# What is the area enclosed by r=8sin(3theta-(2pi)/4) +4theta between theta in [pi/8,(pi)/4]?

May 31, 2018

$\textcolor{b l u e}{A = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{4}} \left[8 {\left(\theta\right)}^{2} - 32 \theta \cos \left(3 \theta\right) + 32 {\cos}^{2} \left(3 \theta\right)\right] \cdot d \left(\theta\right) = 4.68224}$

#### Explanation:

the area of the polar curve given by:

color(red)[A=1/2int_(theta_1)^(theta_2)r^2*d(theta)

The interval of the integral $\theta \in \left[\frac{\pi}{8} , \frac{\pi}{4}\right]$

now lets setup the integral in our interval:

$A = \frac{1}{2} {\int}_{\frac{\pi}{8}}^{\frac{\pi}{4}} {\left(8 \sin \left(3 \theta - \frac{2 \pi}{4}\right) + 4 \theta\right)}^{2} \cdot d \left(\theta\right)$

$A = \frac{1}{2} {\int}_{\frac{\pi}{8}}^{\frac{\pi}{4}} {\left(4 \theta - 8 \cos \left(3 \theta\right)\right)}^{2} \cdot d \left(\theta\right)$

$A = \frac{1}{2} {\int}_{\frac{\pi}{8}}^{\frac{\pi}{4}} {\left(4 \theta - 8 \cos \left(3 \theta\right)\right)}^{2} \cdot d \left(\theta\right)$

$A = {\int}_{\frac{\pi}{8}}^{\frac{\pi}{4}} \left[8 {\left(\theta\right)}^{2} - 32 \theta \cos \left(3 \theta\right) + 32 {\cos}^{2} \left(3 \theta\right)\right] \cdot d \left(\theta\right)$

$= {\left[\frac{24 \sin \left(6 \theta\right) - 96 x \sin \left(3 \theta\right) - 32 \cos \left(3 \theta\right) + 24 {\theta}^{3} + 144 \theta}{9}\right]}_{\frac{\pi}{8}}^{\frac{\pi}{4}}$

$\left[\frac{3 \cdot {2}^{\frac{21}{2}} \pi \sin \left(\frac{3 \pi}{8}\right) + {2}^{\frac{27}{2}} \cdot \cos \left(\frac{3 \pi}{8}\right) + 21 \cdot {2}^{\frac{5}{2}} {\pi}^{3} + \left(9 \cdot {2}^{\frac{19}{2}} - 6144\right) \pi - 3 \cdot {2}^{\frac{23}{2}} + 2048}{9 \cdot {2}^{\frac{17}{2}}}\right] = 4.68224$

Approximation:
$A = 4.68224$

see below the sketch of the polar curve: see below the region bounded by the curve from $\frac{\pi}{8}$ to$\frac{\pi}{4}$ $\text{I hope that helpful}$