I have to find the area of the region that lies between the curves $r=1+\sin\theta; r=1+\cos\theta$ . The answer the book gave was
$\frac {3\pi}{2}-2\sqrt{2}$ .
I tried generating the curve for $r=1+\cos\theta$ from $\frac {\pi}{4}$ to $\pi$, which generated the proper part of the curve, and then generating the curve for $r=1+\sin\theta$ from $\frac {-3\pi}{4}$ to $\frac {\pi}{4}$ which generated the correct part of that curve. I then attempted to put that into the formula given for area between polar curves, $A=2(\int_\alpha^\beta \frac {1}{2}f(\theta)^2 d\theta + \int_\gamma^\alpha \frac {1}{2}g(\theta)^2 d\theta)$ and used the bounds I previously used to generate my curve as the bounds on the integrals, respectively. However, I still am unable to get the answer the book gave. I have thought of trying to divide the section into two and find the area of one half and then doubling my result to find the area of the whole, but I am unsure of how to go about that. Does anyone have any suggestions? The point of intersection I found for the curves is $(1+ \frac {\sqrt{2}}{2} , \frac {\pi}{4})$.
[Math] Find the area of the entire region that lies between $r=1+\sin\theta; r=1+\cos\theta$
calculusintegrationpolar coordinates
Related Solutions
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.
The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1. $$
Best Answer
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.