1. ## Polar Coordinate 02

Find the area enclose by one loop of the curve
r= 2 Cos[x] - Sec[x]

2. Originally Posted by camherokid
Find the area enclose by one loop of the curve
r= 2 Cos[x] - Sec[x]
We need the limits of integration first. since the loop goes through the origin, we need to find the values of x for which the curve passes through the origin, so set r = 0

we have, $2 \cos x - \frac {1}{ \cos x} = 0$

$\Rightarrow 2 \cos^2 x - 1 = 0$

$\Rightarrow \cos x = \pm \frac {1}{ \sqrt { 2 }}$

$\Rightarrow x = \frac {\pi}{4}, \frac {3 \pi}{4}, \frac { 5\pi}{4}, \frac {7 \pi}{4}$

looking at the graph, we see we want the region between $\frac {\pi}{4}$ and $\frac {7 \pi}{4}$

we want to go anticlockwise from $\frac {7 \pi}{4}$ to $\frac {\pi}{4}$ to get the desired area, but we must go from a smaller to a larger angle. so rewrite $\frac {7 \pi}{4}$ as $- \frac { \pi}{4}$ and we are in business

So our desired area is given by

$\int_{- \pi / 4}^{ \pi / 4}\frac {1}{2} r^2 ~dx$

$= \int_{- \pi / 4}^{ \pi / 4}\frac {1}{2} \left( 2 \cos x - \sec x \right)^2 ~dx$

3. That looks good, Jhevon. You could also just write it as:

$\int_{0}^{\frac{\pi}{4}}(2cos\theta-sec\theta)^{2}d{\theta}$

Same thing.

4. Originally Posted by galactus
That looks good, Jhevon. You could also just write it as:

$\int_{0}^{\frac{\pi}{4}}(2cos\theta-sec\theta)^{2}d{\theta}$

Same thing.
yes, that is true. thanks