Answer
$ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = 69 $
Work Step by Step
Step 1: To solve this exercise, we will use the extension of Green's theorem. Taking this theorem into account, the line integral can be written as: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = \int_{C1} \mathbf{F} \cdot d\mathbf{r} - \int_{C2} \mathbf{F} \cdot d\mathbf{r} \] where \(C1\) is the external boundary (counterclockwise orientation) and \(C2\) is the inner boundary (counterclockwise orientation). For this exercise, we parametrize \(C1\) and \(C2\) in pieces as follows: - \(C1(1)\): \[ \begin{align*} x &= t, & 0 &\leq t \leq 5 & \text{(line segment from } (0,0) \text{ to } (5,0)) \\ y &= 0 \end{align*} \] - \(C1(2)\): \[ \begin{align*} x &= 5, & 0 &\leq t \leq 5 & \text{(line segment from } (5,0) \text{ to } (5,5)) \\ y &= t \end{align*} \] - \(C1(3)\): \[ \begin{align*} x &= -t, & -5 &\leq t \leq 0 & \text{(line segment from } (5,5) \text{ to } (0,5)) \\ y &= 5 \end{align*} \] - \(C1(4)\): \[ \begin{align*} x &= 0, & -5 &\leq t \leq 0 & \text{(line segment from } (0,5) \text{ to } (0,0)) \\ y &= -t \end{align*} \] - \(C2(1)\): \[ \begin{align*} x &= t, & 1 &\leq t \leq 3 & \text{(line segment from } (1,1) \text{ to } (3,1)) \\ y &= 1 \end{align*} \] - \(C2(2)\): \[ \begin{align*} x &= 3, & 1 &\leq t \leq 2 & \text{(line segment from } (3,1) \text{ to } (3,2)) \\ y &= t \end{align*} \] - \(C2(3)\): \[ \begin{align*} x &= -t, & -3 &\leq t \leq -1 & \text{(line segment from } (3,2) \text{ to } (1,2)) \\ y &= 2 \end{align*} \] - \(C2(4)\): \[ \begin{align*} x &= 1, & -2 &\leq t \leq -1 & \text{(line segment from } (1,2) \text{ to } (1,1)) \\ y &= -t \end{align*} \] Step 2: Then: \[ \oint_{C1} \mathbf{F} \cdot d\mathbf{r} = \int_{C1(1)} \mathbf{F} \cdot d\mathbf{r} + \int_{C1(2)} \mathbf{F} \cdot d\mathbf{r} + \int_{C1(3)} \mathbf{F} \cdot d\mathbf{r} + \int_{C1(4)} \mathbf{F} \cdot d\mathbf{r} = -\int_{-5}^0 (t^2 + 5)\,dt + \int_0^5 t^2\,dt + \int_0^5 (20 - \cos(t))\,dt - \int_{-5}^0 \cos(t)\,dt =75. \] \[ \oint_{C2} \mathbf{F} \cdot d\mathbf{r} = \int_{C2(1)} \mathbf{F} \cdot d\mathbf{r} + \int_{C2(2)} \mathbf{F} \cdot d\mathbf{r} + \int_{C2(3)} \mathbf{F} \cdot d\mathbf{r} + \int_{C2(4)} \mathbf{F} \cdot d\mathbf{r} = \int_1^3 (t^2 + 1)\,dt - \int_{-3}^{-1} (t^2 + 2)\,dt - \int_{-2}^{-1} (4 - \cos(t))\,dt + \int_1^2 (12 - \cos(t))\,dt = 6 \] Finally, we subtract these integrals resulting in: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = 75 - 6 = 69 \] So: Result: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = 69 \]
