areas shared by polar regions\nfind the areas of the regions in exercises 7 - 9.\n7. shared by the circles r…

Explanation:

Step1: Find intersection points

Set the two polar - equations equal to each other. For the circles $r = 2\cos\theta$ and $r = 2\sin\theta$, we have $2\cos\theta=2\sin\theta$, so $\tan\theta = 1$, and $\theta=\frac{\pi}{4}$.

Step2: Use the formula for the area in polar coordinates

The formula for the area $A$ in polar coordinates is $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. The area of the region shared by the two circles is composed of two parts. The area of the first - part: $A_1=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(2\sin\theta)^{2}d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}4\sin^{2}\theta d\theta$. Using the identity $\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}$, we get $A_1=\int_{0}^{\frac{\pi}{4}}(1 - \cos(2\theta))d\theta=\left[\theta-\frac{1}{2}\sin(2\theta)\right]{0}^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{1}{2}$. The area of the second - part: $A_2=\frac{1}{2}\int{\frac{\pi}{4}}^{\frac{\pi}{2}}(2\cos\theta)^{2}d\theta=\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}4\cos^{2}\theta d\theta$. Using the identity $\cos^{2}\theta=\frac{1+\cos(2\theta)}{2}$, we get $A_2=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(1 + \cos(2\theta))d\theta=\left[\theta+\frac{1}{2}\sin(2\theta)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}=\frac{\pi}{4}-\frac{1}{2}$.

Step3: Calculate the total area

$A = A_1+A_2=\frac{\pi}{2}-1$.

Answer:

$\frac{\pi}{2}-1$