# What is the area enclosed by r=-cos(theta-(7pi)/4)  between theta in [4pi/3,(5pi)/3]?

Feb 17, 2017

$= \frac{\pi}{12} - \frac{1}{8} \sqrt{1.5} = 0.1087$, nearly.

#### Explanation:

The graph is a circle of radius 1/2, through pole, with center on

$\theta = \frac{5}{4} \pi$.

Despite that the non-negative r = sqrt(x^2+y^2) is $- \cos {15}^{o} < 0$, at

the higher limit $\theta = \frac{5}{3} \pi$, I work out for the answer.

graph{(sqrt2(x^2+y^2)+x+y)(y+sqrt3x)(y-sqrt3x)=0 [-2.5, 2.5, -1.25, 1.25]}

Area= $\frac{1}{2} \int {r}^{2} d \theta$, with$r = - \cos \left(\theta - \frac{7}{4} \pi\right) \mathmr{and} \theta$ f

from $\frac{4}{3} \pi$ to $\frac{5}{3} \pi$

$= \frac{1}{2} \int {\cos}^{2} \left(\theta - \frac{7}{4} \pi\right) d \theta$, for the limis

=1/4int(1+cos(2theta-7/4pi) d theta, for the limite

$= \frac{1}{4} \left[\theta + \frac{1}{2} \sin \left(2 \theta - \frac{7}{4} \pi\right)\right]$, between $\theta = \frac{4}{3} \pi \mathmr{and} \frac{5}{3} \pi$

=1/4[(5/3pi-4/3pi)+1/2(sin(19/12pi)-sin(11/12pi)]

$= \frac{1}{4} \left(\frac{\pi}{3} + \cos \left(\frac{5}{4} \pi\right) \sin \left(\frac{\pi}{3}\right)\right)$

$= \frac{\pi}{12} + \frac{1}{4} \left(\left(- \frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right)\right)$

$= \frac{\pi}{12} - \frac{1}{8} \sqrt{1.5}$