calculusintegrationpolar coordinates
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.
If not, how can I find this area?
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.
The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \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
Please always try and draw a sketch. It helps in ensuring you have the right bounds and also understand the easy way to find the area.
Now in the sketch, you can see that there are two circles -
i) Centered at $(0,0)$ with radius $1$
ii) Centered at $(3/2, 0)$ with radius $3/2$
As they are both symmetric to X-axis, they will intersect at the same angle in both first and fourth quadrant. Say that angle is $\alpha$. The Intersection points of both circles will be given by equating -
$3 \cos \alpha = 1$
$\alpha = cos^{-1} ({\frac{1}{3}}) \approx \frac{2\pi}{5}$ (I have taken as $2 \pi / 5$ but it is closer $1.231$. Use $1.231$ for more accurate area).
We have to find the area between two curves (thru points $O, A, B, C$).
a) Integrating the curve $r = 1$ over $\angle AOC$ will give us area bound by radii $OA, OC$ and arc $ABC$ (sector $OAC$ for circle $r = 1$).
b) Integrating the curve $r = 3 \cos \theta$ over angle between $Y$ axis and $OA$ and between $Y$ axis and $OC$ will give us area bound by chord $OA, OC$ and arc $AOC$.
If we add both, we get the area we desire.
a) $A_1 = \displaystyle \frac{1}{2} \int_{-\alpha}^{\alpha}d\theta = \frac{2\pi}{5}$
b) $A_2 = \displaystyle 2 \times \frac{1}{2} \int_{\alpha}^{\pi/2}(3 \cos \theta)^2 d\theta$
$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} 2 \cos^2 \theta \, d\theta$
$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} (1 + \cos2 \theta) \, d\theta$
$ = \displaystyle \frac{9}{2} [\theta + \frac{1}{2} \sin 2 \theta)]_{2\pi/5}^{\pi/2}$
$ = \displaystyle \frac{9}{2} [\frac{\pi}{2} - \frac{2\pi}{5} - \frac{1}{2}\sin \frac{4\pi}{5}]$
$A = A_1 + A_2 \approx 1.257 + 0.063 = 1.32$