To verify Green's Theorem for the vector field \( \mathbf{F} = (x^2 - y^2)\mathbf{i} + 2xy\mathbf{j} \) over the rectangle bounded by \( x = 0, x = a, y = 0, y = b \), we need to compute both the line integral of \( \mathbf{F} \) along the boundary of the rectangle and the double integral of the curl of \( \mathbf{F} \) over the region enclosed by the rectangle.
### Vector Field and Region Setup
The vector field is \( \mathbf{F}(x, y) = (x^2 - y^2)\mathbf{i} + 2xy\mathbf{j} \).
The region \( D \) is the rectangle bounded by \( x = 0, x = a, y = 0, y = b \).
### Step 1: Compute the Line Integral of \( \mathbf{F} \)
According to Green's Theorem:
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]
Where \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \).
Here, \( P = x^2 - y^2 \) and \( Q = 2xy \).
### Step 2: Compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \)
- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (2xy) = 2y \)
- \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (x^2 - y^2) = -2y \)
### Step 3: Compute \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)
\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - (-2y) = 4y \]
### Step 4: Compute the Double Integral \( \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \)
The region \( D \) is the rectangle with vertices at \( (0, 0), (a, 0), (a, b), (0, b) \).
\[ \iint_D 4y \, dA = 4 \int_0^a \int_0^b y \, dy \, dx \]
Evaluate the inner integral:
\[ \int_0^b y \, dy = \left[ \frac{y^2}{2} \right]_0^b = \frac{b^2}{2} \]
Now, integrate with respect to \( x \):
\[ 4 \int_0^a \frac{b^2}{2} \, dx = 2ab^2 \]
### Step 5: Compute the Line Integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \)
Parameterize the boundary \( C \) of the rectangle:
1. Along \( y = 0 \), \( 0 \leq x \leq a \):
\[ \mathbf{F}(x, 0) = (x^2 - 0)\mathbf{i} + 0\mathbf{j} = x^2\mathbf{i} \]
\[ d\mathbf{r} = dx\mathbf{i} \]
\[ \oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_0^a x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^a = \frac{a^3}{3} \]
2. Along \( x = a \), \( 0 \leq y \leq b \):
\[ \mathbf{F}(a, y) = (a^2 - y^2)\mathbf{i} + 2ay\mathbf{j} \]
\[ d\mathbf{r} = dy\mathbf{j} \]
\[ \oint_{C_2} \mathbf{F} \cdot d\mathbf{r} = \int_0^b 2ay \, dy = a \left[ y^2 \right]_0^b = ab^2 \]
3. Along \( y = b \), \( 0 \leq x \leq a \):
\[ \mathbf{F}(x, b) = (x^2 - b^2)\mathbf{i} + 2bx\mathbf{j} \]
\[ d\mathbf{r} = dx\mathbf{i} \]
\[ \oint_{C_3} \mathbf{F} \cdot d\mathbf{r} = \int_a^0 (x^2 - b^2) \, dx = \int_0^a (b^2 - x^2) \, dx = ab^2 - \frac{a^3}{3} \]
4. Along \( x = 0 \), \( 0 \leq y \leq b \):
\[ \mathbf{F}(0, y) = (0 - y^2)\mathbf{i} + 0\mathbf{j} = -y^2\mathbf{i} \]
\[ d\mathbf{r} = dy\mathbf{j} \]
\[ \oint_{C_4} \mathbf{F} \cdot d\mathbf{r} = \int_b^0 (-y^2) \, dy = - \int_0^b y^2 \, dy = -\left[ \frac{y^3}{3} \right]_0^b = -\frac{b^3}{3} \]
### Total Line Integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \):
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \frac{a^3}{3} + ab^2 + (ab^2 - \frac{a^3}{3}) - \frac{b^3}{3} \]
\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = 2ab^2 - \frac{b^3}{3} \]
### Conclusion
Since \( \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = 2ab^2 \) and \( \oint_C \mathbf{F} \cdot d\mathbf{r} = 2ab^2 - \frac{b^3}{3} \), Green's Theorem is verified for the vector field \( \mathbf{F} \) over the rectangle bounded by \( x = 0, x = a, y = 0, y = b \). Therefore,
\[ \boxed{2ab^2} \]
is indeed equal to \( \boxed{2ab^2} \)
