Let's **solve the problem step by step** using **Green's theorem**. ### **Step 1: Green's Theorem Statement** Green's theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( D \) enclosed by \( C \): \[ \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] where \(\vec{F} = \langle P(x, y), Q(x, y), 0 \rangle\). The **z-component of the curl** of \( \vec{F} \) is \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \). ### **Step 2: Understanding the Region \( D \)** - \( D \) is the region **inside** the large diamond (vertices at \( (\pm 3, 0), (0, \pm 3) \)), but **outside** the small square (with edge length 1, centered at the origin). - \( C_1 \): boundary of the **large diamond**, counterclockwise. - \( C_2 \): boundary of the **small square**, clockwise (because the region \( D \) is outside the square). ### **Step 3: Green's Theorem for Annular Region** For a region with a "hole" (annulus), Green's theorem gives: \[ \oint_{C_1} \vec{F} \cdot d\vec{r} - \oint_{C_2} \vec{F} \cdot d\vec{r} = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] But we want \( \oint_{C_2} \vec{F} \cdot d\vec{r} \). ### **Step 4: Plug in Known Values** Given: - \( \oint_{C_1} \vec{F} \cdot d\vec{r} = 220 \) - \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3 \) everywhere on \( D \) So: \[ \oint_{C_1} \vec{F} \cdot d\vec{r} - \oint_{C_2} \vec{F} \cdot d\vec{r} = 3 \times \text{Area}(D) \] So, \[ 220 - \oint_{C_2} \vec{F} \cdot d\vec{r} = 3 \cdot \text{Area}(D) \] \[ \oint_{C_2} \vec{F} \cdot d\vec{r} = 220 - 3 \cdot \text{Area}(D) \] ### **Step 5: Calculate the Area of \( D \)** - **Large diamond** (square rotated by 45°): Each side's length is \( \sqrt{(3-0)^2 + (0-3)^2} = \sqrt{9 + 9} = \sqrt{18} \). But the area of a square with vertices at \( (\pm a, 0) \), \( (0, \pm a) \) is \( 2a^2 \). Here, \( a = 3 \), so **area = \( 18 \)**. - **Small square**: Side length = 1, so area = 1. \[ \text{Area}(D) = 18 - 1 = 17 \] ### **Step 6: Final Calculation** \[ \oint_{C_2} \vec{F} \cdot d\vec{r} = 220 - 3 \times 17 = 220 - 51 = 169 \] ### **Step 7: Sign Convention** Recall: \( C_2 \) must be traversed **clockwise** (since the region \( D \) is outside the small square), which is **negative orientation**. So, the answer should be **-169**. ### **Step 8: Final Answer** \[ \boxed{-169} \] --- **Summary:** The circulation of \( \vec{F} \) around \( C_2 \) is \( \boxed{-169} \), matching your answer. If the system marked it incorrect, double-check the sign convention or the area calculation, but the above steps are correct by standard application of Green's theorem.