Area of the region bounded by the line $y=−x$ and the parabola $r= \frac{1}{1+\cos(\theta)}$ 
Express the area of the region bounded by the line  $y=−x$  and the parabola  $r= \frac{1}{1+\cos(\theta)}$  as an integral in polar coordinates. (Choose limits of integration in the range  $(−\pi,\pi)$ . The integral must evaluate to be positive. Leave expression in the integral form.)

I know that I have to subtract the function on top with the function of the bottom, however I think both equations must both be either in polar coordinates or cartesian coordinates. Any hints on how to do this?
 A: Hint: Your statement that we should "subtract the function on top with the function of the bottom" is incorrect in the context of polar integration. Since we take $r$ as a function of $\theta$ in this context, we subtract the (differential area bounded by the) function with the larger $r$ with the (differential area bounded by the) function with a smaller $r$, which is to say we integrate the "outside function" minus the "inside function". 
For this problem, however, there is only one function that gives us the $r$-limits of the region; the curve $y = -x$ only serves to give us the maximal and minimal values of $\theta$.  So, we will end up with an integrand that only involves the function $r(\theta) = \frac 1{1 + \cos \theta}$.
It is important to note that the region lies to the right side of the line $y = -x$.
A: 
In polar coordinates, the line $ y=-x$ becomes $\tan\theta =-1$, which defines the angular limits $[-\frac\pi4,\frac{3\pi}4]$ for the enclosed area. Then, the area is integrated as
$$A=\int_{-\frac\pi4}^{\frac{3\pi}4 }\frac12r^2(\theta)d\theta
= \frac12\int_{-\frac\pi4}^{\frac{3\pi}4 }\frac1{(1+\cos\theta)^2}d\theta =\frac{4\sqrt2}3
$$
