# Find the area of the region that lies inside both curves. r2 = 50 sin(2θ), r = 5

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.

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.