To evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ using Green's Theorem, we need to apply the theorem, which states:

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

where $\mathbf{F}(x, y) = (P(x, y), Q(x, y))$ and $D$ is the region enclosed by $C$. In our case:

$P(x, y) = y \cos x - xy \sin x$ $Q(x, y) = xy + x \cos x$

Step 1: Calculate $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$

1. Calculate $\frac{\partial Q}{\partial x}$:

   $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (xy + x \cos x) = y + \cos x - x \sin x$

2. Calculate $\frac{\partial P}{\partial y}$:

   $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (y \cos x - xy \sin x) = \cos x - x \sin x$

Step 2: Substitute into the Green's Theorem formula

Now we substitute into the Green's Theorem formula:

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

Substituting $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$:

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \left( y + \cos x - x \sin x \right) - \left( \cos x - x \sin x \right) = y$

So we get:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D y \, dA$

Step 3: Set up the double integral over the triangular region

The region $D$ is the triangle with vertices at $(0, 0)$, $(0, 10)$, and $(2, 0)$. To set up the double integral, we can integrate $y$ with respect to $x$ and $y$ over this region.

In this triangle, $x$ ranges from 0 to 2, and for a fixed $x$, $y$ ranges from 0 to $10 - 5x$ (the equation of the line from $(0, 10)$ to $(2, 0)$).

$\iint_D y \, dA = \int_0^2 \int_0^{10 - 5x} y \, dy \, dx$

Step 4: Evaluate the integral

1. Integrate with respect to $y$:

   $\int_0^2 \int_0^{10 - 5x} y \, dy \, dx = \int_0^2 \left[ \frac{y^2}{2} \right]_0^{10 - 5x} \, dx$

   $= \int_0^2 \frac{(10 - 5x)^2}{2} \, dx$

2. Simplify and integrate with respect to $x$:

   Expanding $(10 - 5x)^2$:

   $= \int_0^2 \frac{100 - 100x + 25x^2}{2} \, dx$

   $= \int_0^2 \left( 50 - 50x + \frac{25x^2}{2} \right) \, dx$

3. Now integrate each term:

   $= \left[ 50x - 25x^2 + \frac{25x^3}{6} \right]_0^2$

   Substituting $x = 2$:

   $= 50(2) - 25(2)^2 + \frac{25(2)^3}{6}$

   $= 100 - 100 + \frac{200}{6} = \frac{200}{6} = \frac{100}{3}$

Final Answer:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \frac{100}{3}$

Would you like further clarification on any part of this solution?

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Additional Related Questions:

1. How does Green’s Theorem relate to the Fundamental Theorem of Calculus?
2. What happens if the orientation of the curve $C$ is reversed?
3. Can Green’s Theorem be applied to regions with holes?
4. How would the result change if $\mathbf{F}$ had a different orientation?
5. How can Green’s Theorem be extended to three-dimensional regions?

Tip: When using Green's Theorem, ensure the curve's orientation is counterclockwise to match the positive orientation for the theorem's application.