## Problem on a double integral over a cardioid

Let $\rho(r,\theta) = 1/r$ and let $R$ be the cardioid given by $r=a(1+\cos \theta)$ for positive $a$. Evaluate $$\iint_R \rho \ dA.$$
• ## Solution

For convenience, we let $M = \iint_R \rho \ dA$.
Recall that
To do that, we need to choose proper variables.
The integral has become $$M = \iint_R \frac{1}{r} r dr d\theta.$$
Recall that
Before we decide which way to use, we draw $R$:
Because we are given the shape of the cardioid in terms of $r(\theta)$, we choose the second way of specifying $R$. Our double integral will then have outer variable $\theta$ and inner variable $r$.
We identify that $\theta$ varies from $0$ to $2 \pi$. For a given $\theta$, $r$ varies from $0$ to $a(1+\cos(\theta))$.
We can now finish computing the double integral: \begin{align} M &= \int_{\theta=0}^{2\pi} \int_{r=0}^{a(1+\cos \theta)} \frac{1}{r} r dr d\theta \\ &= \int_{\theta=0}^{2\pi} \int_{r = 0}^{a(1+\cos \theta)} dr d\theta. \end{align}
Evaluating the inner integral gives \begin{align}
M &= \int_0^{2\pi} (a + a \cos \theta) d\theta \\
\end{align}
which becomes $$M= 2\pi a.$$