es

Green's theorems

Rectangular domain

Semi-infinite strip

Curvilinear coordinates

General case

Green's Theorems in Polar coordinates

Besides rectangular coordinates, it is convenient to consider polar coordinates on ℝ². They originated in the work by the ancient Greek astronomer and astrologer Hipparchus (190–120 BC). The full history of the subject is described in Harvard professor Julian Lowell Coolidge's Origin of Polar Coordinates.

In order to introduce polar coordinates, start with a point O in the plane ℝ² called the pole (we will always identify this point with the origin). From the pole, draw a ray, called the initial ray (we will always draw this ray horizontally, identifying it with the positive abscissa). A point P in the plane is determined by the distance r that P is from O, and the angle θ formed between the initial ray and the segment \( \displaystyle \quad \overline{OP} \quad \) (measured counter-clockwise). We record the distance and angle as an ordered pair (r, θ), which is known as polar notation. This representation is illustrated in the following figure:

point = Graphics[{PointSize[0.02], Pink, Point[{1/2, Sqrt[3]/2}]}]; or = Graphics[{PointSize[0.02], Blue, Point[{0, 0}]}]; xar = Graphics[{Blue, Arrowheads[0.1], Arrow[{{0, 0}, {1, 0}}]}]; line = Graphics[{Black, Thick, Line[{{0, 0}, {1/2, Sqrt[3]/2}}]}]; txt = Graphics[{Black, Text[Style["O", FontSize -> 18, Bold], {-.03, -.03(*-0.5*)}], Text[Style["r", FontSize -> 18, Bold], {0.2, 0.5}], Text[Style["P = P(r, \[Theta])", FontSize -> 18, Bold], {0.5, 0.94}], Text[Style["\[Theta]", FontSize -> 18, Bold], {0.56, 0.45}]}]; (*circle=Graphics[Circle[{0,0},0.6,{0,Pi/3}]];*) ar = Graphics[{Black, Arrowheads[0.04], Arrow[{{0.32, 0.5059}, {0.3, 0.519}}]}]; Labeled[ Show[txt, xar, line, point, or, circle, ar], "Figure 1: Polar coordinate system"]
Figure 1: Polar coordinate system.

Conversion between rectangular and polar coordinates are based on the relations:

\[ \begin{split} x &= r\,\cos\theta , \\ y &= r\,\sin\theta , \qquad 0 < r <\infty , \quad -\pi < \theta \le \pi , \end{split} \]
and
\[ r = +\sqrt{x^2 + y^2} , \qquad \tan\theta = \frac{y}{x} . \]
Note that angle θ is measures in radians (no unit in SI); it is considered positive when θ goes in counterclockwise direction with respect to fixed horizontal axis, called abscissa (usually marked as x-axis). In opposite of θ, variable r is measured in meters (in SI system) and it is restricted to 0 < r < ∞; the point r = 0 corresponds the origin.

In order to preserve a unique representation of a point on the plane in polar coordinates as ordered pair (r, θ), the corresponding angle θ is taken within the interval of length 2π (if measured in radians) or 360 (if angle θ is measured in degrees). We assume that θ ∈ (−π, π] because we use radians; however, some authors prefer to use semi-open interval [0, 2π) or [0, 360°) in case of degrees. So the angle θ is defined according to the formula:

\[ \theta = \begin{cases} \mbox{arccos} \left( \frac{x}{r} \right) , & \quad \mbox{if}\ y \ge 0 \mbox{ and } r \ne 0, \\ - \mbox{arccos} \left( \frac{x}{r} \right) , & \quad \mbox{if}\ y < 0, \\ \mbox{undefined} , & \quad \mbox{if}\ r = 0. \end{cases} \]

We summarize possible outputs in the following table.

