# Find the area of the region enclosed by one loop of the curve. r = sin(12θ).

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:

$A =\int_a^b (f(x) – g(x)) dx$

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 $d \theta$:

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

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

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

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

$A = \dfrac{\pi}{48}$

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:

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