Let's solve the problem step by step as requested. --- ### **Given:** - \(\vec{F} = -y\,\hat{i} + x\,\hat{j} - 2\,\hat{k}\) - Surface \(S\): \(z^2 = x^2 + y^2,\, 0 \leq z \leq 4\) (a cone from \(z=0\) to \(z=4\)) - \(C\) is the closed boundary curve of the surface \(S\) - The normal vector is oriented **downward** - Evaluate the line integral \(\oint_C \vec{F} \cdot d\vec{r}\) - By (a) directly, (b) using Stokes' Theorem --- ## **Step 1: Understand the Curve \(C\)** - The cone \(z^2 = x^2 + y^2\), for \(0 \leq z \leq 4\), is a surface that starts at the origin and ends at the circle \(z=4\). - The **boundary** of this surface is the circle at \(z=4\): \(x^2 + y^2 = 16\), \(z=4\). So, \(C\): circle of radius 4 in the plane \(z=4\). --- ## **(a) Direct Evaluation** We parameterize \(C\): Let \(x = 4\cos\theta,\; y = 4\sin\theta,\; z = 4\), \(\theta \in [0, 2\pi]\) Compute \(d\vec{r}\): \[ d\vec{r} = \frac{d\vec{r}}{d\theta} d\theta = (-4\sin\theta\,\hat{i} + 4\cos\theta\,\hat{j} + 0\,\hat{k}) d\theta \] At each point on \(C\): \[ \vec{F} = -y\,\hat{i} + x\,\hat{j} - 2\,\hat{k} = -4\sin\theta\,\hat{i} + 4\cos\theta\,\hat{j} - 2\,\hat{k} \] Now, the dot product: \[ \vec{F} \cdot d\vec{r} = [-4\sin\theta\,\hat{i} + 4\cos\theta\,\hat{j} - 2\,\hat{k}] \cdot [-4\sin\theta\,\hat{i} + 4\cos\theta\,\hat{j}] \] (Only \(i\) and \(j\) terms matter; \(k\) term vanishes because \(d\vec{r}_k = 0\)). So, \[ \vec{F} \cdot d\vec{r} = (-4\sin\theta)(-4\sin\theta) + (4\cos\theta)(4\cos\theta) = 16\sin^2\theta + 16\cos^2\theta = 16(\sin^2\theta + \cos^2\theta) = 16 \] So the integral becomes: \[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} 16\, d\theta = 16 \cdot 2\pi = \boxed{32\pi} \] --- ## **(b) Using Stokes' Theorem** Stokes' Theorem: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \] ### **Step 1: Compute \(\nabla \times \vec{F}\):** \[ \vec{F} = -y\,\hat{i} + x\,\hat{j} - 2\,\hat{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ -y & x & -2 \end{vmatrix} \] Calculating: - \(i\) component: \( \frac{\partial (-2)}{\partial y} - \frac{\partial x}{\partial z} = 0 - 0 = 0 \) - \(j\) component: \( -\left( \frac{\partial (-2)}{\partial x} - \frac{\partial (-y)}{\partial z} \right) = - (0 - 0) = 0 \) - \(k\) component: \( \frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y} = 1 - (-1) = 2 \) So, \[ \nabla \times \vec{F} = 0\,\hat{i} + 0\,\hat{j} + 2\,\hat{k} \] --- ### **Step 2: Choose the Surface \(S\)** Since the boundary is the circle at \(z=4\), and the normal is oriented **downward**, we should take the surface as the disk \(x^2 + y^2 \leq 16\) at \(z=4\), with normal pointing in the \(-z\) direction. Let’s parameterize the disk: - \(x = r\cos\theta, y = r\sin\theta, z = 4\) - \(0 \leq r \leq 4\), \(0 \leq \theta < 2\pi\) The area element (downward): \[ d\vec{S} = -\hat{k}\, dA = -\hat{k}\, r\,dr\,d\theta \] --- ### **Step 3: Calculate the Surface Integral** \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S (2\,\hat{k}) \cdot (-\hat{k})\, r\,dr\,d\theta = \iint_S -2\, r\,dr\,d\theta \] Integrate over the disk: \[ \int_0^{2\pi} \int_0^4 -2r\, dr\, d\theta = \int_0^{2\pi} \left[ -2 \cdot \frac{1}{2} r^2 \Big|_{0}^{4} \right] d\theta = \int_0^{2\pi} [ - (4^2 - 0)] d\theta = \int_0^{2\pi} (-16) d\theta = -16 \cdot 2\pi = -32\pi \] But **wait:** In the direct calculation we got \(+32\pi\). This sign difference is **due to orientation**. Recall: **Stokes' theorem** uses the right-hand rule: The positive orientation of \(C\) is such that, if you walk along \(C\) in the positive direction with your head pointing in the direction of the normal (\(-z\)), the surface is always on your left. But in our parameterization in (a), as \(\theta\) increases from 0 to \(2\pi\), the circle is traversed **counterclockwise** when viewed from above (i.e., in \(+\hat{k}\) direction), but our normal is \(-\hat{k}\). So, **the orientation of \(C\) in (a) matches the opposite of the Stokes' theorem orientation.** Thus, **the sign matches** (the surface normal is downward, so the positive orientation is clockwise when viewed from above). In (a), we actually computed the counterclockwise (upward normal) integral; for downward, it should be negative. **Final answer:** \[ \boxed{\oint_C \vec{F} \cdot d\vec{r} = -32\pi} \] --- ## **Summary Table** | Method | Answer | |------------------------|-----------| | (a) Direct | \(-32\pi\)| | (b) Stokes' Theorem | \(-32\pi\)| --- ### **Explanation of Sign** - If you use the parameterization \(x = 4\cos\theta, y=4\sin\theta, z=4, 0 \leq \theta < 2\pi\), you traverse counterclockwise viewed from above, which corresponds to the **upward normal**. - For the downward normal, you need to traverse **clockwise** (\(\theta\) from \(2\pi\) to \(0\)), giving a negative sign. - Thus, the correct answer for the **downward normal** is \(\boxed{-32\pi}\). --- ## **Final boxed answer:** \[ \boxed{ \oint_C \vec{F} \cdot d\vec{r} = -32\pi } \]