Answer on Question #75967 – Math – Calculus
Question
Obtain the curl of the following vector field:
A = e r + r cos θ e θ + r e φ A = e _ {r} + r \cos \theta e _ {\theta} + r e _ {\varphi} A = e r + r cos θ e θ + r e φ
Solution
Obtain the curl of the vector field
curl A ⃗ = ∇ ⃗ × A ⃗ = ( e r ∂ ∂ r + e θ r ∂ ∂ θ + e φ r sin θ ∂ ∂ φ ) × ( A r e r + A θ e θ + A φ e φ ) = = e r r sin θ [ ∂ ∂ θ ( A φ sin θ ) − ∂ A θ ∂ φ ] + e θ r sin θ [ ∂ A r ∂ φ − sin θ ∂ ∂ r ( r A φ ) ] + + e φ r [ ∂ ∂ r ( r A θ ) − ∂ A r ∂ θ ] = e r r sin θ [ r cos θ − 0 ] + e θ r sin θ [ 0 − sin θ ( 2 r ) ] + + e φ r [ 2 r cos θ − 0 ] = cot θ e r − 2 e θ + 2 cos θ e φ . \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} curl A = ∇ × A = ( e r ∂ r ∂ + r e θ ∂ θ ∂ + r sin θ e φ ∂ φ ∂ ) × ( A r e r + A θ e θ + A φ e φ ) = = r sin θ e r [ ∂ θ ∂ ( A φ sin θ ) − ∂ φ ∂ A θ ] + r sin θ e θ [ ∂ φ ∂ A r − sin θ ∂ r ∂ ( r A φ ) ] + + r e φ [ ∂ r ∂ ( r A θ ) − ∂ θ ∂ A r ] = r sin θ e r [ r cos θ − 0 ] + r sin θ e θ [ 0 − sin θ ( 2 r )] + + r e φ [ 2 r cos θ − 0 ] = cot θ e r − 2 e θ + 2 cos θ e φ .
Answer: curl A ⃗ = ∇ ⃗ × A ⃗ = cot θ e r − 2 e θ + 2 cos θ e φ \operatorname{curl} \vec{A} = \vec{\nabla} \times \vec{A} = \cot \theta e_r - 2e_\theta + 2\cos \theta e_\varphi curl A = ∇ × A = cot θ e r − 2 e θ + 2 cos θ e φ
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#340153 on Dec 2023
Comments
You did not explain which places of the solution are not clear. In the solution, the curl is calculated as the cross (vector) product with a help of the corresponding determinant. One need to take A_r= 1, A_phi=r, A_theta=rcos(theta) in this problem.
I want step by step explaination?