Why is the formula for the area of a cardioid $ \int_a^b \frac{1}{2} r^2 d \theta$

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Solution 1

There are at least two way to look at this formula. The simplest, in my opinion, is approximating the area inside the arc into small circular arcs.

$\hspace{3.5cm}$enter image description here

The area inside each arc of radius $r$ and angle $\Delta\theta$ is $\frac12r^2\Delta\theta$ (this is proportional to $\pi r^2$ for the entire circle). Summing these approximations as $\Delta\theta\to0$ yields $$ \int\frac12r^2\mathrm{d}\theta $$


Another approach is to look at the Jacobian of the conversion from rectangular to polar coordinates. $$ \mbox{ $\begin{align} x&=r\cos(\theta)\\ y&=r\sin(\theta) \end{align} $} \quad\text{and}\quad \mbox{$ \begin{align} \mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\ \mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta \end{align} $} $$ That is $$ \begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix} =\begin{bmatrix}\cos(\theta)&-r\sin(\theta)\\\sin(\theta)&r\cos(\theta)\end{bmatrix} \begin{bmatrix}\mathrm{d}r\\\mathrm{d}\theta\end{bmatrix} $$ The determinant of the Jacobian yields $$ \begin{align} \mathrm{d}x\,\mathrm{d}y &=r(\cos^2(\theta)+\sin^2(\theta))\,\mathrm{d}r\,\mathrm{d}\theta\\ &=r\,\mathrm{d}r\,\mathrm{d}\theta \end{align} $$ if we integrate in $r$ from the origin first, this yields the area to be $$ \int\frac12r^2\,\mathrm{d}\theta $$

Solution 2

Answer based on my answer to the question Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$.

1. Area in Cartesian and in polar coordinates. Let $T$ be a bounded closed region of the $xy$-plane. If we decompose $T$ into $n$ rectangular cells the area $A$ of $R$ is the limit

$$ A=\lim_{n\to\infty}\; \sum_{i=1}^{n }\Delta A_{i}$$

which by definition of a double integral is equal to

$$A=\iint_T \mathrm{d}x\;\mathrm{d}y,\tag{1}$$

where $\Delta A_{i}$ is the area of a generic rectangular cell. The limits of integration are to be found in terms of $x,y$.

enter image description here

Figure 1: Generic $i^{th}$-cell in polar coordinates with the shape of a circle sector

Let's compute the same area using polar coordinates $r,\theta$. If we decompose $T$ into cells with a shape of sectors of a circle defined by two radii whose difference is $\Delta r_{i}$ for the generic $i^{th}$ cell and two rays making an angle $\Delta \theta _{i}$ with each other, the area of the cell (see figure 1 above), using the formula of a circle sector, is

$$\frac{1}{2}\left[ \left( r_{i}+\frac{1}{2}\Delta r_{i}\right) ^{2}-\left( r_{i}-\frac{1}{2}\Delta r_{i}\right) ^{2}\right] \Delta \theta _{i}=r_{i}\Delta r_{i}\Delta \theta _{i},$$

where $r_{i}$ is the radius of the middle point of the cell. The same area $A$ can be expressed as the limit

$$A=\lim_{n\to\infty}\;\sum_{i=1}^{n }r_{i}\Delta r_{i}\Delta \theta _{i},$$

which by definition of a double integral is equal to

$$A=\iint_T r\;\mathrm{d}r\;\mathrm{d}\theta. \tag{2}$$

The limits of integration of $T$ are expressed in terms of $r$ and $\theta$. The relation between Cartesian and polar coordinates is

$$x=r\cos\theta,\qquad y=r\sin\theta.$$

In 2. and 3. we evaluate $(2)$ when $T$ is a circle and a cardioid.

