# Find the area of Region of Polar Coordinate We have the polar equations r = 1/2 and r =...

## Question:

Find the area of Region of Polar Coordinate

We have the polar equations r = 1/2 and r = cos({eq}\theta {/eq})

The shaded region we're interested in is outside r=1/2 and inside r=({eq}\theta {/eq}).

The bounds are {eq}\pi/ 3 {/eq} and {eq}5 \pi /3 {/eq}.

The equation is 2 times the integral of {eq}(1/2)(cos(\theta))^2 - (1/2)^2 ) d(\theta) {/eq}.

## Integration:

A mechanical quantity that represents the joining of the differential of the function or equation in the definite form to analyse the mathematical problem is known as integration. It used in engineering applications.

GIVEN DATA:

• The polar equation is: {eq}r = \dfrac{1}{2} {/eq}
• {eq}r = \cos \theta {/eq}
• {eq}\theta = \dfrac{\pi }{3} {/eq}
• {eq}\theta = \dfrac{{5\pi }}{3} {/eq}
• The equation for area of region is: {eq}A = 2\int {\left( {\dfrac{1}{2}{{\left( {\cos \theta } \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right)d\theta } {/eq}

The domain is

{eq}\dfrac{\pi }{3} \le \theta \le \dfrac{{5\pi }}{3} {/eq}

Substitute the value in equation of area of region with respective region

{eq}\begin{align*} A &= 2\int_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}} {\left( {\dfrac{1}{2}{{\left( {\cos \theta } \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right)d\theta } \\ &= 2\int_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}} {\left( {\dfrac{1}{2}\left( {\dfrac{{1 + \cos 2\theta }}{2}} \right) - \left( {\dfrac{1}{4}} \right)} \right)d\theta } \\ &= 2\left[ {\dfrac{1}{4}\left( {\theta + \dfrac{{\sin 2\theta }}{2}} \right) - \dfrac{\theta }{4}} \right]_{\dfrac{\pi }{3}}^{\dfrac{{5\pi }}{3}}\\ &= 2\left[ {\dfrac{1}{4}\left( {\left( {\dfrac{{5\pi }}{3} + \dfrac{{\sin 2\left( {\dfrac{{5\pi }}{3}} \right)}}{2}} \right) - \left( {\dfrac{\pi }{3} + \dfrac{{\sin 2\left( {\dfrac{\pi }{3}} \right)}}{2}} \right)} \right) - \left( {\dfrac{{\left( {\dfrac{{5\pi }}{3}} \right)}}{4} - \dfrac{{\left( {\dfrac{\pi }{3}} \right)}}{4}} \right)} \right]\\ &= 2\left[ {\dfrac{1}{4}\left( {\dfrac{{5\pi }}{3} - \dfrac{\pi }{3} + 0.0908 - 0.0182} \right) - \left( {\dfrac{{5\pi }}{{12}} - \dfrac{\pi }{{12}}} \right)} \right]\\ &= 0.0363 \end{align*} {/eq}

Thus the area of region of polar coordinate is {eq}0.0363 {/eq}