2. Using Green's theorem find the value of \( \oint_{\boldsymbol{c}} \boldsymbol{F} . d r \) Where \( \boldsymbol{F}(x, y)= \) \( \left(2 e^{-5 x}-y^{2}\right) \boldsymbol{i}+\left(y^{3}+2 x^{2}\right) \boldsymbol{j} \) and C is the closed circle with parametric equations \( x=4 \operatorname{cost} \), and \( \boldsymbol{y}=4 \operatorname{sint} \).

To evaluate the line integral \( \oint_{\boldsymbol{C}} \boldsymbol{F} \cdot d\boldsymbol{r} \) using Green's Theorem, follow these steps: ### Given: - **Vector Field**: \( \boldsymbol{F}(x, y) = \left(2 e^{-5 x} - y^{2}\right) \boldsymbol{i} + \left(y^{3} + 2 x^{2}\right) \boldsymbol{j} \) - **Curve \( \boldsymbol{C} \)**: A closed circle with parametric equations \( x = 4 \cos t \), \( y = 4 \sin t \) ### Green's Theorem: Green's Theorem relates a line integral around a simple closed curve \( \boldsymbol{C} \) to a double integral over the region \( D \) enclosed by \( \boldsymbol{C} \): \[ \oint_{\boldsymbol{C}} \boldsymbol{F} \cdot d\boldsymbol{r} = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy \] where \( \boldsymbol{F} = P \boldsymbol{i} + Q \boldsymbol{j} \). ### Compute Partial Derivatives: Identify \( P \) and \( Q \) from \( \boldsymbol{F} \): \[ P = 2 e^{-5 x} - y^{2}, \quad Q = y^{3} + 2 x^{2} \] Calculate the necessary partial derivatives: \[ \frac{\partial Q}{\partial x} = 4x \] \[ \frac{\partial P}{\partial y} = -2y \] So, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 4x + 2y \] ### Set Up the Double Integral: The region \( D \) is a disk of radius 4 centered at the origin. It's convenient to use polar coordinates for this integral: \[ \iint_{D} (4x + 2y) \, dx\, dy \] Convert to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \): \[ 4x + 2y = 4r \cos \theta + 2r \sin \theta \] The area element \( dx\, dy = r \, dr\, d\theta \). Thus, the integral becomes: \[ \int_{0}^{4} \int_{0}^{2\pi} \left(4r \cos \theta + 2r \sin \theta\right) r \, d\theta\, dr = \int_{0}^{4} \int_{0}^{2\pi} \left(4r^{2} \cos \theta + 2r^{2} \sin \theta\right) d\theta\, dr \] ### Evaluate the Integral: Notice that both integrals of \( \cos \theta \) and \( \sin \theta \) over \( 0 \) to \( 2\pi \) are zero: \[ \int_{0}^{2\pi} \cos \theta \, d\theta = 0, \quad \int_{0}^{2\pi} \sin \theta \, d\theta = 0 \] Therefore: \[ \oint_{\boldsymbol{C}} \boldsymbol{F} \cdot d\boldsymbol{r} = 0 \] ### Conclusion: The value of the line integral around the closed circle \( \boldsymbol{C} \) is **zero**. **Answer:** 0