Quadrant Range Value
1st     (x > 0, y > 0) 0 < θ < π/2 θ = α
2nd     (x < 0, y > 0) π/2 < θ < π θ = π − α
3rd     (x < 0, y < 0) −π < θ < −π/2 θ = −π + α
4th     (x > 0, y < 0) −π/2 < θ < 0 θ = −α

Defining a new coordinate system allows us to create a new kind of function, a polar function. Rectangular coordinates lent themselves well to creating functions that related x and y, such as y = sin(x). Polar coordinates allow us to create functions that relate r and θ. Normally these functions look like r = f(θ), although we can create functions of the form θ = g(r).

A polar curve can be parametrized by θ or by any parameter. The arc length of a polar curve given by r = r(θ) is

\[ s = \int_{\theta_1}^{\theta_2} {\text d}\theta \,\sqrt{r^2 + \left( \frac{{\text d}r}{{\text d}\theta} \right)^2} . \]
The area enclosed by a polar curve r = r(θ) is
\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 {\text d}\theta . \]
The rotor (curl) of a plane vector field is
\[ \mbox{curl} \left( A_r , A_{\theta} , 0 \right) = \frac{1}{r} \left( \frac{\partial \left( r\,A_{\theta} \right)}{\partial r} - \frac{\partial A_r}{\partial \theta} \right) \mathbf{k} , \]
where k = (0, 0, 1) is the unit vector along applicate axis. A radius vector r of any point (r, θ) on the plane ℝ² is expressed through the single vector:
\[ {\bf r} = \sqrt{x^2 + y^2}\,\hat{\bf r} = r(\theta )\,\hat{\bf r} , \]
where unit vector (of unit length) is
\[ \hat{\bf r} = \frac{x\,\hat{\bf x} + y\,\hat{\bf y}}{r} = \frac{\frac{{\text d}{\bf r}}{{\text d}r}}{\left\vert \frac{{\text d}{\bf r}}{{\text d}r} \right\vert} = \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix} . \]
Here we represent \( \displaystyle \quad \hat{\bf r} \ \) as a column vector. Its derivative with respect to t (parameter)) is expressed through two orthogonal vectors
\[ \dot{\bf r} = \begin{bmatrix} -r\,\sin\theta\,\dot{\theta} + \cos\theta\,\dot{r} \\ r\,\cos\theta \,\dot{\theta} + \sin\theta\,\dot{r} \end{bmatrix} = r\,\dot{\theta}\,\hat{\theta} + \dot{r}\,\hat{\bf r} , \]
where
\[ \hat{\theta} = \frac{\frac{{\text d}{\bf r}}{{\text d}\theta}}{\left\vert \frac{{\text d}{\bf r}}{{\text d}\theta} \right\vert} = \begin{bmatrix} -\sin\theta \\ \cos\theta \end{bmatrix} . \]

The basic rectangular equations of the form x = h and y = k create vertical and horizontal lines, respectively; the basic polar equations r = h and θ = α create circles and lines through the pole, respectively. A typical example of this form is a wedge:

\[ \theta = \theta_1 , \qquad \theta = \theta_2 \quad (\alpha = \theta_2 - \theta_1 ) , \qquad 0 \le r \le a . \]
or = Graphics[{PointSize[0.02], Blue, Point[{0, 0}]}]; xar = Graphics[{Blue, Arrowheads[0.1], Arrow[{{0, 0}, {1, 0}}]}]; line = Graphics[{Black, Thick, Line[{{0, 0}, {1/2, Sqrt[3]/2}}]}]; circle = Graphics[Circle[{0, 0}, 0.6, {0, Pi/3}]]; ar = Graphics[{Black, Arrowheads[0.04], Arrow[{{0.32, 0.5059}, {0.3, 0.519}}]}]; ine2 = Graphics[{Black, Thick, Line[{{0, 0}, {Sqrt[3]/2, 1/2}}]}]; ar1 = Graphics[{Black, Arrowheads[0.04], Arrow[{{0.535, 0.27}, {0.519, 0.3}}]}]; txt = Graphics[{Black, Text[Style["O", FontSize -> 18, Bold], {0, -0.1}], Text[Style[Subscript[\[Theta], 1], FontSize -> 18, Bold], {0.66, 0.2}], Text[Style["\[Alpha]", FontSize -> 18, Bold], {0.5, 0.45}]}]; Show[txt, xar, line, line2, or, circle, ar, ar1]
Figure 2: Wedge domain.

