Given the vector field $\mathbf{F} = \langle x - y, -x + 5y \rangle$ and the region $R$ as the parallelogram $\{ (x,y) : 1 - x \leq y \leq 6 - x, 0 \leq x \leq 2 \}$, we need to compute:

(a) the circulation on the boundary of the region (b) the outward flux across the boundary of the region

(a) Circulation on the Boundary

The circulation of $\mathbf{F}$ around the boundary of $R$ can be computed using the line integral:

$\oint_{\partial R} \mathbf{F} \cdot d\mathbf{r}$

However, by Green's Theorem, this circulation can be transformed into a double integral over the region $R$:

$\oint_{\partial R} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$

where $\mathbf{F} = \langle P, Q \rangle$, with $P = x - y$ and $Q = -x + 5y$.

Calculating the partial derivatives: $\frac{\partial Q}{\partial x} = \frac{\partial (-x + 5y)}{\partial x} = -1$ $\frac{\partial P}{\partial y} = \frac{\partial (x - y)}{\partial y} = -1$

Thus, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -1 - (-1) = 0$

So, the circulation integral becomes: $\oint_{\partial R} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} 0 \, dA = 0$

(b) Outward Flux Across the Boundary

The outward flux of $\mathbf{F}$ across the boundary of $R$ can be computed using the surface integral:

$\oint_{\partial R} \mathbf{F} \cdot \mathbf{n} \, ds$

By the Divergence Theorem, this can be transformed into a double integral over the region $R$:

$\oint_{\partial R} \mathbf{F} \cdot \mathbf{n} \, ds = \iint_{R} (\nabla \cdot \mathbf{F}) \, dA$

where $\nabla \cdot \mathbf{F}$ is the divergence of $\mathbf{F}$:

$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$

Calculating the partial derivatives: $\frac{\partial P}{\partial x} = \frac{\partial (x - y)}{\partial x} = 1$ $\frac{\partial Q}{\partial y} = \frac{\partial (-x + 5y)}{\partial y} = 5$

Thus, $\nabla \cdot \mathbf{F} = 1 + 5 = 6$

So, the flux integral becomes: $\iint_{R} 6 \, dA = 6 \iint_{R} 1 \, dA$

The area of the region $R$ can be calculated by considering the parallelogram. The region $R$ is bounded by the lines $y = 1 - x$, $y = 6 - x$, $x = 0$, and $x = 2$.

The area of the parallelogram is given by the base times the height. Here, the base length $\Delta x = 2 - 0 = 2$. The height is the distance between the lines $y = 1 - x$ and $y = 6 - x$ at any $x$.

The vertical distance between these lines is: $(6 - x) - (1 - x) = 5$

Therefore, the area $A$ of the parallelogram is: $A = 2 \times 5 = 10$

Thus, the outward flux is: $\iint_{R} 6 \, dA = 6 \times 10 = 60$

Summary

• (a) The circulation on the boundary of the given region is $0$.
• (b) The outward flux across the boundary of the given region is $60$.

Would you like further details or have any questions? Here are some related questions you can ask:

1. How do you determine the boundary curves for a given region?
2. What are the applications of Green's Theorem in physics?
3. How do you compute line integrals for vector fields?
4. Can you provide an example of a vector field with non-zero circulation?
5. What are the key differences between circulation and flux in vector calculus?

Tip: When using Green's Theorem, always check if the vector field and the region meet the necessary conditions for its application, such as the vector field being continuously differentiable on the region.