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Chapter 9: Vector Calculus

Section 9.3: Differential Operators

Example 9.3.10

Derive the expression for the curl in cylindrical coordinates.

Solution

Mathematical Solution

An extension of the results in Example 9.2.9 gives

$i = \cos (θ) e_{r} - \sin (θ) e_{θ}$

$j = \sin (θ) e_{r} + \cos (θ) e_{θ}$

$k = e_{z}$

and

$e_{r} = \cos (θ) i + \sin (θ) j$

$e_{θ} = - \sin (θ) i + \cos (θ) j$

$e_{z} = k$

so that the Cartesian form of $F = f (r, θ, z) e_{r} + g (r, θ, z) e_{θ} + h (r, θ, z) e_{x}$ becomes

$G = [\begin{array}{c} (x f - y g) / r \\ (y f + x g) / r \\ h \end{array}] \equiv [\begin{array}{c} A \\ B \\ C \end{array}]$

The curl of the Cartesian vector G is then

$\nabla \times G = [\begin{array}{c} C_{y} - B_{z} \\ A_{z} - C_{x} \\ B_{x} - A_{y} \end{array}] \equiv [\begin{array}{c} u \\ v \\ w \end{array}]$

where, by the chain rule,

$C_{y} = h_{r} \frac{y}{r} + h_{θ} \frac{x}{r^{2}} = h_{r} \sin (θ) + h_{θ} \frac{\cos (θ)}{r}$

$B_{z} = f_{z} \frac{y}{r} + g_{z} \frac{x}{r} = f_{z} \sin (θ) + g_{z} \frac{\cos (θ)}{r}$

$C_{y} - B_{z} = h_{r} \sin (θ) + h_{θ} \frac{\cos (θ)}{r} - f_{z} \sin (θ) - g_{z} \frac{\cos (θ)}{r} \equiv u$

$A_{z} = f_{z} \frac{x}{r} - g_{z} \frac{y}{r} = f_{z} \cos (θ) - g_{z} \sin (θ)$

$C_{x} = h_{r} \frac{x}{r} - h_{θ} \frac{y}{r^{2}} = h_{r} \cos (θ) - h_{θ} \frac{\sin (θ)}{r}$

$A_{z} - C_{x} = f_{z} \cos (θ) - g_{z} \sin (θ) - h_{r} \cos (θ) + h_{θ} \frac{\sin (θ)}{r} \equiv v$

$B_{x} = \frac{r {(y f + x g)}_{x} - (y f + x g) \frac{x}{r}}{r^{2}} = \frac{y (f_{r} \frac{x}{r} - f_{θ} \frac{y}{r^{2}}) + g + x (g_{r} \frac{x}{r} - g_{θ} \frac{y}{r^{2}})}{r} - \frac{x y f + x^{2} g}{r^{3}}$

$A_{y} = \frac{r {(x f - y g)}_{y} - (x f - y g) \frac{y}{r}}{r^{2}} = \frac{x (f_{r} \frac{y}{r} + f_{θ} \frac{x}{r^{2}}) - g - y (g_{r} \frac{y}{r} + g_{θ} \frac{x}{r^{2}})}{r} - \frac{x y f - y^{2} g}{r^{3}}$

$B_{x} - A_{y}$	$= - \frac{1}{r} f_{θ} \frac{x^{2} + y^{2}}{r^{2}} + \frac{2 g}{r} + g_{r} \frac{x^{2} + y^{2}}{r^{2}} - g \frac{x^{2} + y^{2}}{r^{3}}$
	$= - \frac{f_{θ}}{r} + \frac{2 r}{r} + g_{r} - \frac{g}{r}$
	$= - \frac{f_{θ}}{r} + \frac{g}{r} + g \equiv w$

The Cartesian vector $u i + v j + w k$ becomes the cylindrical vector

$u [\begin{array}{c} \cos (θ) \\ - \sin (θ) \\ 0 \end{array}] + v [\begin{array}{c} \sin (θ) \\ \cos (θ) \\ 0 \end{array}] + w [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}]$ = $[\begin{array}{c} u \cos (θ) + v \sin (θ) \\ - u \sin (θ) + v \cos (θ) \\ w \end{array}]$

where

$u \cos (θ) + v \sin (θ)$	$= [h_{r} \sin (θ) + h_{θ} \frac{\cos (θ)}{r} - f_{z} \sin (θ) - g_{z} \frac{\cos (θ)}{r}] \cos (θ) + [f_{z} \cos (θ) - g_{z} \sin (θ) - h_{r} \cos (θ) + h_{θ} \frac{\sin (θ)}{r}] \sin (θ)$
	$= \frac{h_{θ}}{r} - g_{z}$
$- u \sin (θ) + v \cos (θ)$	$= [f_{z} \cos (θ) - g_{z} \sin (θ) - h_{r} \cos (θ) + h_{θ} \frac{\sin (θ)}{r}] \cos (θ) - [h_{r} \sin (θ) + h_{θ} \frac{\cos (θ)}{r} - f_{z} \sin (θ) - g_{z} \frac{\cos (θ)}{r}] \sin (θ)$
	$= f_{z} - h_{r}$

so $\nabla \times F = [\begin{array}{c} \frac{h_{θ}}{r} - g_{z} \\ f_{z} - h_{r} \\ - \frac{f_{θ}}{r} + \frac{g}{r} + g_{r} \end{array}]$ .