We choose a finite part of the wedge bounded by 0 < r ≤ 𝑎 for some positive 𝑎:

\[ \mbox{Wedge} = \left\{ \left( r, \theta \right) : \ 0 < r \le a, \quad \theta_1 \le \theta \le \theta_2\right\} . \]
wedge = Append[ Table[{r Cos[t], r Sin[t]} /. r -> 1, {t, Pi/6, Pi/3, 0.01}], {0, 0}]; ar1 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{0.1, 0.01}, {0.7, 0.35}}]}]; ar2 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{0.85, 0.57}, {0.63, 0.81}}]}]; ar3 = Graphics[{Blue, Thick, Arrowheads[0.04], Arrow[{{0.3, 0.57}, {0.01, 0.071}}]}]; txt = Graphics[{Black, Text[Style["O", FontSize -> 18, Bold], {-0.03, -0.03}], Text[Style["A", FontSize -> 18, Bold], {0.9, 0.5}], Text[Style["B", FontSize -> 18, Bold], {0.5, 0.9}]}]; Labeled[ Show[Graphics[{Red, Polygon[wedge]}, Axes -> True], ar1, ar2, ar3, txt], "Figure3: Finite piece of the Wedge"]
Figure 3: Finite piece of the Wedge.

The boundary ∂Wedge of our domain consists of two lines, OA and BO, and a part of circle r = 𝑎 (θ₁ ≤ θ ≤ θ₂). We want to integrate an arbitrary vector field

\[ \mathbf{F} = F_r \hat{\bf e}_r + F_{\theta} \hat{\bf e}_{\theta} = \left( F_r , F_{\theta} \right) \in \mathbb{R}^2 , \]
over the closed curve ∂Wedge. The integral over line OA is
\[ I_{OA} = \int_O^A \mathbf{F} \bullet {\text d}\mathbf{r} = \int_O^A F_r (r, \theta_1 )\,{\text d}s = \int_0^a F_r (r, \theta_1 )\,{\text d}r . \]
Similarly,
\[ I_{BO} = \int_B^O \mathbf{F} \bullet {\text d}\mathbf{r} = \int_B^O F_r (r, \theta_2 )\,{\text d}s = -\int_0^a F_r (r, \theta_2 )\,{\text d}r . \]
Their sum becomes
\[ I_{OA} + I_{BO} = \int_0^a {\text d}r \left( F_r (r, \theta_1 ) - F_r (r, \theta_2 ) \right) = -\int_0^a {\text d}r \left( F_r (r, \theta_2 ) - F_r (r, \theta_1 ) \right) . \]
Using the Fundamental Theorem of Calculus, we get
\[ F_r (r, \theta_2 ) - F_r (r, \theta_1 ) = \int_{\theta_1}^{\theta_2} \frac{\partial F_r}{\partial \theta} \,{\text d} \theta . \]
This allows us to represent the sum of these two line integrals as the double integral:
\[ I_{OA} + I_{BO} = - \iint_W \frac{\partial F_r}{\partial \theta} \,{\text d}r \,{\text d} \theta = - \iint_W \frac{1}{r}\,\frac{\partial F_r}{\partial \theta} \,{\text d}A , \]
where dA = dr rdθ is the element of area in polar coordinates. Here is Wolfram code replicating the above.
f[r_, \[Theta]_] := {fr[r, \[Theta]], f\[Theta][r, \[Theta]]}; iOA = Integrate[fr[r, \[Theta]1], {r, 0, a}]; iBO = -Integrate[fr[r, \[Theta]2], {r, 0, a}]
\( \displaystyle \quad -\int_0^a f_r (r, \theta_2 )\,{\text d}r \)
sumIntegrals = Integrate[fr[r, \[Theta]1] - fr[r, \[Theta]2], {r, 0, a}]
\( \displaystyle \quad \int_0^a \left( f_r (r, \theta_1 ) - f_r (r, \theta_2 )\right) {\text d}r \)

