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
Ways of computing a double integral

The typical way to evaluate a double integral is to express it as an iterated integral.
To do that, we need to choose proper variables.
Double integrals in polar coordinates

Because $\rho$ and $R$ are simple to describe with $r$ and $\theta$, we write the iterated integral in polar coordinates.
Double integrals in polar coordinates

Recall that in polar coordinates:$$dA = r \ dr \ d\theta.$$
The integral has become $$M = \iint_R \frac{1}{r} r dr d\theta.$$
Recall that
Setting up an iterated integral

To express $\iint_R \rho \ dA$ as an iterated integral, $R$ must be expressed in one of two ways:
$r$ varies from $r_\text{min}$ to $r_\text{max}$, and $\theta$ varies from $\theta_\text{min}(r)$ to $\theta_\text{max}(r)$.

$\theta$ varies from $\theta_\text{min}$ to $\theta_\text{max}$, and $r$ varies from $r_\text{min}(\theta)$ to $r_\text{max}(\theta)$.
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 $$ M= 2\pi a.$$

Problem on a double integral over a cardioid

Solution

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