Let's solve the line integral using **Green's Theorem** step by step. #### **Given:** \[ \oint_C \Big( \sin(y) + x^2 + 4xy - y^2 \Big)dx + \Big( x\cos(y) + y^2 + 2x^2 \Big)dy \] Let \( M(x, y) = \sin(y) + x^2 + 4xy - y^2 \) and \( N(x, y) = x\cos(y) + y^2 + 2x^2 \). --- ### **Step 1: State Green's Theorem** Green's Theorem: \[ \oint_C M\,dx + N\,dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dA \] --- ### **Step 2: Compute the Partial Derivatives** #### **a) \( \frac{\partial N}{\partial x} \)** \[ N(x, y) = x\cos(y) + y^2 + 2x^2 \] \[ \frac{\partial N}{\partial x} = \cos(y) + 0 + 4x = \cos(y) + 4x \] #### **b) \( \frac{\partial M}{\partial y} \)** \[ M(x, y) = \sin(y) + x^2 + 4xy - y^2 \] \[ \frac{\partial M}{\partial y} = \cos(y) + 0 + 4x - 2y = \cos(y) + 4x - 2y \] --- ### **Step 3: Find the Difference** \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = (\cos(y) + 4x) - (\cos(y) + 4x - 2y) = 2y \] --- ### **Step 4: Set Up the Double Integral** \[ \oint_C M\,dx + N\,dy = \iint_R 2y \, dA \] --- ### **Step 5: Determine the Region \( R \)** Since the figure for question 4 is not provided, let's **assume** (as is typical in such problems) that \( C \) is the boundary of a simple region like a rectangle or a circle. Let's **assume** \( C \) is the unit circle: \( x^2 + y^2 = 1 \). --- ### **Step 6: Evaluate the Double Integral over the Unit Circle** Convert to polar coordinates: - \( x = r\cos\theta \) - \( y = r\sin\theta \) - \( dA = r \, dr \, d\theta \) - \( r \) from \( 0 \) to \( 1 \), \( \theta \) from \( 0 \) to \( 2\pi \) \[ \iint_R 2y\, dA = \iint_{\text{unit circle}} 2y \, dA = \int_0^{2\pi} \int_0^1 2(r\sin\theta) \cdot r \, dr \, d\theta \] \[ = \int_0^{2\pi} \int_0^1 2r^2\sin\theta \, dr \, d\theta \] Integrate with respect to \( r \): \[ \int_0^1 2r^2 dr = 2 \cdot \left[ \frac{r^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3} \] So, \[ \int_0^{2\pi} \frac{2}{3}\sin\theta \, d\theta = \frac{2}{3} \int_0^{2\pi} \sin\theta\, d\theta \] \[ \int_0^{2\pi} \sin\theta \, d\theta = 0 \] --- ### **Step 7: Final Answer** \[ \boxed{0} \] **Explanation:** The line integral around the closed curve \( C \) equals zero because the double integral of \( 2y \) over a symmetric region about the x-axis (such as a circle) is zero, as the positive and negative contributions cancel out. --- **If your region \( C \) is not a circle, please provide its equation or description for a specific answer!**

Let's carefully evaluate the line integral using Green's theorem, providing detailed calculations and reasoning. --- ### **Given:** The line integral: \[ \oint_C \left[ M(x,y) \, dx + N(x,y) \, dy \right] \] where \[ M(x,y) = \sin(y) + x^2 + 4xy - y^2 \] \[ N(x,y) = x \cos(y) + y^2 + 2x^2 \] --- ### **Step 1: Recall Green's Theorem** Green's theorem states: \[ \oint_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \] Our goal is to compute the double integral of the difference of partial derivatives over the region \( R \) enclosed by \( C \). --- ### **Step 2: Compute Partial Derivatives** **a) \( \frac{\partial N}{\partial x} \):** \[ N(x,y) = x \cos(y) + y^2 + 2x^2 \] \[ \frac{\partial N}{\partial x} = \cos(y) + 0 + 4x = \cos(y) + 4x \] --- **b) \( \frac{\partial M}{\partial y} \):** \[ M(x,y) = \sin(y) + x^2 + 4xy - y^2 \] \[ \frac{\partial M}{\partial y} = \cos(y) + 0 + 4x - 2y \] --- ### **Step 3: Find the integrand** \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = (\cos(y) + 4x) - (\cos(y) + 4x - 2y) = 2y \] **Remark:** The \( \cos(y) \) and \( 4x \) terms cancel out, leaving only \( 2y \). --- ### **Step 4: Set up the Double Integral** \[ \boxed{ \text{Line integral} = \iint_R 2y \, dA } \] The problem references a figure (from question 4) for the curve \( C \). Since the figure isn't provided here, **we'll assume** the region \( R \) is a **unit circle** centered at the origin: \[ x^2 + y^2 \leq 1 \] --- ### **Step 5: Convert to Polar Coordinates** In polar coordinates: \[ x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta \] The limits: \[ r: 0 \to 1, \quad \theta: 0 \to 2\pi \] Express the integrand: \[ 2y = 2 r \sin \theta \] --- ### **Step 6: Write the integral in polar coordinates** \[ \iint_R 2y \, dA = \int_{0}^{2\pi} \int_{0}^{1} 2 r \sin \theta \cdot r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{1} 2 r^2 \sin \theta \, dr \, d\theta \] --- ### **Step 7: Integrate with respect to \( r \)** \[ \int_{0}^{1} 2 r^2 \, dr = 2 \left[ \frac{r^3}{3} \right]_0^1 = 2 \times \frac{1}{3} = \frac{2}{3} \] --- ### **Step 8: Integrate with respect to \( \theta \)** \[ \int_0^{2\pi} \sin \theta \, d\theta \] Recall: \[ \int_0^{2\pi} \sin \theta \, d\theta = 0 \] because the positive and negative contributions cancel over a full period. Multiplying: \[ \frac{2}{3} \times 0 = 0 \] --- ### **### **Final answer:** \[ \boxed{ \text{Value of the line integral} = 0 } \] --- ### **Summary:** - We used Green's theorem to convert the line integral into a double integral over the region \( R \). - The integrand simplifies to \( 2y \). - Assuming the region is a unit circle, the integral over the symmetric region yields zero because the integral of \( y \) over a symmetric circle cancels out. **Note:** If the actual region \( R \) differs, please specify for a precise calculation.