# What is the area enclosed by r=sintheta/theta-theta^3-theta  between theta in [pi/12,pi/3]?

May 26, 2017

Signed area $\approx - 0.08885 \ldots$

#### Explanation:

Use the Riemann integral.
Area under $f \left(x\right)$ from $\left[a , b\right]$ is ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

So, the area under $r \left(\theta\right) = \frac{\sin \theta}{\theta} - {\theta}^{3} - \theta$ where $\theta \in \left[\frac{\pi}{12} , \frac{\pi}{3}\right]$ is ${\int}_{\frac{\pi}{12}}^{\frac{\pi}{3}} r \left(\theta\right) \setminus \quad d \theta$

Let's integrate the indefinite and ignore the constant:

$\int \setminus \quad r \left(\theta\right) \setminus \quad d \theta$
$= \int \setminus \quad \frac{\sin \theta}{\theta} - {\theta}^{3} - \theta \setminus \quad d \theta$
$= \int \setminus \quad \frac{\sin \theta}{\theta} \setminus \quad d \theta - \int \setminus \quad {\theta}^{3} \setminus \quad d \theta - \int \setminus \quad \theta \setminus \quad d \theta$

Now, $\int \setminus \quad \sin \frac{\theta}{\theta} \setminus \quad d \theta = \text{Si} \left(\theta\right) + C$ which is a very interesting function. We have entered the lair of Trigonometric Integrals. There is no closed form of $\text{Si}$, so we would have to deal with $\text{Si"(pi/3)-"Si} \left(\frac{\pi}{12}\right) \approx 0.724654 \ldots$

Anyways, we now integrate the (trivial) rest, again, ignoring the constant

$\int \setminus \quad {\theta}^{3} \setminus \quad d \theta = \frac{{\theta}^{4}}{4}$

$\int \setminus \quad \theta \setminus \quad d \theta = \frac{{\theta}^{2}}{2}$

Finally, we piece everything together:
${\int}_{\frac{\pi}{12}}^{\frac{\pi}{3}} r \left(\theta\right) \setminus \quad d \theta = | \left[\text{Si} \left(\theta\right) - \frac{\left({\theta}^{2}\right) \left({\theta}^{2} + 2\right)}{4}\right] {|}_{\frac{\pi}{12}}^{\frac{\pi}{3}}$
$\approx 0.724564 \ldots - \frac{{\left(\frac{\pi}{3}\right)}^{2} \left({\left(\frac{\pi}{3}\right)}^{2} + 2\right)}{4} + \frac{{\left(\frac{\pi}{12}\right)}^{2} \left({\left(\frac{\pi}{12}\right)}^{2} + 2\right)}{4}$
$\approx 0.724564 \ldots - 0.813512 \ldots$

$\approx - 0.08885 \ldots$

$\therefore$ The (unsigned) area under $r \left(\theta\right)$ is approx. $0.0889$