2. Example. Area of a circle with radius $R$. Observing that $r\in [0,R]$ and $\theta\in[0,2\pi]$ the integral $(2)$ reduces to

$$\begin{eqnarray*}A_{\text{circle}}=\int_{0}^{2\pi }\left( \int_{0}^{R}r\ \mathrm{d}r\right) \mathrm{d}\theta &=&\int_{0}^{2\pi }\frac{R^{2}}{2}\ \mathrm{d}\theta\tag{$\ast$}\\ &=&\pi R^{2}.\end{eqnarray*}$$

Equation $(\ast)$ corresponds to the formula in the question. In this case $R$ is a constant.

3. Area of a Cardioid. To compute the area of a cardioid with equation $r=4a\cos ^{2}\frac{\theta }{2}$ (see figure 2 below) we need to find the limits of integration. From the figure we see that $r\in [0,4a\cos ^{2}\frac{\theta }{2}]$ and $\theta\in[0,2\pi]$. The integral $(2)$ reduces to

$$\begin{eqnarray*} A_{\text{cardioid}}=\int_{0}^{2\pi }\left( \int_{0}^{4a\cos ^{2}\frac{\theta }{2}}r\ \mathrm{d} r\right)\ \mathrm{d}\theta &=&\int_{0}^{2\pi }\left( \left. \frac{r^{2}}{2}\right\vert _{0}^{4a\cos ^{2}\frac{\theta }{2}}\right)\ \mathrm{d}\theta\tag{$\ast\ast$} \\ &=&\int_{0}^{2\pi }\left( 8a^{2}\cos ^{4}\frac{\theta}{2} \right)\ \mathrm{d}\theta \\ &=&6\pi a^{2}. \end{eqnarray*}$$

Equation $(\ast\ast)$ corresponds to the formula in the question. Now $r$ is not constant, rather a function of $\theta$.

enter image description here

$$\text{Figure 2: cardioid }r=4a\cos ^{2}\frac{\theta }{2}$$

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Added: The factor $r$ in $(2)$ is the Jacobian of the transformation. Here is an evaluation: $$\begin{eqnarray*} \frac{\partial \left( x,y\right) }{\partial \left( r,\theta \right) } &=&\det \begin{pmatrix} \partial x/\partial r & \partial x/\partial \theta \\ \partial y/\partial r & \partial y/\partial \theta \end{pmatrix} \\ &=&\det \begin{pmatrix} \cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{pmatrix} \\ &=&r\cos ^{2}\theta +r\sin ^{2}\theta \\ &=&r. \end{eqnarray*}$$

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Updated on May 31, 2020

Comments

  • Person
    Person over 3 years

    I've seen this expression in many places :$\int_a^b \frac{1}{2} r^2 d \theta$ and was wondering if someone can explain where this came from? I've noticed that it's sometimes explained in conjunction with double integrals but I'm not very familiar with double integrals (as in what it means using a picture).

    • imranfat
      imranfat over 10 years
      This formula is not just for the area for a cardioid, but a formula in general to calculate area of a polar equation. The derivation of this formula is based on adding up thinly sliced circle sectors drawn from the Origin to the curve, in the form of a Riemann sum. Look up the formula of the sector of a circle, where the sector spans a central angle measured in radians, and see the similarity between both formulas
    • Person
      Person over 10 years
      your explanation makes sense
    • Erick Wong
      Erick Wong over 10 years
      The area of a "pie wedge" of radius $r$ and angle $d\theta$ is exactly $\tfrac12 r^2 d\theta$.
  • Person
    Person over 10 years
    I just have an issue with the geometry of a double integral. What is $\rho d\rho$? I need a solid picture in my head to see what you're saying and what does it mean to integrate over two variables pictorially?
  • John Douma
    John Douma over 10 years
    Look at this They use $r$ instead of $\rho$. I used $\rho$ so not to confuse it with the $r$ in the final answer.
  • Muphrid
    Muphrid over 10 years
    It's may be best to group up $\rho \, d\theta$ together; this is the length of the circular arc spanning an angle $d\theta$.