Answer
So, the line integral result is: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = -24\pi \]
Work Step by Step
Step 1: To solve this exercise, we will use the extension of Green's theorem (see page 1091). Taking into account this theorem, 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). The vector field is given by: \[ \mathbf{F}(\mathbf{r}) = (e^{-x} + 3y)\mathbf{i} + x\mathbf{j} \] We use the following parametrizations for the curves: - \(C1\): \[ \begin{align*} x &= 4\cos(\theta), & y &= 4\sin(\theta), & 0 &\leq \theta \leq 2\pi \end{align*} \] - \(C2\): \[ \begin{align*} x &= 1 + 2\cos(\theta), & y &= 2\sin(\theta), & 0 &\leq \theta \leq 2\pi \end{align*} \] Step 2: Then: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = \int_{C1} \mathbf{F} \cdot d\mathbf{r} - \int_{C2} \mathbf{F} \cdot d\mathbf{r} = \int_{C1} (\mathbf{F}(\mathbf{r}) - \mathbf{F}(\mathbf{r}_0)) \cdot d\mathbf{r} - \int_{C2} (\mathbf{F}(\mathbf{r}) - \mathbf{F}(\mathbf{r}_0)) \cdot d\mathbf{r} \] \[ = \int_{C1} \left(\begin{pmatrix} e^{-x} + 3y \\ x \end{pmatrix} - \begin{pmatrix} e^{-x_0} + 3y_0 \\ x_0 \end{pmatrix}\right) \cdot d\mathbf{r} - \int_{C2} \left(\begin{pmatrix} e^{-x} + 3y \\ x \end{pmatrix} - \begin{pmatrix} e^{-x_0} + 3y_0 \\ x_0 \end{pmatrix}\right) \cdot d\mathbf{r} \] \[ = \int_{0}^{2\pi} \left(\begin{pmatrix} e^{-4\cos(\theta)} + 12\sin(\theta) \\ 4\cos(\theta) \end{pmatrix} \cdot \begin{pmatrix} -4\sin(\theta) \\ 4\cos(\theta) \end{pmatrix} + \begin{pmatrix} 16\cos^2(\theta) \\ 0 \end{pmatrix} \right) d\theta \] \[ - \int_{0}^{2\pi} \left(\begin{pmatrix} e^{-1 - 2\cos(\theta)} + 6\sin(\theta) \\ 1 + 2\cos(\theta) \end{pmatrix} \cdot \begin{pmatrix} -2\sin(\theta) \\ -2\sin(\theta) \end{pmatrix} + \begin{pmatrix} 2(2\cos(\theta) + 1)\cos(\theta) \\ 0 \end{pmatrix} \right) d\theta \] \[ = -4\int_{0}^{2\pi} (\sin^2(\theta) - 4\cos^2(\theta)) d\theta + 16\int_{0}^{2\pi} \cos^2(\theta) d\theta - 2\int_{0}^{2\pi} (\cos(\theta) + 1) d\theta - 12\int_{0}^{2\pi} \sin^2(\theta) d\theta - 4\int_{0}^{2\pi} \cos^2(\theta) d\theta - 2\int_{0}^{2\pi} \cos(\theta) d\theta \] \[ = -32\pi - (-8\pi) = -24\pi \] So, the line integral result is: \[ \oint_{1} \mathbf{F} \cdot d\mathbf{r} = -24\pi \]
