Question

Obtain the curl of the following vector field :
A=(eˆr+rcosθeˆθ+reˆφ)

Answer 1

Answer on Question #75967 – Math – Calculus

Question

Obtain the curl of the following vector field:

$$A = e _ {r} + r \cos \theta e _ {\theta} + r e _ {\varphi}$$

Solution

Obtain the curl of the vector field

$$\begin{aligned} \operatorname{curl} \vec {A} &= \vec {\nabla} \times \vec {A} = \left(e _ {r} \frac {\partial}{\partial r} + \frac {e _ {\theta}}{r} \frac {\partial}{\partial \theta} + \frac {e _ {\varphi}}{r \sin \theta} \frac {\partial}{\partial \varphi}\right) \times \left(A _ {r} e _ {r} + A _ {\theta} e _ {\theta} + A _ {\varphi} e _ {\varphi}\right) = \\ &= \frac {e _ {r}}{r \sin \theta} \left[ \frac {\partial}{\partial \theta} \left(A _ {\varphi} \sin \theta\right) - \frac {\partial A _ {\theta}}{\partial \varphi} \right] + \frac {e _ {\theta}}{r \sin \theta} \left[ \frac {\partial A _ {r}}{\partial \varphi} - \sin \theta \frac {\partial}{\partial r} \left(r A _ {\varphi}\right) \right] + \\ &+ \frac {e _ {\varphi}}{r} \left[ \frac {\partial}{\partial r} \left(r A _ {\theta}\right) - \frac {\partial A _ {r}}{\partial \theta} \right] = \frac {e _ {r}}{r \sin \theta} [ r \cos \theta - 0 ] + \frac {e _ {\theta}}{r \sin \theta} [ 0 - \sin \theta (2 r) ] + \\ &+ \frac {e _ {\varphi}}{r} [ 2 r \cos \theta - 0 ] = \cot \theta e _ {r} - 2 e _ {\theta} + 2 \cos \theta e _ {\varphi}. \end{aligned}$$

Answer: $\operatorname{curl} \vec{A} = \vec{\nabla} \times \vec{A} = \cot \theta e_r - 2e_\theta + 2\cos \theta e_\varphi$.

Answer provided by https://www.AssignmentExpert.com