Maple Solution - Interactive

The Context Panel can only be invoked on the displayed form of objects. The displays in this interactive derivation are very large; a few small adjustments to notation help alleviate this problem. But the computation can seem overwhelming because of the visual clutter.

•	Define a sequence of the cylindrical coordinate variables.

$P ≔ r, θ, z &colon;$

•	Suppress the appearance of the arguments $r, θ, z$ .

$Typesetting :- Suppress (\{f (p), g (P), h (P)\}) &colon;$

•	Tools≻Load Package: Student Vector Calculus

Loading Student:-VectorCalculus

•	Tools≻Tasks≻Browse: Calculus - Vector≻Vector Algebra and Settings≻Display Format for Vectors

•	Press the Access Settings button and select "Display as Column Vector"

Display Format for Vectors

Define F, a vector field in cylindrical coordinates

•	Write the free vector. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻ Conversions≻Apply Co-ordinate System≻ Conversions≻To Vector Field≻Assign to a Name≻F

$⟨f, g, h⟩$ = $[\begin{array}{c} f \\ g \\ h \end{array}]$ $\overset{apply coordinates}{\to}$ $[\begin{array}{c} f \\ g \\ h \end{array}]$ $\overset{to Vector Field}{\to}$ $[\begin{array}{c} f \\ g \\ h \end{array}]$ $\overset{assign to a name}{\to}$ $F$

Convert the cylindrical vector field F to a Cartesian vector field G

•	Write F. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Vector Calculus≻ Conversions≻Change Co-ordinate System

•	Context Panel: Assign to a Name≻G

$F$ = $[\begin{array}{c} f \\ g \\ h \end{array}]$ $\overset{change coordinates}{\to}$ $[\begin{array}{c} \frac{f x}{\sqrt{x^{2} + y^{2}}} - \frac{g (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{\sqrt{x^{2} + y^{2}}} \\ \frac{f y}{\sqrt{x^{2} + y^{2}}} + \frac{g (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) x}{\sqrt{x^{2} + y^{2}}} \\ h (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) \end{array}]$ $\overset{assign to a name}{\to}$ $G$

Compute the curl of G, which possible since the form of the curl in Cartesian coordinates is known.

Convert this curl from Cartesian to cylindrical coordinates.

•	Common Symbols palette: Apply $\nabla \times$ to G and press the Enter key.

•	Context Panel: Student Vector Calculus≻Conversions≻Change Co-ordinate System (See Figure 9.3.10(a).)

•	Context Panel: Simplify≻Assuming Positive

•	Context Panel: Simplify≻With Side Relations (See Figure 9.3.10(b).)

•	Context Panel: Apply a Command≻convert, diff (See Figure 9.3.10(c).)

•	Context Panel: Expand


Figure 9.3.10(a) Change coordinate system	Figure 9.3.10(b) Simplify with side relations	Figure 9.3.10(c) Apply a command

$\nabla \times G$ =

$[\begin{array}{c} \frac{D_{1} (h) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{\sqrt{x^{2} + y^{2}}} + \frac{D_{2} (h) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z)}{x (1 + \frac{y^{2}}{x^{2}})} - \frac{D_{3} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) x}{\sqrt{x^{2} + y^{2}}} \\ - \frac{D_{3} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{\sqrt{x^{2} + y^{2}}} - \frac{D_{1} (h) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) x}{\sqrt{x^{2} + y^{2}}} + \frac{D_{2} (h) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{x^{2} (1 + \frac{y^{2}}{x^{2}})} \\ \frac{(\frac{D_{1} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) x}{\sqrt{x^{2} + y^{2}}} - \frac{D_{2} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{x^{2} (1 + \frac{y^{2}}{x^{2}})}) x}{\sqrt{x^{2} + y^{2}}} + \frac{2 g (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z)}{\sqrt{x^{2} + y^{2}}} - \frac{g (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) x^{2}}{{(x^{2} + y^{2})}^{\frac{3}{2}}} + \frac{(\frac{D_{1} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y}{\sqrt{x^{2} + y^{2}}} + \frac{D_{2} (g) (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z)}{x (1 + \frac{y^{2}}{x^{2}})}) y}{\sqrt{x^{2} + y^{2}}} - \frac{g (\sqrt{x^{2} + y^{2}}, \arctan (y, x), z) y^{2}}{{(x^{2} + y^{2})}^{\frac{3}{2}}} \end{array}]$

