1. **Problem Statement:** We are given two spheres and bounding planes: - Outer sphere: $x^2 + y^2 + z^2 = 16$ - Inner sphere: $z^2 + y^2 = 4 - x^2$ - Bounding planes: $y \geq x$, $z \geq 0$, and $z \leq 4$ We need to: (i) Derive the integral for $\iiint_Q \frac{\sqrt[3]{x^2 + y^2 + z^2}}{2} \, dV$ using spherical coordinates. (ii) Determine the volume of the solid $Q$. 2. **Spherical Coordinates Setup:** Recall spherical coordinates: $$ x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi $$ where $\rho \geq 0$ is the radius, $0 \leq \phi \leq \pi$ is the polar angle from the positive $z$-axis, and $0 \leq \theta < 2\pi$ is the azimuthal angle in the $xy$-plane from the positive $x$-axis. The volume element is: $$ dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$ 3. **Region $Q$ Description:** - Outer sphere: $\rho = 4$ (since $x^2 + y^2 + z^2 = 16$) - Inner sphere: rewrite $z^2 + y^2 = 4 - x^2$ as $x^2 + y^2 + z^2 = 4$ which is a sphere of radius 2. So the inner sphere is $\rho = 2$. The solid $Q$ lies between the spheres of radius 2 and 4. 4. **Bounding planes in spherical coordinates:** - $y \geq x$ means $\sin\phi \sin\theta \geq \sin\phi \cos\theta$ or $\sin\theta \geq \cos\theta$. Dividing by $\cos\theta$ (assuming $\cos\theta > 0$), we get $\tan\theta \geq 1$ so $\theta \geq \pi/4$. - $z \geq 0$ means $\rho \cos\phi \geq 0$ so $0 \leq \phi \leq \pi/2$. - $z \leq 4$ is automatically satisfied since $\rho \leq 4$. 5. **Limits of integration:** - $\rho$: from 2 to 4 (between inner and outer spheres) - $\phi$: from 0 to $\pi/2$ (upper hemisphere) - $\theta$: from $\pi/4$ to $\pi/2$ (due to $y \geq x$ and first octant) 6. **Integral for (i):** The integrand is: $$ \frac{\sqrt[3]{x^2 + y^2 + z^2}}{2} = \frac{\sqrt[3]{\rho^2}}{2} = \frac{\rho^{2/3}}{2} $$ The integral is: $$ \iiint_Q \frac{\sqrt[3]{x^2 + y^2 + z^2}}{2} \, dV = \int_{\theta=\pi/4}^{\pi/2} \int_{\phi=0}^{\pi/2} \int_{\rho=2}^4 \frac{\rho^{2/3}}{2} \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$ Simplify the integrand: $$ \frac{\rho^{2/3}}{2} \cdot \rho^2 = \frac{\rho^{2/3 + 2}}{2} = \frac{\rho^{8/3}}{2} $$ So the integral becomes: $$ \int_{\pi/4}^{\pi/2} \int_0^{\pi/2} \int_2^4 \frac{\rho^{8/3}}{2} \sin\phi \, d\rho \, d\phi \, d\theta $$ 7. **Integral for (ii) Volume:** Volume is: $$ \iiint_Q dV = \int_{\pi/4}^{\pi/2} \int_0^{\pi/2} \int_2^4 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$ 8. **Evaluate the integrals:** (i) Integral for the function: - Integrate w.r.t $\rho$: $$ \int_2^4 \frac{\rho^{8/3}}{2} d\rho = \frac{1}{2} \int_2^4 \rho^{8/3} d\rho = \frac{1}{2} \left[ \frac{\rho^{11/3}}{11/3} \right]_2^4 = \frac{3}{22} \left(4^{11/3} - 2^{11/3}\right) $$ - Integrate w.r.t $\phi$: $$ \int_0^{\pi/2} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/2} = 1 $$ - Integrate w.r.t $\theta$: $$ \int_{\pi/4}^{\pi/2} d\theta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $$ So the value is: $$ \frac{3}{22} \left(4^{11/3} - 2^{11/3}\right) \times 1 \times \frac{\pi}{4} = \frac{3\pi}{88} \left(4^{11/3} - 2^{11/3}\right) $$ (ii) Volume integral: - Integrate w.r.t $\rho$: $$ \int_2^4 \rho^2 d\rho = \left[ \frac{\rho^3}{3} \right]_2^4 = \frac{64}{3} - \frac{8}{3} = \frac{56}{3} $$ - Integrate w.r.t $\phi$: $$ \int_0^{\pi/2} \sin\phi \, d\phi = 1 $$ - Integrate w.r.t $\theta$: $$ \int_{\pi/4}^{\pi/2} d\theta = \frac{\pi}{4} $$ So volume is: $$ \frac{56}{3} \times 1 \times \frac{\pi}{4} = \frac{14\pi}{3} $$ **Final answers:** (i) $$\iiint_Q \frac{\sqrt[3]{x^2 + y^2 + z^2}}{2} dV = \frac{3\pi}{88} \left(4^{11/3} - 2^{11/3}\right)$$ (ii) $$\text{Volume}(Q) = \frac{14\pi}{3}$$