Now we integrate over the arc of circle AB; note that dr = 0 on this piece because variable r is constant. So

\[ I_{AB} = \int_A^B \mathbf{F} \bullet {\text d}\,\mathbf{r} = \int_A^B F_{\theta} (a, \theta )\,{\text d}s = \int_{\theta_1}^{\theta_2} F_{\theta} (a, \theta )\,{\text d}\left( a\theta\right) . \]
frcDiff = Integrate[ D[fr[r, \[Theta]], \[Theta]], {\[Theta], \[Theta]1, \[Theta]2}]
iAB = Integrate[ f\[Theta][a, \[Theta]], {\[Theta], \[Theta]1, \[Theta]2}]
Again, using the Fundamental Theorem of Calculus, we represent the integrand as the definite integral:
\[ a\,F_{\theta} (a, \theta ) = \int_0^a \frac{\partial \left( r\,F_{\theta} (r, \theta )\right)}{\partial r}\,{\text d}r \]
subject that
\begin{equation} \label{EqPolar.1} \lim_{r\to 0} r\,F_{\theta} (r, \theta ) = 0. \end{equation}
Summing all integrals together, we obtain
doubleIntegral = Integrate[(D[r f\[Theta][r, \[Theta]], r] - D[fr[r, \[Theta]], \[Theta]]), {r, 0, a}, {\[Theta], \[Theta]1, \[Theta]2}]
\[ \oint_{\partial W} \mathbf{F} \bullet {\text d}\mathbf{r} = \iint_W \left( \frac{\partial \left( r\,F_{\theta} (r, \theta )\right)}{\partial r} - \frac{\partial F_r}{\partial \theta} \right) {\text d}r\,{\text d}\theta \]
assuming that the corner condition \eqref{EqPolar.1} holds. We can rewrite it in terms of rectangular components because
\[ \mathbf{F} = F_x \hat{\bf x} + F_y \hat{\bf y} = F_r \hat{\bf r} + F_{\theta} \hat{\bf \theta} \]
Since
\[ \begin{split} \hat{\bf r} &= \cos\theta\,\hat{\bf x} + \sin\theta \,\hat{\bf y} = \cos\theta\,\hat{\bf i} + \sin\theta \,\hat{\bf j} , \\ \hat{\bf \theta} &= -\sin\theta \,\hat{\bf i} + \cos\theta \,\hat{\bf j} , \end{split} \]
because it is a custom to use i and j as unit vectors in Cartesian coordinates. This yields
\[ F_x = F_r \cos\theta - \sin\theta\,F_{\theta} , \quad F_y = F_r \sin\theta + F_{\theta} \cos\theta . \]
Therefore,
\[ F_r = F_x \cos\theta + F_y \sin\theta , \qquad F_{\theta} = F_y \cos\theta - F_x \sin\theta = F_y \,\frac{x}{r} - F_x \,\frac{y}{r} . \]
fx = fr[r, \[Theta]] Cos[\[Theta]] - f\[Theta][r, \[Theta]] Sin[\[Theta]] fy = fr[r, \[Theta]] Sin[\[Theta]] + f\[Theta][r, \[Theta]] Cos[\[Theta]]
Cos[\[Theta]] fr[r, \[Theta]] - f\[Theta][r, \[Theta]] Sin[\[Theta]]
Cos[\[Theta]] f\[Theta][r, \[Theta]] + fr[r, \[Theta]] Sin[\[Theta]]
frBack = fx Cos[\[Theta]] + fy Sin[\[Theta]] f\[Theta]Back = fy Cos[\[Theta]] - fx Sin[\[Theta]]
Sin[\[Theta]] (Cos[\[Theta]] f\[Theta][r, \[Theta]] + fr[r, \[Theta]] Sin[\[Theta]]) + Cos[\[Theta]] (Cos[\[Theta]] fr[r, \[Theta]] - f\[Theta][r, \[Theta]] Sin[\[Theta]])
Cos[\[Theta]] (Cos[\[Theta]] f\[Theta][r, \[Theta]] + fr[r, \[Theta]] Sin[\[Theta]]) - Sin[\[Theta]] (Cos[\[Theta]] fr[r, \[Theta]] - f\[Theta][r, \[Theta]] Sin[\[Theta]])
Solve[{fx == fr*c - ft*s, fy == fr*s + ft*c}, {fr, ft}]
\( \displaystyle \quad \left( fr \to \frac{-c\,fx - fy\,s }{c^2 + s^2} \quad ft \to \frac{-c \, fy + fx\, s}{c^2 + s^2} \right) \)