$\overset{change coordinates}{\to}$

$[\begin{array}{c} (\frac{D_{1} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z)}{r \cos (θ) (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})} - \frac{D_{3} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}}) \cos (θ) + (- \frac{D_{3} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{D_{1} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) \sin (θ)}{r {\cos (θ)}^{2} (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}) \sin (θ) \\ - (\frac{D_{1} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z)}{r \cos (θ) (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})} - \frac{D_{3} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}}) \sin (θ) + (- \frac{D_{3} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{D_{1} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (h) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) \sin (θ)}{r {\cos (θ)}^{2} (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}) \cos (θ) \\ \frac{(\frac{D_{1} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{D_{2} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) \sin (θ)}{r {\cos (θ)}^{2} (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}) r \cos (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{2 g (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{g (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r^{2} {\cos (θ)}^{2}}{{(r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2})}^{\frac{3}{2}}} + \frac{(\frac{D_{1} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} + \frac{D_{2} (g) (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z)}{r \cos (θ) (1 + \frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}})}) r \sin (θ)}{\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}} - \frac{g (\sqrt{r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2}}, \arctan (r \sin (θ), r \cos (θ)), z) r^{2} {\sin (θ)}^{2}}{{(r^{2} {\cos (θ)}^{2} + r^{2} {\sin (θ)}^{2})}^{\frac{3}{2}}} \end{array}]$

$\overset{assuming positive}{\to}$

$[\begin{array}{c} \frac{- D_{3} (g) (r, \arctan (\sin (θ), \cos (θ)), z) r + D_{2} (h) (r, \arctan (\sin (θ), \cos (θ)), z)}{r} \\ - D_{1} (h) (r, \arctan (\sin (θ), \cos (θ)), z) \\ \frac{D_{1} (g) (r, \arctan (\sin (θ), \cos (θ)), z) r + g (r, \arctan (\sin (θ), \cos (θ)), z)}{r} \end{array}]$

$\overset{simplify siderels}{=}$

$[\begin{array}{c} \frac{- D_{3} (g) (r, θ, z) r + D_{2} (h) (r, θ, z)}{r} \\ - D_{1} (h) (r, θ, z) \\ \frac{D_{1} (g) (r, θ, z) r + g}{r} \end{array}]$

$\to$

$[\begin{array}{c} \frac{- (\frac{\partial g}{\partial z}) r + \frac{\partial h}{\partial θ}}{r} \\ - \frac{\partial h}{\partial r} \\ \frac{(\frac{\partial g}{\partial r}) r + g}{r} \end{array}]$

$\overset{expand (mapped)}{\to}$

$[\begin{array}{c} - \frac{\partial g}{\partial z} + \frac{\frac{\partial h}{\partial θ}}{r} \\ - \frac{\partial h}{\partial r} \\ \frac{\partial g}{\partial r} + \frac{g}{r} \end{array}]$

•	The side-relation in Figure 9.3.10(b) is $\arctan (\sin (θ), \cos (θ)) = θ$ , where $θ$ is spelled out as "theta".

Obtain $\nabla \times F$ the curl of a cylindrical vector field in cylindrical coordinates, and compare

$\nabla \times F$ = $[\begin{array}{c} \frac{- (\frac{\partial g}{\partial z}) r + \frac{\partial h}{\partial θ}}{r} \\ - \frac{\partial h}{\partial r} \\ \frac{g + (\frac{\partial g}{\partial r}) r}{r} \end{array}]$ $\overset{expand (mapped)}{\to}$ $[\begin{array}{c} - \frac{\partial g}{\partial z} + \frac{\frac{\partial h}{\partial θ}}{r} \\ - \frac{\partial h}{\partial r} \\ \frac{\partial g}{\partial r} + \frac{g}{r} \end{array}]$

Maple Solution - Coded

Initialize

•	Load the Student VectorCalculus package and execute the BasisFormat command.

$with (Student :- VectorCalculus) &colon; BasisFormat (false) &colon;$

Install a notation-improving device

•	Implement the declare command in the PDEtools package.

$PDEtools [declare] (f (r, θ, z), g (r, θ, z), h (r, θ, z), quiet)$

At the present time, the Typesetting tools designed to simplify notation do not work in either of the two VectorCalculus packages. Hence, the resort to the alternate, and older, notational device that suppresses the display of the independent variables and displays partial derivatives as subscripts.

Use the VectorField command to define F as a vector field in cylindrical coordinates

$F ≔ VectorField (⟨f (r, θ, z), g (r, θ, z), h (r, θ, z)⟩, cylindrical [r, θ, z])$

Use the MapToBasis command to change to Cartesian coordinates

$G ≔ MapToBasis (F, cartesian [x, y, z])$

Use the Curl command to obtain the curl of this Cartesian vector field

$curlG ≔ simplify (Curl (G))$

Express the components of this Cartesian vector in cylindrical coordinates

•	Use the eval command to make the appropriate substitutions, and then apply the simplify command.

$q ≔ simplify (eval (curlG, [x = r \cos (θ), y = r \sin (θ)])) assuming r > 0, θ ∷ RealRange (Open (- π), π)$

Use the MapToBasis and simplify commands to change this vector to cylindrical coordinates

$Q ≔ simplify (MapToBasis (q, cylindrical [r, θ, z]))$

Use the convert command to change to partial-derivative notation

$convert (Q, diff)$

Apply the Curl command directly to the vector field F

$Curl (F)$

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