Aim of this **question** is to understand how the definite **integrals** can be applied to **calculate** the area enclosed by the one **curve** of the loop and area **in-between** the 2 two curves by **applying** the **calculus** methods.

Between two points **area** under a curve can be **found** by doing a definite **integral** of **range** **a** to **b**. **Area** under the **curve** y = f(x) between the **range** **a** and **b** is **calculated** as:

\[ A = \int_a^b f(x) dx \]

**Area** between the two **curves** can be found, if there **functions** and the **limits** are known. Area that **falls** between **function** $g(x)$ and **function** $f(x)$ from **range** $a$ to $b$ is **calculated** as:

\[ A =\int_a^b (f(x) – g(x)) dx \]

## Expert Answer

Given the **curve** is $r = sin(12 \theta)$

The range of the $\theta$ for one loop is $0 \leq \theta \geq \dfrac{\pi}{12}$

The formula of **Area** $(A)$ is given as:

\[ A = \underset{\theta}{\int} \dfrac{1}{2} r^2 d\theta \]

Inserting the **limits** and the $r$:

\[ A = \int_0^{\dfrac{\pi}{12}} \space \dfrac{1}{2} (sin(12 \theta))^2 d\theta \]

\[ A = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space sin^2(12 \theta) d\theta \]

Using the formula:

\[ sin^2x = \dfrac{1-cos2x}{2} \]

\[ A = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space \dfrac{1-cos(24 \theta)}{2} d\theta \]

\[ = \dfrac{1}{2} \int_0^{\dfrac{\pi}{12}} \space \dfrac{1-cos(24 \theta)}{2} d\theta \]

\[ = \dfrac{1}{2} \left[ \int_0^{\dfrac{\pi}{12}} \space \dfrac{1}{2} d \theta \space – \space \int_0^{\dfrac{\pi}{12}} \space \left( \dfrac{1-cos(24 \theta)}{2} \right) d\theta \right] \]

Integrating with respect $d \theta$:

\[ A = \dfrac{1}{2} \left[ \left( \dfrac{\theta}{2} \right)_0^{\dfrac{\pi}{12}} \space – \space \left( \dfrac{1-sin(24 \theta)}{2(24)} \right)_0^{\dfrac{\pi}{12}} \right] \]

\[ = \dfrac{1}{2} \left[ \left( \dfrac{\pi/12}{2} – \dfrac{0}{2} \right) \space – \space \left( \dfrac{1-sin(24 \dfrac{\pi}{12})}{48} \space – \space \dfrac{1-sin(24 (0))}{48} \right) \right] \]

\[ = \dfrac{1}{2} \left[ \left( \dfrac{\pi}{24} \right) \space – \space \left( \dfrac{\pi}{24} – \dfrac{\pi}{24} \right) \right] \]

\[ = \dfrac{1}{2} \left[ \left( \dfrac{\pi}{24} \right) \right] \]

\[ A = \dfrac{\pi}{48} \]

## Numerical Answer:

Area of the **region** enclosed by one **loop** of the **curve** $r = sin(12 \theta) is \dfrac{\pi}{48} $.

## Example:

Find the **area** of the region that **falls** in-between the two curves**.**

\[r= 4sin\theta, \space \space r= 2 \]

The given **curves** are $r = 4sin \theta$ and $r = 2$.

\[ 4 sin \theta = 2 \]

\[ sin \theta = \dfrac{1}{2} \]

\[ \theta = sin^{-1} \left( \dfrac{1}{2} \right) \]

$\theta = \dfrac{\pi}{6}$ and $\theta = \dfrac{5 \pi}{6}$

Inserting **limits** and $r$ in the formula of area:

\[ A = \dfrac{1}{2} \int_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} ((4sin(\theta))^2 – 2^2) d \theta \]

\[ = \dfrac{1}{2} \int_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} (16sin^2(\theta) – 4) d \theta \]

\[ = 4.\dfrac{1}{2} \int_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} (4sin^2(\theta) – 1) d \theta \]

\[ = 2 \int_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} (4. \dfrac{1}{2} (1-cos2 \theta ) – 1) d \theta \]

\[A = 2 \int_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} (1-2cos2 \theta) d \theta \]

**Integrating** $A$ with respect to $d \theta$:

\[ A = 2 \left[ \theta – 2. \dfrac{1}{2} sin 2 \theta \right]_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} \]

\[ A = 2 \left[ \theta – sin 2 \theta \right]_{\dfrac{\pi}{6}}^{ \dfrac{5\pi}{6}} \]

By **Solving** the above expression, **Area** comes out to be:

\[A = \dfrac{4 \pi}{3} + 2 \sqrt{3} \]