# Area in polar coordinates

• Nov 14th 2008, 03:32 PM
Apprentice123
Area in polar coordinates
Calculate in polar coordinates the area of the region outside of $\displaystyle P=1$, $\displaystyle P=2cos(\theta)$ and in inside $\displaystyle P=2$

This is correct:

$\displaystyle A=2\int_{0}^{\frac{\pi}{3}} \int_{2cos\theta}^{2} PdPd\theta + 2\int_{\frac{\pi}{3}}^{\pi} \int_{1}^{2} PdPd\theta$

and the answer is: $\displaystyle (\frac{8\pi}{3}-\frac{\sqrt{3}}{2})$
• Nov 14th 2008, 06:30 PM
Soroban
Hello, Apprentice123!

What does $\displaystyle P$ stand for in your equation?

Quote:

Calculate in polar coordinates the area of the region
outside of $\displaystyle r_1=1,\;\;r_2 =2\cos\theta$, and inside $\displaystyle r_3=2$

I don't see the need for double integrals . . .

. . $\displaystyle \begin{array}{c}r_1\text{ is a circle: center (0,0), radius 1} \\ r_2 \text{ is a circle: center (1,0), radius 1} \\ r_3\text{ is a circle: center (0,0), radius 2} \end{array}$

The region is symmetric to the $\displaystyle x$-axis.
. . We can determine the area of the upper half and multiply by 2.

From $\displaystyle \theta = 0$ to $\displaystyle \theta = \tfrac{\pi}{3}$, the area is between $\displaystyle r_3\text{ and }r_2$

. . $\displaystyle A_1 \;=\;\tfrac{1}{2}\int^{\frac{\pi}{3}}_0\bigg[ 2^2 - (2\cos\theta)^2\bigg]\,d\theta$

From $\displaystyle \theta = \tfrac{\pi}{3}$ to $\displaystyle \theta = \pi$, the area is between $\displaystyle r_3\text{ and }r_1$

. . $\displaystyle A_2 \;=\;\tfrac{1}{2}\int^{\pi}_{\frac{\pi}{3}} \bigg[2^2 - 1^2\bigg]\,d\theta$

Add the two areas and multiply by 2 . . .