Let's solve **Question 1** step by step: --- ### **Step 1: Understand the Surface \( S \)** The conical surface is defined as: \[ S = \{(x, y, z) \in \mathbb{R}^3 \mid z = \sqrt{x^2 + y^2},\ z \leq 1\} \] This is a cone opening upwards, with height going from \( 0 \) to \( 1 \). --- ### **Step 2: Find the Parametrization** Let’s use polar coordinates: - Let \( x = r \cos\theta \) - Let \( y = r \sin\theta \) - Since \( z = \sqrt{x^2 + y^2} = r \), so \( z = r \) The bounds: - \( r \) goes from \( 0 \) to \( 1 \) (since \( z \leq 1 \)), - \( \theta \) goes from \( 0 \) to \( 2\pi \). So, the parametrization is: \[ \vec{r}(r, \theta) = (r \cos\theta, r \sin\theta, r),\quad 0 \leq r \leq 1,\, 0 \leq \theta < 2\pi \] Now, let’s match this form to the given options: --- ### **Step 3: Check Each Option** #### (a) \(\vec{r}_1(u, v) = (u \cos(v), u \sin(v), u)\) - \(u\) plays the role of \(r\), \(v\) is \(\theta\). - Bounds: \(0 \leq u \leq 1\), \(0 \leq v < 2\pi\). - **This matches exactly.** - **Correct!** #### (b) \(\vec{r}_2(u, v) = (u \cos(v), u \sin(v), u^2)\) - Here, \(z = u^2\), but \(z\) should equal \( \sqrt{x^2 + y^2} = u \). - **Incorrect.** #### (c) \(\vec{r}_3(u, v) = (u^2 \cos(v), u^2 \sin(v), u)\) - \(x = u^2 \cos v\), \(y = u^2 \sin v\), \(z = u\). - \(z = \sqrt{x^2 + y^2} = \sqrt{u^4} = |u^2|\), so \(z = u\) only if \(u \geq 0\). - But for \(0 \leq u \leq 1\), this is valid. - But, \(x^2 + y^2 = u^4\), so \(z = \sqrt{u^4} = u^2\) which does not match \(z = u\). - **Incorrect.** #### (d) \(\vec{r}_4(u, v) = ((1-u) \cos(v), (1-u)\sin(v), u)\) - \(x = (1-u)\cos v\), \(y = (1-u)\sin v\), \(z=u\). - \(z = u\), \(x^2 + y^2 = (1-u)^2\), so \(z = \sqrt{x^2 + y^2}\) gives \(u = |1-u|\), only holds for \(u=0\) or \(u=1/2\). - **Incorrect.** #### (e) \(\vec{r}_5(u,v) = (u, v, \sqrt{u^2 + v^2})\) - \(x = u\), \(y = v\), \(z = \sqrt{u^2 + v^2}\). - This is a valid parametrization for the cone. - But check the bounds: \(0 \leq \sqrt{u^2 + v^2} \leq 1\) or equivalently \(u^2 + v^2 \leq 1\). - The given bounds: \( -1 \leq u \leq 1 \) and \( -\sqrt{1-u^2} \leq v \leq \sqrt{1-u^2} \), which covers the disk \(u^2 + v^2 \leq 1\). - **Correct!** #### (f) \(\vec{r}_6(u,v) = (u, v, u^2 + v^2)\) - \(z = u^2 + v^2\), but the cone has \(z = \sqrt{u^2 + v^2}\). - **Incorrect.** --- ### **Step 4: Final Answers** The correct parametrizations are: **(a) \(\vec{r}_1(u,v) = (u\cos(v), u\sin(v), u)\)** **(e) \(\vec{r}_5(u,v) = (u, v, \sqrt{u^2 + v^2})\)** --- ## **Final Answer** \[ \boxed{a \text{ and } e} \] are the correct parametrizations of the conical surface.

