The aim of this question is to understand the application of integration for finding **the area under the curves** or the **area bounded by two curves**.

To solve this question, we first combine both curves by substituting the value of $r$ from one curve to the other. This gives us a **single mathematical equation**. Once we have this equation, we simply find the **integration of the function** to find the area under this combined mathematical function that (actually) represents the **region bounded by both the curves**.

## Expert Answer

Given that:

\[r^2 = 50sin2\theta\]

\[r = 5\]

Combining both equations, we get:

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

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

\[\Rightarrow \theta = \frac{sin^{-1}(\frac{25}{50})}{2}\]

\[\theta = \frac{sin^{-1}(0.5)}{2}\]

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

These are the values that represent the **bounds on the area**.

To find the **area bounded** by this **region,** we need to perform the following **integration**:

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

Simplifying:

\[A = 2 \bigg \{ \int_{0}^{\frac{\pi}{12}} 50sin(2\theta) d\theta + \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} (25) d\theta \bigg \}\]

Applying the power rule of integration, we get:

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

Simplifying:

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

\[A = 2 \bigg \{ [-(25)cos(2\theta)]_{0}^{\frac{\pi}{12}} + [25(\theta)]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg \}\]

\[A = 2 \bigg \{ -25[cos(2\theta)]_{0}^{\frac{\pi}{12}} + 25[\theta]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg \}\]

\[A = 2 \times 25 \bigg \{ -[cos(2\theta)]_{0}^{\frac{\pi}{12}} + [\theta]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg \}\]

\[A = 50 \bigg \{ -[cos(2\theta)]_{0}^{\frac{\pi}{12}} + [\theta]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg \}\]

Evaluating the **definite integrals** using the bounds, we get:

\[A = 50 \bigg \{ -[cos(2\times \frac{\pi}{12}) – cos(2\times 0)] + [\frac{\pi}{4} – \frac{\pi}{12}] \bigg \}\]

\[A = 50 \bigg \{ -[cos(\frac{\pi}{6}) – cos(0)] + [\frac{3\pi-\pi}{12}] \bigg \}\]

Substituting the values of **trigonometric function**, we get:

\[A = 50 \bigg \{ -[\frac{\sqrt{3}}{2} – 1] + [\frac{2\pi}{12}] \bigg \}\]

Simplifying:

\[A = 50 \bigg \{ -[\frac{\sqrt{3}}{2} – 1] + [\frac{\pi}{6}] \bigg \}\]

\[A = 50 \bigg \{ -\frac{\sqrt{3}}{2} + 1 + \frac{\pi}{6} \bigg \}\]

\[A = -50 \times \frac{\sqrt{3}}{2} + 50 \times 1 + 50 \times \frac{\pi}{6}\]

## Numerical Result

**The area bounded by two curves **is calculated as:

\[A = -25 \times \sqrt{3} + 50 + 25 \frac{\pi}{3}\]

## Example

Find the **area bounded** by following **two curves.**

\[r = 20sin2\theta\]

\[r = 10\]

Combining both equations, we get:

\[10 = 20sin(2\theta) \]

\[\Rightarrow \theta = \frac{sin^{-1}(0.5)}{2}\]

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

Performing **Integration:**

\[A = 2 \bigg \{ 2 \times \frac{1}{2} \int_{0}^{\frac{\pi}{12}} \bigg (\sqrt{20sin(2\theta)}\bigg )^2 d\theta + 2 \times \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg ( 10 \bigg ) \bigg \}\]

\[A = 2 \bigg \{ [-10cos(2\theta)]_{0}^{\frac{\pi}{12}} + [10(\theta)]_{\frac{\pi}{12}}^{\frac{\pi}{4}} \bigg \}\]

\[A = 2 \bigg \{ -10[cos(2\times \frac{\pi}{12}) – cos(2\times 0)] + 10[\frac{\pi}{4} – \frac{\pi}{12}] \bigg \}\]

\[A = 2 \bigg \{ -10[\frac{\sqrt{3}}{2} – 1] + 10[\frac{\pi}{6}] \bigg \}\]

\[A = -10 \sqrt{3} + 20 + 10 \frac{\pi}{3}\]

Which is the value of the required **area.**