Here is code that implements the above manipulations to replicate the wedge graphic we started with and the vector field in which it lies. First we define specific functions for fr and fθ:

fr[r_, \[Theta]_] := r Cos[\[Theta]] f\[Theta][r_, \[Theta]_] := r Sin[\[Theta]]
We then define the vector field
f[r_, \[Theta]_] := {fr[r, \[Theta]], f\[Theta][r, \[Theta]]}
Visualize the vector field
vectorField = VectorPlot[ f[Sqrt[x^2 + y^2], ArcTan[x, y]], {x, -1.5, 1.5}, {y, -1.5, 1.5}, VectorScale -> {Small, Automatic, None}, PlotRange -> {{-1.2, 1.2}, {-1.2, 1.2}}, Axes -> True]

Figure 4: Vector field.

Then define the wedge region
wedgeRegion = Polygon[Append[ Table[{Cos[t], Sin[t]}, {t, Pi/6, Pi/3, 0.01}], {0, 0}]]
Finally, overlay the wedge region on the vector field
Show[vectorField, Graphics[{Red, Opacity[0.3], wedgeRegion}], PlotLabel -> "Vector Field with Wedge Region"]
Figure 5: Wedge domain within vector field.

Before formulating Green's theorem, we remind some definitions. A rectifiable curve is a curve having finite length. If curve does not cross itself, it is called a simple curve.

Theorem 1 (Green’s Theorem in polar coordinates): Let R be an open, simply connected region with a boundary curve C = ∂R that is a piecewise smooth, simple closed curve oriented counterclockwise. Let 𝐅 = (𝑃, 𝑄) be a vector field with component functions that have continuous partial derivatives on the closer of R. If the origin belongs to the boundary ∂R of domain R and components of the vector field F satisfy the corner condition
\[ \lim_{r\to 0} r\,F_{\theta} (r, \theta ) = \lim_{(x,y) \to 0} \left( x\,Q - y\,P \right) = 0 , \tag{1} \]
then
\begin{equation} \label{EqGreen.2} \oint_{\partial R} P\,{\text d}r + Q\,r\,{\text d} \theta = \iint_R \left( \frac{\partial \left( r\,Q \right)}{\partial r} - \frac{\partial P}{\partial \theta} \right) {\text d}r\,{\text d}\theta . \end{equation}
Note that
\[ \iint_R \left( \frac{\partial \left( r\,Q \right)}{\partial r} - \frac{\partial P}{\partial \theta} \right) {\text d}r\,{\text d}\theta = \iint_R \mbox{curl}\left( \mathbf{F} \right) \bullet {\bf k}\, {\text d} A , \]
where dA = dr rdθ is the area element, and k = (0, 0, 1) is the unit applicate vector in ℝ³.    
Since Green's theorem was determined previously in rectangular coordinates, we just need to reformulate it in polar coordinates.

