To use Green's Theorem, we need to express the line integral $\oint_{\gamma} \vec{F} \cdot d\vec{r}$ as a double integral over the region $R$ enclosed by the curve $\gamma$. Green's Theorem states:

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

where $\vec{F}(x, y) = \langle P, Q \rangle$ and $P = (x + y)^2$ and $Q = x^2 + y^2$.

Step 1: Calculate the Partial Derivatives

First, we need to compute $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$:

1. $P = (x + y)^2$

   • $\frac{\partial P}{\partial y} = 2(x + y)$

2. $Q = x^2 + y^2$

   • $\frac{\partial Q}{\partial x} = 2x$

Now we substitute these into Green's Theorem:

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2(x + y) = 2x - 2x - 2y = -2y$

Step 2: Set Up the Double Integral

Next, we define the region $R$. The vertices of the triangle are $O(0, 0)$, $A(3, 0)$, and $B(1, 1)$. We will integrate with respect to $x$ first and then $y$.

1. The bounds for $y$ range from $0$ to $1$ (from vertex $O$ to vertex $B$).
2. For each $y$, $x$ ranges from the left edge of the triangle (line segment $OB$) to the right edge (line segment $OA$):
   • The line segment $OB$ can be described by the equation $y = x$ (or $x = y$).
   • The line segment $OA$ is a horizontal line at $y = 0$ and ends at $x = 3$.

Thus, the bounds for $x$ are:

• Lower bound: $x = y$ (for line segment $OB$)
• Upper bound: $x = 3 - 3y$ (for line segment $OA$)

Step 3: Write the Integral

Now we can set up the double integral:

$\oint_{\gamma} \vec{F} \cdot d\vec{r} = \iint_{R} -2y \, dA = \int_{0}^{1} \int_{y}^{3 - 3y} -2y \, dx \, dy$

Step 4: Evaluate the Inner Integral

Evaluating the inner integral:

[ \int_{y}^{3 - 3y} -2y , dx = -2y \left[ x \right]_{y}^{3 - 3y} = -2y \left[(3 - 3y) - y\right] = -2y (3 - 4y) = -6y + 8y^2 ]

Step 5: Evaluate the Outer Integral

Now we evaluate the outer integral:

[ \int_{0}^{1} (-6y + 8y^2) , dy = \left[ -3y^2 + \frac{8}{3}y^3 \right]_{0}^{1} = \left[ -3(1)^2 + \frac{8}{3}(1)^3 \right] - \left[ -3(0)^2 + \frac{8}{3}(0)^3 \right] ]

This simplifies to:

$-3 + \frac{8}{3} = -\frac{9}{3} + \frac{8}{3} = -\frac{1}{3}$

Final Result

Thus, the value of the line integral is:

$\oint_{\gamma} \vec{F} \cdot d\vec{r} = -\frac{1}{3}$

If you have any questions or need further details, let me know! Here are some related questions you might consider:

1. What is the geometric interpretation of Green's Theorem?
2. How would the result change if the orientation of the boundary were reversed?
3. Can you apply Green's Theorem to more complex regions?
4. What are the conditions for applying Green's Theorem?
5. How do you verify the correctness of your calculations in this context?

Tip: Always visualize the region you're integrating over; drawing the region helps clarify the bounds of integration.