_0^{\pi/2} \int_2^4 \frac{\rho^{10/3}}{2} \sin\phi \, d\rho \, d\phi \, d\theta $$ 7. **Integral for (ii) volume:** Volume is: $$ \iiint_Q dV = \int_{\pi/4}^{5\pi/4} \int_0^{\pi/2} \int_2^4 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$ --- **Final answers:** (i) Integral for given function: $$ \boxed{\int_{\pi/4}^{5\pi/4} \int_0^{\pi/2} \int_2^4 \frac{\rho^{10/3}}{2} \sin\phi \, d\rho \, d\phi \, d\theta} $$ (ii) Volume integral: $$ \boxed{\int_{\pi/4}^{5\pi/4} \int_0^{\pi/2} \int_2^4 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta} $$