Trying to compute area inside a region question: Use double integral to compute the area of the region inside the cardioid $r=1+\cos(\theta)$ and to the right of the line $x=3/4$.
So my first integral will be from $3/4$ to $2\pi$ and my second one will be $0$ to $1+\cos(\theta)$ with a function of $r dr$.
is that correct?
 A: Almost certainly not correct. It looks as if you're saying you plan to compute
$$
\int_{\theta = 3/4}^{2\pi} \int_{r =0}^{1 + \cos \theta} r ~dr ~ d\theta.
$$
The inner integral looks fine, but the outer integral... no. Where did the $\theta = 3/4$ come from? Where did the $2\pi$ come from? If you have a point given by  $r = 1 + cos(\theta)$, then the $x$ coordinate is 
$$
r \cos \theta = (1 + \cos(\theta)) \cos(\theta).
$$ 
You want to know the values of $\theta$ that make this equal 3/4, because those will be your limits of integration. So write
$$
(1 + \cos(\theta)) \cos(\theta) = 3/4 \\
\cos (\theta) + \cos^2(\theta) - 3/4 = 0
$$
Now let $u = \cos(\theta)$ and you've got
$$
u^2 + u - 3/4 = 0
$$
or 
$$4u^2 + 4u - 3 = 0$$
whose solutions are 
$$
u= \frac{-4 \pm \sqrt{16 +48}}{8} \\
= \frac{-4 \pm \sqrt{64}}{8}\\ 
= \frac{-4 \pm 8}{8}\\
= 1/2, -3/2
$$
There's no value of $\theta$ for which $\cos(\theta) = -3/2$, but there are two values for which $\cos(\theta) = 1/2$, namely $\pm \frac{\pi}{3}$. So your limits of integration become $\theta = -\pi/3 \text{ to } \pi/3$: 
$$
\int_{\theta = -\pi/3}^{\pi/3} \int_{r =0}^{1 + \cos \theta} r ~dr ~ d\theta.
$$
And now I'm going to let you do the integration yourself. 
A: Another way to get the values for $\;\theta\;$:
Note that passing to polar coordinates:
$$x=\frac34\iff r\cos\theta=\frac34$$
and thus calculating the intersection poins with the given function
$$\frac3{4\cos\theta}=1+\cos\theta\iff4\cos^2\theta+4\cos\theta-3=0$$
anmd solving this quadratic we get
$$\cos\theta=\begin{cases}-\frac{12}8=-\frac32\\{}\\\frac48=\frac12\end{cases}$$
Since the first option is clearly absurd, we get
$$\cos\theta=\frac12\iff\theta=\pm\frac\pi3$$
Now just note you can take the integral over $\;2\int\limits_0^{\frac\pi3} d\theta\;$ since the area wanted is symmetric...
