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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?

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    $\begingroup$ Your parabola has equation $x=\frac{1}{2}(1-y^2)$. ;-) $\endgroup$ – Pacciu May 24 '20 at 23:56
  • $\begingroup$ @Pacciu Would you mind explaining why is this? $\endgroup$ – user792292 May 25 '20 at 0:12
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    $\begingroup$ 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). $\endgroup$ – Pacciu May 25 '20 at 0:21
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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

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

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  • $\begingroup$ Which tool did you used for graphing? $\endgroup$ – user792292 May 25 '20 at 16:28
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    $\begingroup$ @Diego - Mathematica $\endgroup$ – Quanto May 25 '20 at 16:30
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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 $y = -x$.

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