Let F be a vector field on the plane: \[ \mathbf{F} = F_r \hat{\bf e}_r + F_{\theta}\hat{\bf e}_{\theta} , \] which we want to integrate around a closed path ∂R containing a simple domain R. First, we make connection polar coordinate representation with its Cartesian one: \begin{align*} \mathbf{F} &= P(x,y)\,{\bf i} + Q(x,y)\,{\bf j} = F_r \hat{\bf e}_r + F_{\theta}\hat{\bf e}_{\theta} \\ &= F_r \left( \cos\theta\,{\bf i} + \sin\theta\,{\bf j} \right) + F_{\theta} \left( -\sin\theta\,{\bf i} + \cos\theta\, {\bf j} \right) \\ &= \left( F_r \cos\theta - F_{\theta} \sin\theta \right) {\bf i} + \left( F_r \sin\theta + F_{\theta} \cos\theta \right) {\bf j} . \end{align*} We know that partial derivatives in rectangular coordinates are expressed through partial derivatives of polar coordinates as \[ \frac{\partial}{\partial x} = \cos\theta\,\frac{\partial}{\partial r} - \frac{\sin\theta}{r} \,\frac{\partial}{\partial \theta} , \qquad \frac{\partial}{\partial y} = \sin\theta\,\frac{\partial}{\partial r} + \frac{\cos\theta}{r} \,\frac{\partial}{\partial \theta} . \] Therefore, \begin{align*} \frac{\partial Q}{\partial x} &= \left( \cos\theta\,\frac{\partial}{\partial r} - \frac{\sin\theta}{r} \,\frac{\partial}{\partial \theta} \right) \left[ F_r \sin\theta + F_{\theta} \cos\theta \right] \\ &= \cos\theta \left( \frac{\partial F_r}{\partial r} \cdot \sin\theta + \frac{\partial F_{\theta}}{\partial r} \cdot \cos\theta\right) \\ & \quad - \frac{\sin\theta}{r} \left( \frac{\partial F_r}{\partial \theta} \cdot \sin\theta + F_r \cos\theta + \frac{\partial F_{\theta}}{\partial \theta} \cdot \cos\theta - F_{\theta} \sin\theta \right) \end{align*} Similarly, \begin{align*} - \frac{\partial P}{\partial y} &= - \left( \sin\theta\,\frac{\partial}{\partial r} + \frac{\cos\theta}{r}\,\sin\theta\,\frac{\partial}{\partial \theta} \right) \left[ F_r \cos\theta - F_{\theta} \sin\theta \right] \\ &= - \sin\theta \left( \frac{\partial F_r}{\partial r} \cdot \cos\theta - \frac{\partial F_{\theta}}{\partial r} \cdot \sin\theta \right) \\ & \quad - \frac{\cos\theta}{r} \left( \frac{\partial F_r}{\partial \theta} \cdot \cos\theta - F_r \sin\theta - \frac{\partial F_{\theta}}{\partial \theta} \cdot \sin\theta - F_{\theta} \cos\theta \right) . \end{align*} Summing up, \begin{align*} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} &= \frac{1}{r} \left( \frac{\partial}{\partial r} \left( r\,F_{\theta} \right) - \frac{\partial F_r}{\partial \theta} \right) = \mbox{curl}(\mathbf{F}) \bullet \mathbf{k} , \end{align*} where k = (0, 0, 1) is a unit vector along applicate axis (usually denoted by z).

