1. **Problem Statement:** Find the volume of the solid region $D$ bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the plane $z=1$ using spherical coordinates. 2. **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. 3. **Region boundaries:** - The cone $z=\sqrt{x^2+y^2}$ in spherical coordinates becomes: $$\rho \cos\phi = \rho \sin\phi \implies \cos\phi = \sin\phi \implies \phi = \frac{\pi}{4}$$ - The plane $z=1$ becomes: $$\rho \cos\phi = 1 \implies \rho = \frac{1}{\cos\phi}$$ 4. **Limits for $\theta$:** Since the region is symmetric around the $z$-axis, $\theta$ ranges from $0$ to $2\pi$. 5. **Limits for $\phi$:** The region is bounded below by the cone $\phi=\pi/4$ and above by the $z$-axis $\phi=0$, so: $$0 \leq \phi \leq \frac{\pi}{4}$$ 6. **Limits for $\rho$:** For fixed $\phi$, $\rho$ goes from the origin to the plane: $$0 \leq \rho \leq \frac{1}{\cos\phi}$$ 7. **Volume element in spherical coordinates:** $$dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$$ --- ### (a) Order $d\rho \, d\phi \, d\theta$: $$V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^{\frac{1}{\cos\phi}} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta$$ --- ### (b) Order $d\phi \, d\rho \, d\theta$: For fixed $\rho$, $\phi$ ranges from $0$ to $\pi/4$ but we must check if the upper bound $\rho = \frac{1}{\cos\phi}$ restricts $\phi$. Rewrite $\rho \leq \frac{1}{\cos\phi} \implies \cos\phi \leq \frac{1}{\rho}$. Since $\cos\phi$ decreases as $\phi$ increases from $0$ to $\pi/4$, for fixed $\rho$, the upper limit for $\phi$ is: $$\phi \leq \cos^{-1}\left(\frac{1}{\rho}\right)$$ But $\rho$ must be at least 1 because $\cos\phi \leq 1$, so $\rho$ ranges from 1 to $\sqrt{2}$ (since at $\phi=\pi/4$, $\rho=1/\cos(\pi/4)=\sqrt{2}$). Thus: $$V = \int_0^{2\pi} \int_1^{\sqrt{2}} \int_0^{\cos^{-1}(1/\rho)} \rho^2 \sin\phi \, d\phi \, d\rho \, d\theta$$ --- ### Final answers: (a) $$\boxed{\int_0^{2\pi} \int_0^{\pi/4} \int_0^{\frac{1}{\cos\phi}} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta}$$ (b) $$\boxed{\int_0^{2\pi} \int_1^{\sqrt{2}} \int_0^{\cos^{-1}(1/\rho)} \rho^2 \sin\phi \, d\phi \, d\rho \, d\theta}$$