Let's solve each part step by step using **Stokes' Theorem**. --- # Step-by-Step Solution ## **Stokes' Theorem Statement** Stokes' Theorem relates a surface integral of the curl of a vector field over a surface \( S \) to a line integral of the vector field over the boundary curve \( C \) of the surface: \[ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \] where: - \( S \) is the surface, - \( C = \partial S \) is the boundary curve of \( S \) (with positive orientation: counterclockwise when viewed from above). --- ## **1. \( \vec{F} = \langle x, y, z \rangle \) and \( S \) is upper half of ellipsoid** ### **Given:** - Surface \( S \) is the upper half of the ellipsoid: \(\frac{x^2}{16} + \frac{y^2}{25} + z^2 = 1\) - \( \vec{F} = \langle x, y, z \rangle \) ### **Boundary Curve \( C \):** At the "equator" (\( z = 0 \)): \[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \] This is an ellipse in the \( xy \)-plane. A parameterization of \( C \): \[ \vec{r}(t) = \langle 4\cos t, 5\sin t, 0 \rangle, \quad 0 \leq t \leq 2\pi \] ### **Stokes' Theorem:** \[ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \] ### **Parameterization for Line Integral:** - \( \vec{F}(\vec{r}(t)) = \langle 4\cos t, 5\sin t, 0 \rangle \) - \( \vec{r}'(t) = \langle -4\sin t, 5\cos t, 0 \rangle \) So the line integral becomes: \[ \oint_{C} \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt = \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle dt \] **Final Answer for (1):** \[ \boxed{ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle \, dt } \] --- ## **2. \( \vec{F} = \langle x+y, y+z, z+x \rangle \), tilted disk** ### **Given:** - \( \vec{F} = \langle x+y, y+z, z+x \rangle \) - \( \vec{r}(t) = \langle 3\cos t, 7\sin t, \sqrt{40}\cos t \rangle \), \( 0 \leq t \leq 2\pi \) ### **Boundary Curve \( C \):** Parameterization is already given. ### **Stokes' Theorem:** \[ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \] ### **Parameterization for Line Integral:** - \( \vec{F}(\vec{r}(t)) = \langle 3\cos t + 7\sin t,\ 7\sin t + \sqrt{40}\cos t,\ \sqrt{40}\cos t + 3\cos t \rangle \) - \( \vec{r}'(t) = \langle -3\sin t, 7\cos t, -\sqrt{40}\sin t \rangle \) So the line integral is: \[ \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \] **Final Answer for (2):** \[ \boxed{ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt } \] where \[ \vec{F}(\vec{r}(t)) = \langle 3\cos t + 7\sin t,\ 7\sin t + \sqrt{40}\cos t,\ \sqrt{40}\cos t + 3\cos t \rangle \] \[ \vec{r}'(t) = \langle -3\sin t, 7\cos t, -\sqrt{40}\sin t \rangle \] --- ## **Summary Table** | Part | Stokes' Theorem | Parameterization | Line Integral | |------|------------------|-----------------|--------------| | 1 | \( \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \) | \( \vec{r}(t) = \langle 4\cos t, 5\sin t, 0 \rangle \) | \( \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle \, dt \) | | 2 | \( \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \) | \( \vec{r}(t) = \langle 3\cos t, 7\sin t, \sqrt{40}\cos t \rangle \) | \( \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \) | --- **Do not integrate; just leave the answers as above as per the instructions.**

# Stokes' Theorem Solutions Using **Stokes' Theorem**, we rewrite the surface integral as a line integral for two given vector fields and surfaces. ## 1. Ellipsoid Case ### Given - Vector field: \( \vec{F} = \langle x, y, z \rangle \) - Surface \( S \): Upper half of the ellipsoid defined by \(\frac{x^2}{16} + \frac{y^2}{25} + z^2 = 1\) ### Boundary Curve \( C \) At \( z = 0 \): \[ \frac{x^2}{16} + \frac{y^2}{25} = 1 \] This represents an ellipse in the \( xy \)-plane. ### Parameterization of \( C \) \[ \vec{r}(t) = \langle 4\cos t, 5\sin t, 0 \rangle, \quad 0 \leq t < 2\pi \] ### Application of Stokes' Theorem Using Stokes' Theorem: \[ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \] ### Line Integral Calculation 1. **Calculate \( \vec{F}(\vec{r}(t)) \)**: \[ \vec{F}(\vec{r}(t)) = \langle 4\cos t, 5\sin t, 0 \rangle \] 2. **Calculate \( \vec{r}'(t) \)**: \[ \vec{r}'(t) = \langle -4\sin t, 5\cos t, 0 \rangle \] 3. **Line Integral**: \[ \oint_{C} \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle \, dt \] ### Final Answer for Part 1 \[ \boxed{ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle \, dt } \] --- ## 2. Tilted Disk Case ### Given - Vector field: \( \vec{F} = \langle x+y, y+z, z+x \rangle \) - Surface \( S \): Tilted disk defined by \( \vec{r}(t) = \langle 3\cos t, 7\sin t, \sqrt{40}\cos t \rangle \) ### Boundary Curve \( C \) The parameterization is already given. ### Application of Stokes' Theorem Using Stokes' Theorem again: \[ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \] ### Line Integral Calculation 1. **Calculate \( \vec{F}(\vec{r}(t)) \)**: \[ \vec{F}(\vec{r}(t)) = \langle 3\cos t + 7\sin t, 7\sin t + \sqrt{40}\cos t, \sqrt{40}\cos t + 3\cos t \rangle \] 2. **Calculate \( \vec{r}'(t) \)**: \[ \vec{r}'(t) = \langle -3\sin t, 7\cos t, -\sqrt{40}\sin t \rangle \] 3. **Line Integral**: \[ \oint_{C} \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \] ### Final Answer for Part 2 \[ \boxed{ \iint_{S} (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt } \] --- ## Summary Table | Part | Stokes' Theorem | Parameterization | Line Integral | |------|------------------|-----------------|--------------| | 1 | \( \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \) | \( \vec{r}(t) = \langle 4\cos t, 5\sin t, 0 \rangle \) | \( \int_0^{2\pi} \langle 4\cos t, 5\sin t, 0 \rangle \cdot \langle -4\sin t, 5\cos t, 0 \rangle \, dt \) | | 2 | \( \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \) | \( \vec{r}(t) = \langle 3\cos t, 7\sin t, \sqrt{40}\cos t \rangle \) | \( \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \) | **Note:** The integrals are set up but not evaluated, as per the instructions.