   

Example 1: We consider the vector field \[ \mathbf{F} = \frac{1}{x^2 + y^2}\, \left( 1, x \right) \] that does not satisfy the corner condition. We integrate it along a unit circle centered at the origin. Its circulation is finite because the field F is finite on the unit circle. So \[ \oint_C \mathbf{F} \bullet {\text d}\mathbf{r} , \] which we evaluate using parametrization by polar coordinates. \[ \oint_C \mathbf{F} \bullet {\text d}\mathbf{r} = \int_0^{2\pi} \frac{\cos\theta}{1^2}\,{\text d}\theta = 0 . \]
Integrate[Cos[t], {t, 0, 2*Pi}]
0
However, the double integral \[ \iint_{r \le 1} \mbox{curl}\left( \mathbf{F} \right) {\text d}A \] does not exist.    ■
End of Example 1
   
Example 2: We consider the line integral \[ \oint_C x^2 y\,{\text d}x - 2x\,\cos y\,{\text d}y , \] where 𝐶 is the boundary of the plane domain Ω formed by the circle x² + y² = 1 and below parabola y = x².

We use Green's theorem.

P[x_, y_] := x^2 y; Q[x_, y_] := -2 x Cos[y]; region = ImplicitRegion[x^2 + y^2 < 1 && y < x^2, {x, y}];
First, we visualize the region of integration:
Region @ region
Example 2: Domain of integration.

Now we perform integration:
NIntegrate[D[Q[x, y], x] - D[P[x, y], y], {x, y} ∈ region]
-4.39276
or
NIntegrate[(D[Q[x, y], x] - D[P[x, y], y])* Boole[x^2 + y^2 < 1 && y < x^2], {x, -1, 1}, {y, -1, 1}]
   ■
End of Example 2
   
Example 3:
PolarPlot[ Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t] + 7/5) - 2 Sin[t] + 2, {t, 0, 2 Pi}, PlotStyle -> Red, Epilog -> {Red, Opacity[0.3], Polygon[Table[{r Cos[t], r Sin[t]} /. r -> Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t] + 7/5) - 2 Sin[t] + 2, {t, 0, 2 Pi, 0.01}]]} ]
Figure 3: The heart domain.

polarFunction[t_] := Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t] + 7/5) - 2*Sin[t] + 2 (* Convert to Cartesian coordinates *) cartesianPoints = Table[ {polarFunction[t] Cos[t], polarFunction[t] Sin[t]}, {t, \[Pi]/2, 3 \[Pi]/2, 0.01} ]; (* Rotate the points by 90 degrees clockwise *) rotatedPoints = cartesianPoints /. {x_, y_} :> {y, -x};
Plot the rotated points
ListLinePlot[rotatedPoints, PlotStyle -> Red, Axes -> False, PlotRange -> All, AspectRatio -> Automatic, Filling -> Axis]
Half-Heart, tail left
     
Half-Heart, tail right

Define the parametric equations for the rotated heart - like curve

x[t_] := Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t] + 7/5) - 2 Sin[t] + 2 y[t_] := -((Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t] + 7/5) - 2 Sin[t] + 2) Cos[t])
Compute the derivatives
xPrime[t_] = D[x[t], t] yPrime[t_] = D[y[t], t]
Compute the area using NIntegrate
area = NIntegrate[(-xPrime[t]/2) y[t] + (yPrime[t]/2) x[t], {t, \[Pi]/ 2, 3 \[Pi]/2}]
-3.41572
Negative answer indicates that our orientaion is wrong---it is clockwise. So we go from t to −t, which leads to positive answer:
NIntegrate[(-xPrime[t]/2) y[t] + (yPrime[t]/2) x[t], {t, 3 \[Pi]/2, \[Pi]/2}]
3.41572
   ■
End of Example 3

 

 

  4. Coolidge, J.L., The Origin of Polar Coordinates, American Mathematical Monthly. 1952, volume 59, issue 2, pp. 78–85. doi:10.2307/2307104.
  6. Fleming, W., Functions of Several Variables, Undergraduate Texts in Mathematics, Second edition, Springer-Verlag, New York–Heidelberg, 1977.
  7. Hubbard, J.H. and Hubbard, B.B., Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach, Matrix Editions; 5th edition, 2015; ISBN-13: ‎ 978-0971576681