Question

Find the divergence of the vector field \(\mathrm{F}\). \(\mathbf{F}(x, y, z)=\sin x \mathbf{i}+\cos y \mathbf{j}+z^{2} \mathbf{k}\)

Short answer

Therefore the divergence of the given vector field \(\mathbf{F}(x, y, z)=\sin x \mathbf{i}+\cos y \mathbf{j}+z^{2} \mathbf{k}\) is \(\nabla \cdot \mathbf{F} = \cos x - \sin y + 2z\)

Step-by-step solution

Identify Components of the Vector Field

The first step is to identify the components of the vector field. For the given vector, \(\mathbf{F}(x, y, z) = \sin x \mathbf{i} + \cos y \mathbf{j} + z^{2} \mathbf{k}\), the components are \(F_{1} = \sin x\), \(F_{2} = \cos y\) and \(F_{3} = z^{2}\).

Equation for Divergence

Next, the divergence equation should be provided which is given by \(\nabla \cdot \mathbf{F} = \frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}+ \frac{\partial F_{3}}{\partial z}\).

Apply the Divergence Equation

Then the partial derivatives of the components of the vector field should be calculated. Using the above formula: \(\frac{\partial (\sin x)}{\partial x} = \cos x\), \(\frac{\partial (\cos y)}{\partial y} = -\sin y\), and \(\frac{\partial (z^{2})}{\partial z} = 2z\).

Calculate Divergence

Finally, adding the calculated partial derivatives will give the divergence of the vector field which is: \(\nabla \cdot \mathbf{F} = \cos x - \sin y + 2z\).