To evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ using Stokes' Theorem, we first need to understand the relationship between the line integral and the surface integral provided by Stokes' Theorem. Stokes' Theorem states that:

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

where $\mathbf{F} = (4xy \sin z, 2x^2 \sin z, 2x^2 y \cos z)$ and $C$ is the boundary of the plane $z = 15 - 5x - 3y$ in the first octant.

Step 1: Compute $\nabla \times \mathbf{F}$

First, we find the curl of $\mathbf{F}$:

$\nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)$

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:

1. $\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}$:

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

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

Thus,

$\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = 2x^2 \cos z - 2x^2 \cos z = 0$

2. $\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}$:

$\frac{\partial (4xy \sin z)}{\partial z} = 4xy \cos z$

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

Thus,

$\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = 4xy \cos z - 4xy \cos z = 0$

3. $\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}$:

$\frac{\partial (2x^2 \sin z)}{\partial x} = 4x \sin z$

$\frac{\partial (4xy \sin z)}{\partial y} = 4x \sin z$

Thus,

$\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 4x \sin z - 4x \sin z = 0$

So, we have:

$\nabla \times \mathbf{F} = (0, 0, 0)$

Step 2: Apply Stokes' Theorem

Since $\nabla \times \mathbf{F} = \mathbf{0}$, we have:

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

Therefore,

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

Conclusion

The line integral of $\mathbf{F}$ over the curve $C$, which is the boundary of the plane $z = 15 - 5x - 3y$ in the first octant, evaluates to zero when using Stokes' Theorem.

Relative Questions

1. What are the conditions for applying Stokes' Theorem?
2. How do you compute the curl of a vector field?
3. What is the physical interpretation of the curl of a vector field?
4. How does the orientation of the curve $C$ affect the result in Stokes' Theorem?
5. Can Stokes' Theorem be applied to non-planar surfaces? If so, how?
6. What happens if $\nabla \times \mathbf{F} \neq 0$?
7. How can Stokes' Theorem be used in electromagnetism?
8. What are some practical applications of Stokes' Theorem in engineering?

Tip

When applying Stokes' Theorem, always ensure that the surface $S$ and its boundary curve $C$ are properly oriented according to the right-hand rule.