Chapter 9: Problem 13

For the given vector field, find the divergence and curl of the field. a. \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}\) b. \(\mathbf{F}=\frac{y}{r} \mathbf{i}-\frac{x}{r} \mathbf{j}\), for \(r=\sqrt{x^{2}+y^{2}}\). c. \(\mathbf{F}=x^{2} y \mathbf{i}+z \mathbf{j}+x y z \mathbf{k} .\)

Short Answer

Expert verified

The divergences are 2, 0, 0 respectively, and the curls are 0, 0, and \((1 - 2xy) \mathbf{k}\) respectively.

Step by step solution

Compute Divergence and Curl for Field F1

For the field \(F = x \mathbf{i} + y \mathbf{j}\), the divergence is computed as the dot product of the del operator and the vector field. Here, \(\nabla \cdot \mathbf{F} = \partial x/ \partial x +\partial y/ \partial y = 1 + 1 = 2\). For the curl, we need to calculate the cross product of the del operator and the field, leading to \(\nabla \times \mathbf{F} = i(\partial / \partial y - \partial / \partial z) - j(\partial / \partial x - \partial / \partial z) + k(\partial / \partial x - \partial / \partial y) \). Computing these differences gives \(\nabla \times \mathbf{F} = 0 \mathbf{i} - 0 \mathbf{j} + 0 \mathbf{k} = 0\)

Compute Divergence and Curl for Field F2

For the field \(F = \frac{y}{r} \mathbf{i} - \frac{x}{r} \mathbf{j}\) where \(r = \sqrt{x^2 + y^2}\), the divergence is \(\partial / \partial x (\frac{y}{r}) + \partial / \partial y (-\frac{x}{r})\). Substituting r and simplifying gives \(0\). The curl is computed similarly, leading to: \(\nabla \times \mathbf{F} = 0\)

Compute Divergence and Curl for Field F3

For the field \(F = x^2y\mathbf{i} + z\mathbf{j} + xyz\mathbf{k}\), the divergence is \(\partial / \partial x (x^2y) + \partial / \partial y (z) + \partial / \partial z (xyz) = 0\). The curl is \(\nabla \times \mathbf{F}\) = \(i(\frac{\partial}{\partial y}(xyz) - \frac{\partial}{\partial z}(z)) - j(\frac{\partial}{\partial x}(xyz) - \frac{\partial}{\partial z}(x^2y)) + k(\frac{\partial}{\partial x}(z) - \frac{\partial}{\partial y}(x^2y))\). Simplifying, we get \(\nabla \times \mathbf{F} = 0 \mathbf{i} - 0 \mathbf{j} + (1 - 2xy) \mathbf{k}\)