To rewrite the given line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ as an area integral over the appropriate region of the $xy$ plane using Stokes' Theorem, we will follow these steps:

1. Restate Stokes' Theorem:

   $\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

2. Identify the Surface $S$ and its Boundary $C$:

   The curve $C$ is the boundary of the plane $z = 15 - 5x - 3y$ in the first octant.

3. Determine the Region $R$ in the $xy$ Plane:

   The plane $z = 15 - 5x - 3y$ intersects the $xy$ plane in the first octant, where $z \geq 0$. Therefore, the region $R$ in the $xy$ plane is bounded by the lines where $z = 0$.

   Setting $z = 0$:

   $0 = 15 - 5x - 3y \implies 5x + 3y = 15$

   Thus, the region $R$ is a triangle with vertices at $(0, 0)$, $(3, 0)$, and $(0, 5)$.

4. Calculate the Curl of $\mathbf{F}$:

   From the previous calculation, we found:

   $\nabla \times \mathbf{F} = \mathbf{0}$

   However, for completeness and clarity, let's recheck and confirm:

   Given $\mathbf{F} = (4xy \sin z, 2x^2 \sin z, 2x^2 y \cos z)$:

   • $F_1 = 4xy \sin z$
   • $F_2 = 2x^2 \sin z$
   • $F_3 = 2x^2 y \cos z$

   Compute each component of the curl again:

   $\left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)$

   $\frac{\partial (2x^2 y \cos z)}{\partial y} = 2x^2 \cos z, \quad \frac{\partial (2x^2 \sin z)}{\partial z} = 2x^2 \cos z$

   $\frac{\partial (4xy \sin z)}{\partial z} = 4xy \cos z, \quad \frac{\partial (2x^2 y \cos z)}{\partial x} = 4xy \cos z$

   $\frac{\partial (2x^2 \sin z)}{\partial x} = 4x \sin z, \quad \frac{\partial (4xy \sin z)}{\partial y} = 4x \sin z$

   So, indeed,

   $\nabla \times \mathbf{F} = \mathbf{0}$

   This simplifies our surface integral as the integrand becomes zero:

5. Set Up the Area Integral:

   Since $\nabla \times \mathbf{F} = \mathbf{0}$, the surface integral becomes:

   $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S \mathbf{0} \cdot d\mathbf{S} = 0$

   Therefore, the area integral over the region $R$ in the $xy$ plane is:

   $\int_C \mathbf{F} \cdot d\mathbf{r} = 0$

Thus, we have rewritten the line integral as an area integral over the appropriate region of the $xy$ plane, and confirmed that the integral evaluates to zero due to the curl of $\mathbf{F}$ being zero.

Relative Questions

1. What are the implications of a vector field having a zero curl?
2. How do you determine the region in the $xy$ plane for a given surface?
3. What are the steps to convert a line integral to a surface integral using Stokes' Theorem?
4. How does the orientation of a surface affect the application of Stokes' Theorem?
5. Can Stokes' Theorem be applied to surfaces in higher dimensions?
6. What other theorems in vector calculus are similar to Stokes' Theorem?
7. How do boundary conditions affect the outcome of a surface integral?
8. What is the significance of the first octant in this problem?

Tip

When working with vector calculus theorems like Stokes' Theorem, always double-check the orientation and boundary conditions to ensure the calculations are set up correctly.