# 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?

• Your parabola has equation $x=\frac{1}{2}(1-y^2)$. ;-) – Pacciu May 24 '20 at 23:56
• @Pacciu Would you mind explaining why is this? – user792292 May 25 '20 at 0:12
• Diego, it is really simple algebra. Try to compute $1-y^2$ with $y=r \sin \theta = \frac{\sin \theta}{1 + \cos \theta}$ (by polar coordinates): you will find it equals $2x$ with $x=r \cos \theta = \frac{\cos \theta}{1+\cos \theta}$ (again by polar coordinates). – Pacciu May 25 '20 at 0:21

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
$$A=\int_{-\frac\pi4}^{\frac{3\pi}4 }\frac12r^2(\theta)d\theta = \frac12\int_{-\frac\pi4}^{\frac{3\pi}4 }\frac1{(1+\cos\theta)^2}d\theta =\frac{4\sqrt2}3$$
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 $$y = -x$$.