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

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 $x,y$.

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 $\theta$.

$$\text{Figure 2: cardioid }r=4a\cos ^{2}\frac{\theta }{2}$$
--
Added: The factor $r$ in $(2)$ is the Jacobian of the transformation. Here is an  evaluation: 
$$\begin{eqnarray*}
\frac{\partial \left( x,y\right) }{\partial \left( r,\theta \right) }
&=&\det 
\begin{pmatrix}
\partial x/\partial r & \partial x/\partial \theta  \\ 
\partial y/\partial r & \partial y/\partial \theta 
\end{pmatrix}
\\
&=&\det 
\begin{pmatrix}
\cos \theta  & -r\sin \theta  \\ 
\sin \theta  & r\cos \theta 
\end{pmatrix}
\\
&=&r\cos ^{2}\theta +r\sin ^{2}\theta  \\
&=&r.
\end{eqnarray*}$$
