Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]

Printable View

- Jul 21st 2007, 04:30 PMcamherokidPolar Coordinate 03
Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x]

- Jul 21st 2007, 05:43 PMSoroban
Hello, camherokid!

Quote:

Find the area inside $\displaystyle r \,= \,3\cos x$ and outside $\displaystyle r \:= \:1 + \cos x$

$\displaystyle r\:=\:3\cos x$ is a circle with center $\displaystyle \left(\frac{3}{2},\,0\right)$ and radius $\displaystyle r = \frac{3}{2}$

$\displaystyle r\:=\:1 + \cos x$ is a cardioid with intercepts: $\displaystyle (2,\,0),\: \left(1,\,\frac{\pi}{2}\right),\:\left(1,\,\frac{3 \pi}{2}\right)$

. . and "dimples in" to the origin from the left.

The polar formula for the area between two curves is: .$\displaystyle A \;=\;\frac{1}{2}\int^{\beta}_{\alpha}\left(r_{_2}^ 2 - r_{_1}^2\right)\,d\theta$

The curves intersect when: .$\displaystyle 3\cos x \;=\;1 + \cos x\quad\Rightarrow\quad 2\cos x \:=\:1\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2}$

. . Hence, they intersect at: .$\displaystyle \theta \:=\:\pm\frac{\pi}{3}$

Therefore: .$\displaystyle A \;=\;\frac{1}{2}\int^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\bigg[(3\cos\theta)^2 - (1 + \cos x)^2\bigg]\,d\theta$

- Jul 21st 2007, 05:52 PMJhevon
We will use the formula:

$\displaystyle A = \int_{ \alpha}^{ \beta} \frac {1}{2} (r_o^2 - r_i^2)~dx$

where $\displaystyle A$ is the area between the curves $\displaystyle r_o$ and $\displaystyle r_i$, $\displaystyle \alpha$ and $\displaystyle \beta$ are the limits of integration (the points of intersection), $\displaystyle r_o$ is the outer curve, and $\displaystyle r_i$ is the inner curve.

First find the points of intersection:

this is where $\displaystyle 3 \cos x = 1 + \cos x$

$\displaystyle \Rightarrow \cos x = \frac {1}{2}$

$\displaystyle \Rightarrow x = \frac {\pi}{3}, \frac {5 \pi}{3}$

we want to go from $\displaystyle \frac {5 \pi}{3}$ to $\displaystyle \frac {\pi}{3}$, but we must go from a smaller angle to a bigger angle. changing $\displaystyle \frac {5 \pi}{3}$ to $\displaystyle - \frac {\pi}{3}$ fixes this problem. so our area is given by:

$\displaystyle A = \int_{- \pi / 3}^{ \pi / 3} \frac {1}{2} \left[ (3 \cos x)^2 - (1 + \cos x )^2 \right]~dx$

EDIT: Beaten by Soroban! any way, that's ok. I'm a bit rusty on polar areas, so it's good to have a confirmation that I did the right thing. Soroban, can you check the other posts camherokid put up today and make sure I didn't make any mistakes - Jul 21st 2007, 06:00 PMcurvature