Certainly! Let's proceed step-by-step with detailed calculations for questions 2, 3, 4, and 5. --- ### **Question 2:** Evaluate the upward flux of \(\mathbf{F} = (-z, -2y, 1 + 3z)\) across the conical surface \(S\) defined by \(z = \sqrt{x^2 + y^2}\), with \(z \leq 1\). --- ### **Step 1: Surface and Parametrization** - The surface is a cone: \(z = r\) in cylindrical coordinates, with \(0 \leq r \leq 1\), \(\theta \in [0, 2\pi)\). - Parametrize \(S\): \[ \mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r) \] - The surface element \(d\mathbf{S}\) points outward (upward), with the normal vector. --- ### **Step 2: Compute \(\mathbf{r}_r\) and \(\mathbf{r}_\theta\)** \[ \mathbf{r}_r = \frac{\partial \mathbf{r}}{\partial r} = (\cos \theta, \sin \theta, 1) \] \[ \mathbf{r}_\theta = \frac{\partial \mathbf{r}}{\partial \theta} = (-r \sin \theta, r \cos \theta, 0) \] ### **Step 3: Compute the surface element \(d\mathbf{S}\)** \[ d\mathbf{S} = \mathbf{r}_r \times \mathbf{r}_\theta \, dr\, d\theta \] Calculate the cross product: \[ \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos \theta & \sin \theta & 1 \\ - r \sin \theta & r \cos \theta & 0 \end{vmatrix} \] Compute components: - \(i\)-component: \[ (\sin \theta)(0) - (1)(r \cos \theta) = - r \cos \theta \] - \(j\)-component: \[ - [(\cos \theta)(0) - (1)(- r \sin \theta)] = - (0 + r \sin \theta) = - r \sin \theta \] - \(k\)-component: \[ (\cos \theta)(r \cos \theta) - (\sin \theta)(- r \sin \theta) = r \cos^2 \theta + r \sin^2 \theta = r (\cos^2 \theta + \sin^2 \theta) = r \] Thus, \[ \mathbf{r}_r \times \mathbf{r}_\theta = (- r \cos \theta, - r \sin \theta, r) \] --- ### **Step 4: Setup the surface integral** The flux: \[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \int_{0}^{2\pi} \int_{0}^{1} \mathbf{F}(\mathbf{r}(r,\theta)) \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) \, dr\, d\theta \] Evaluate \(\mathbf{F}\) at \(\mathbf{r}(r,\theta)\): \[ x = r \cos \theta, \quad y = r \sin \theta, \quad z = r \] \[ \mathbf{F} = (-z, -2 y, 1 + 3 z) = (-r, -2 r \sin \theta, 1 + 3 r) \] Dot product: \[ \mathbf{F} \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) = (-r, -2 r \sin \theta, 1 + 3 r) \cdot (- r \cos \theta, - r \sin \theta, r) \] Calculate: \[ = (-r)(- r \cos \theta) + (-2 r \sin \theta)(- r \sin \theta) + (1 + 3 r)(r) \] \[ = r^2 \cos \theta + 2 r^2 \sin^2 \theta + r + 3 r^2 \] --- ### **Step 5: Integrate over \(r\) and \(\theta\)** The flux: \[ \Phi = \int_{0}^{2\pi} \int_{0}^{1} \left[ r^2 \cos \theta + 2 r^2 \sin^2 \theta + r + 3 r^2 \right] dr\, d\theta \] Split into parts: \[ \Phi = \int_{0}^{2\pi} \left[ \int_{0}^{1} r^2 \cos \theta \, dr + \int_{0}^{1} 2 r^2 \sin^2 \theta \, dr + \int_{0}^{1} r \, dr + \int_{0}^{1} 3 r^2 \, dr \right] d\theta \] Note that \(\cos \theta\) and \(\sin^2 \theta\) are integrated over \(\theta\): - \(\int_0^{2\pi} \cos \theta \, d\theta = 0\) - \(\int_0^{2\pi} \sin^2 \theta \, d\theta = \pi\) (since \(\sin^2 \theta\) averages to \(\pi\) over \(0\) to \(2\pi\), more precisely, \(\int_0^{2\pi} \sin^2 \theta \, d\theta = \pi\)) Calculate the \(r\)-integrals: - \(\int_0^1 r^2 dr = \frac{1}{3}\) - \(\int_0^1 r dr = \frac{1}{2}\) - \(\int_0^1 3 r^2 dr = 3 \times \frac{1}{3} = 1\) Now, the flux becomes: \[ \Phi = \left[ 0 + 2 \times \frac{1}{3} \times \pi + \frac{1}{2} \times 2\pi + 1 \right] \] Breaking down: - First term: \(0\) - Second term: \(2 \times \frac{1}{3} \times \pi = \frac{2}{3} \pi\) - Third term: \(\frac{1}{2} \times 2\pi = \pi\) - Fourth term: \(1\) Total: \[ \boxed{ \Phi = \frac{2}{3} \pi + \pi + 1 = \left(\frac{2}{3} + 1\right) \pi + 1 = \frac{5}{3} \pi + 1 } \] --- ### **Final answer for Question 2:** \[ \boxed{ \boxed{\text{Flux} \approx \frac{5}{3} \pi + 1 \approx 6.24} } \] (Rounding to **nearest hundredth**: approximately **6.24**) --- ## **Question 3:** Evaluate the outward flux of \(\mathbf{F} = \langle 2 r, y, z \rangle\) through the part of the hyperboloid \(z = \sqrt{x^2 + y^2}\) between \(z = \sqrt{3}\) and \(z = \sqrt{v3}\). --- ### **Step 1: Recognize the surface** - The surface is the hyperboloid \(z = \sqrt{x^2 + y^2}\) (a cone), between \(z = \sqrt{3}\) and \(z = \sqrt{3}\). - Note: The bounds specify \(z\) from \(\sqrt{3}\) to \(\sqrt{3}\), which suggests a typo; likely, the bounds are from \(z = \sqrt{3}\) to \(z = \sqrt{v3}\). Assuming the bounds are from \(z=\sqrt{3}\) to \(z=\sqrt{3}\): - Let's interpret as between \(z = \sqrt{3}\) and \(z = \sqrt{v3}\). - The surface is a double cone, with radius \(r = z\). --- ### **Step 2: Parametrize** Use cylindrical coordinates: \[ x = r \cos \theta,\quad y = r \sin \theta,\quad z = r \] - For \(z\) from \(\sqrt{3}\) to \(\sqrt{v3}\), \(r\) varies accordingly. - The surface is the cone \(z = r\), with \(r\) between \(\sqrt{3}\) and \(\sqrt{v3}\). --- ### **Step 3: Surface element** - The surface element: \[ d\mathbf{S} = \mathbf{r}_r \times \mathbf{r}_\theta \] - Parametrize: \[ \mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r) \] - Compute derivatives: \[ \mathbf{r}_r = (\cos \theta, \sin \theta, 1) \] \[ \mathbf{r}_\theta = (-r \sin \theta, r \cos \theta, 0) \] - Cross product: \[ \mathbf{r}_r \times \mathbf{r}_\theta = (- r \cos \theta, - r \sin \theta, r) \] --- ### **Step 4: \(\mathbf{F}\) at \(\mathbf{r}\):** \[ \mathbf{F} = \langle 2 r, y, z \rangle = \langle 2 r, r \sin \theta, r \rangle \] - Dot product: \[ \mathbf{F} \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) = (2 r, r \sin \theta, r) \cdot (- r \cos \theta, - r \sin \theta, r) \] Calculate: \[ = 2 r (- r \cos \theta) + r \sin \theta (- r \sin \theta) + r (r) \] \[ = -2 r^2 \cos \theta - r^2 \sin^2 \theta + r^2 \] --- ### **Step 5: Integrate over \(r\) and \(\theta\)** Flux: \[ \Phi = \int_{0}^{2\pi} \int_{r_1}^{r_2} \left( -2 r^2 \cos \theta - r^2 \sin^2 \theta + r^2 \right) dr\, d\theta \] with bounds: \[ r_1 = \sqrt{3}, \quad r_2 = \sqrt{v3} \] --- ### **Step 6: Integrate over \(\theta\)** - \(\int_0^{2\pi} \cos \theta \, d\theta = 0\) - \(\int_0^{2\pi} \sin^2 \theta \, d\theta = \pi\) Express flux: \[ \Phi = \int_{r_1}^{r_2} \left[ -2 r^2 \times 0 - r^2 \times \pi + r^2 \times 2\pi \right] dr \] Wait, correction: The integral over \(\theta\) of the entire integrand: \[ \int_0^{2\pi} \left( -2 r^2 \cos \theta - r^2 \sin^2 \theta + r^2 \right) d\theta \] - \(\int_0^{2\pi} -2 r^2 \cos \theta \, d\theta = 0\) - \(\int_0^{2\pi} - r^2 \sin^2 \theta \, d\theta = - r^2 \pi\) - \(\int_0^{2\pi} r^2 \, d\theta = 2\pi r^2\) Total: \[ \int_0^{2\pi} (\cdots) d\theta = 0 - r^2 \pi + 2 \pi r^2 = r^2 \pi \] --- ### **Step 7: Final integral over \(r\)** \[ \Phi = \int_{r_1}^{r_2} r^2 \pi dr = \pi \int_{\sqrt{3}}^{\sqrt{v3}} r^2 dr \] Compute: \[ \int r^2 dr = \frac{r^3}{3} \] Evaluate: \[ \Phi = \pi \left[\frac{r^3}{3}\right]_{r=\sqrt{3}}^{r=\sqrt{v3}} = \frac{\pi}{3} \left[ (\sqrt{v3})^3 - (\sqrt{3})^3 \right] \] Simplify: \[ (\sqrt{v3})^3 = (v3)^{3/2} = (v3) \times \sqrt{v3} = 3 \sqrt

Certainly! Here are the detailed versions of questions 3, 4, and 5, along with their solutions: --- ### **Question 3:** **Evaluate the outward flux of** \(\mathbf{F} = \langle 2r, y, z \rangle\) **through the part of the hyperboloid** \(z = \sqrt{x^2 + y^2}\) **between** \(z = \sqrt{3}\) **and** \(z = \sqrt{v3}\). **Round your answer to the nearest tenth.** --- ### **Solution to Question 3:** **Step 1: Surface parametrization** - The surface is the cone \(z = \sqrt{x^2 + y^2}\), which in cylindrical coordinates is \(z = r\), with \(r \ge 0\). - The bounds for \(z\) are from \(z = \sqrt{3}\) to \(z = \sqrt{v3}\), so \(r\) varies between these bounds: \[ r := z, \quad r \in [\sqrt{3}, \sqrt{v3}] \] - The surface can be parametrized as: \[ \mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r), \quad 0 \leq \theta < 2\pi, \quad r \in [\sqrt{3}, \sqrt{v3}] \] --- **Step 2: Compute the surface element** - Derivatives: \[ \mathbf{r}_r = (\cos \theta, \sin \theta, 1) \] \[ \mathbf{r}_\theta = (- r \sin \theta, r \cos \theta, 0) \] - Cross product: \[ \mathbf{r}_r \times \mathbf{r}_\theta = (- r \cos \theta, - r \sin \theta, r) \] - Differential surface element: \[ d\mathbf{S} = (\mathbf{r}_r \times \mathbf{r}_\theta) dr d\theta \] --- **Step 3: Compute \(\mathbf{F}\) at \(\mathbf{r}(r, \theta)\):** \[ \mathbf{F} = \langle 2r, y, z \rangle = \langle 2r, r \sin \theta, r \rangle \] --- **Step 4: Compute flux integral** \[ \text{Flux} = \iint_{S} \mathbf{F} \cdot d\mathbf{S} = \int_0^{2\pi} \int_{r=\sqrt{3}}^{\sqrt{v3}} \mathbf{F} \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) dr d\theta \] Dot product: \[ \mathbf{F} \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) = (2r, r \sin \theta, r) \cdot (- r \cos \theta, - r \sin \theta, r) \] Calculate: \[ = 2r \times (- r \cos \theta) + r \sin \theta \times (- r \sin \theta) + r \times r \] \[ = -2 r^2 \cos \theta - r^2 \sin^2 \theta + r^2 \] --- **Step 5: Integrate over \(\theta\):** \[ \int_0^{2\pi} (-2 r^2 \cos \theta) d\theta = 0 \] \[ \int_0^{2\pi} (- r^2 \sin^2 \theta) d\theta = - r^2 \pi \] \[ \int_0^{2\pi} r^2 d\theta = 2 \pi r^2 \] Sum: \[ \int_0^{2\pi} \mathbf{F} \cdot (\mathbf{r}_r \times \mathbf{r}_\theta) d\theta = 0 - r^2 \pi + 2 \pi r^2 = r^2 \pi \] --- **Step 6: Integrate over \(r\):** \[ \text{Flux} = \int_{r=\sqrt{3}}^{\sqrt{v3}} r^2 \pi dr = \pi \left[ \frac{r^3}{3} \right]_{r=\sqrt{3}}^{r=\sqrt{v3}} \] Calculate: \[ r^3 = (\sqrt{z})^3 = z^{3/2} \] So: \[ \text{Flux} = \frac{\pi}{3} \left[ (\sqrt{v3})^3 - (\sqrt{3})^3 \right] \] \[ (\sqrt{v3})^3 = (v3)^{3/2} = (3 v)^{3/2} = 3^{3/2} v^{3/2} \] \[ (\sqrt{3})^3 = 3^{3/2} \] Therefore: \[ \text{Flux} = \frac{\pi}{3} \times 3^{3/2} \left( v^{3/2} - 1 \right) \] Since \(3^{3/2} = 3 \sqrt{3}\), \[ \boxed{ \text{Flux} = \frac{\pi}{3} \times 3 \sqrt{3} \left( v^{3/2} - 1 \right) = \pi \sqrt{3} (v^{3/2} - 1) } \] --- ### **Final answer:** \[ \boxed{ \text{Flux} = \pi \sqrt{3} \left( v^{3/2} - 1 \right) } \] **Numerical approximation** (for specific \(v\) value if given). --- ### **Question 4:** **Evaluate the surface integral** \[ \iint_S \mathbf{F} \cdot d\mathbf{S} \] **where** \[ \mathbf{F} = (y z, z x, x y) \] **over the surface** \[ H = \left\{ (x, y, z) \mid z = 5 - 27 - 17 \right\} \] **with an orientation where the normal vectors point away from the origin. Round to the nearest hundredth.** --- ### **Note:** *Please clarify the surface and its bounds for a precise solution; the provided description seems incomplete or contains typo. Once clarified, the solution involves parametrizing the surface, computing the flux via the surface integral, and evaluating accordingly.* --- ### **Question 5:** **Calculate the electric flux** of a uniform electric field **\( \mathbf{E} \)** **across the cylindrical surface** \[ S: \quad (r, \theta, z), \quad r = 1/4, \quad 0 \leq z \leq 1/2 \] **where** \[ \mathbf{E} = \frac{155}{r} \hat{r} \quad \text{(or as described)}, \] **oriented outward (normal vectors pointing away from the axis). Assume all measurements in meters. Round to the nearest thousandth.** --- ### **Note:** *Please specify the exact form of \(\mathbf{E}\) and the orientation for a precise calculation.* --- **Summary:** - For questions 3, 4, and 5, please clarify the surface descriptions, bounds, and vector fields if needed. - The above solutions for question 3 demonstrate the detailed approach to flux calculation. --- Let me know if you'd like me to proceed with detailed solutions for questions 4 and 5 once the clarifications are provided!