$r^{2}=50\sin(2\theta),\: r=5$

The **article aims to find the region’s area under the given curves. Area under the curve** is calculated by various methods, the most popular of which is the **antiderivative method** of finding the area.

Area under a curve can be found by knowing the curve’s equation, the **boundaries of the curve**, and the **axis surrounding the curve. **Generally, we have formulas to find **areas of regular shapes like square, rectangle, quadrilateral, polygon, and circle**, but there is no general formula to find the **area under a curve**. The **process of integration helps solve the equation and find the required region**.

**Antiderivative methods** are beneficial for finding regions of irregular planar surfaces. This article discusses how to find the **area between two curves.**

Area under the curve can be calculated in **three simple steps**.

–** First**, we need to know the **equation of the curve** $(y = f(x))$, the limits over which the area is to be calculated, and the **axis bounding the area.**

– **Second**, we need to find the** integration (antiderivative)** of the curve.

– **Finally**, we need to apply an** upper **and **lower bound** to the integral response and **take the difference to get the area under the curve.**

\[Area=\int_{a}^{b} y.dx\]

\[=\int_{a}^{b} f(x)dx\]

\[=[g(x)]_{a}^{b}\]

\[Area=g(b)-g(a)\]

**Area under the curve can be calculated using three ways.** Also, which method is used to find the area under the curve depends on the need and available data inputs to find the area under the curve.

**Expert Answer**

**Step 1:**

Consider the **given curves** $r^{2}=50\sin(2\theta),\: r=5$

The** aim is to find the area of the region that lies under both curves.**

**From the curves:**

\[5^{2}=50\sin(2\theta)\]

\[25=50\sin(2\theta)\]

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

\[2\theta=\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \dfrac{17\pi}{6}\]

\[\theta=\dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}\]

**Step 2:**

The** formula to find the area of the region** under the **curves** is given by:

\[A=\int_{a}^{b}\dfrac{1}{2}[f(\theta)]^2 \:d(\theta)\]

The **required area can be calculated by adding the area inside the cardioid between** $\theta=0$ and $\theta=\dfrac{\pi}{4}$ from the area inside the circle $\theta=0$ to $\theta=\dfrac{\pi}{4}$.

Since the **area is symmetric** about $\theta=\dfrac{\pi}{4}$, the area can be **calculated as:**

\[A=2[2\times \dfrac{1}{2}\int_{0}^{\dfrac{\pi}{12}}(\sqrt(50\sin(2\theta))^{2}d\theta +2\times \frac{1}{2} \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}} 5^{2} d\theta]\]

\[=2[\int-{0}^{\dfrac{\pi}{12}} 50\sin(2\theta)d\theta+\int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}25 \:d\theta]\]

\[=2[-\dfrac{50}{2}\cos(2\theta)|_{0}^{\dfrac{\pi}{12}}+25[|_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}]\]

\[=2[-25(\cos\dfrac{\pi}{6}-\cos(0))+25(\dfrac{2\pi}{12}-\dfrac{\pi}{12})]\]

\[=2[-25(\dfrac{\sqrt 3}{2}-1)+25(\dfrac{2\pi}{12})]\]

\[=2(-\dfrac{25\sqrt 3}{2}+25+\dfrac{25\pi}{6})\]

**Numerical Result**

The **area of the region under the curves** $r^{2}=50\sin(2\theta),\: r=5$ is

\[A=2(-\dfrac{25\sqrt 3}{2}+25+\dfrac{25\pi}{6})\]

**Example**

Calculate the area of the region that lies inside both curves.

$r^{2}=32\sin(2\theta),\: r=4$

**Step 1:**

Consider the **given curves** $r^{2}=32\sin(2\theta),\: r=4$

The** aim is to find the area of the region that lies under both curves.**

**From the curves:**

\[4^{2}=32\sin(2\theta)\]

\[16=32\sin(2\theta)\]

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

\[2\theta=\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \dfrac{17\pi}{6}\]

\[\theta=\dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{13\pi}{12}, \dfrac{17\pi}{12}\]

**Step 2:**

The** formula to find the area of the region** under the **curves** is given by:

\[A=\int_{a}^{b}\dfrac{1}{2}[f(\theta)]^2 \:d(\theta)\]

The **required area can be calculated by adding the area inside the cardioid between** $\theta=0$ and $\theta=\dfrac{\pi}{4}$ from the area inside the circle $\theta=0$ to $\theta=\dfrac{\pi}{4}$.

Since the **area is symmetric** about $\theta=\dfrac{\pi}{4}$, area can be **calculated as:**

\[A=2[2\times \dfrac{1}{2}\int_{0}^{\dfrac{\pi}{12}}(\sqrt(32\sin(2\theta))^{2}d\theta +2\times \frac{1}{2} \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}} 4^{2} d\theta]\]

\[=2[\int-{0}^{\dfrac{\pi}{12}} 32\sin(2\theta)d\theta+\int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}16 \:d\theta]\]

\[=2[-\dfrac{32}{2}\cos(2\theta)|_{0}^{\dfrac{\pi}{12}}+16[|_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}]\]

\[=2[-16(\cos\dfrac{\pi}{6}-\cos(0))+16(\dfrac{2\pi}{12}-\dfrac{\pi}{12})]\]

\[=2[-16(\dfrac{\sqrt 3}{2}-1)+16(\dfrac{2\pi}{12})]\]

\[=2(-\dfrac{16\sqrt 3}{2}+16+\dfrac{16\pi}{6})\]

The **area of the region under the curves** $r^{2}=32\sin(2\theta),\: r=4$ is

\[A=2(-\dfrac{16\sqrt 3}{2}+16+\dfrac{16\pi}{6})\]