Question

Find the curl of the vector field \(F\). \(\mathbf{F}(x, y, z)=(2 y-z) \mathbf{i}+x y z \mathbf{j}+e^{z} \mathbf{k}\)

Short answer

The curl of the vector field \(F\) is \(-xyz\mathbf{i} + \mathbf{j} + (yz-2) \mathbf{k}\)

Step-by-step solution

Identify the components of the vector field

First, identify \(P\), \(Q\), \(R\) as the coefficients of \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) in the vector field \(F\). From the given function \(\mathbf{F}(x, y, z)=(2y-z) \mathbf{i}+xyz \mathbf{j}+e^{z}\mathbf{k}\), we have \(P=2y-z\), \(Q=xyz\), and \(R=e^{z}\).

Calculate the partial derivatives

Next, calculate the partial derivatives \(\frac{\partial R}{\partial y}\), \(\frac{\partial Q}{\partial z}\), \(\frac{\partial P}{\partial z}\), \(\frac{\partial R}{\partial x}\), \(\frac{\partial Q}{\partial x}\), and \(\frac{\partial P}{\partial y}\). Here, \(\frac{\partial R}{\partial y} = 0\), \(\frac{\partial Q}{\partial z} = xy\), \(\frac{\partial P}{\partial z} = -1\), \(\frac{\partial R}{\partial x} = 0\), \(\frac{\partial Q}{\partial x} = yz\), and \(\frac{\partial P}{\partial y} = 2\).

Plug the values into the formula for curl

Finally, substitute these values into the formula for curl and simplify to find the curl of \(F\). This yields \(\text{curl } \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) \mathbf{i} - \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) \mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \mathbf{k} = (0-xyz)\mathbf{i} - (-1-0)\mathbf{j} + (yz-2)\mathbf{k} = -xyz\mathbf{i} + \mathbf{j} + (yz-2) \mathbf{k}\)