## anonymous 4 years ago Find the areas of the region shared by the circles of the polar equations:

1. anonymous

$r = 2 \cos \theta$ $r = 2 \sin \theta$

2. EarthCitizen

$\theta=?$

3. anonymous

What do you mean? The upper and lower bounds?

4. anonymous

I found them to be 0 and pi if that's what you mean.

5. EarthCitizen

yp, are they given ?

6. anonymous

Nope.

7. anonymous

I just used the calculator.

8. anonymous

It's the same for all the questions in my book.

9. EarthCitizen

alryt which is the inner curve and the outer curve or gr8r and smaller curve?

10. anonymous

2 sin theta is tangent to the x-axis and isis positive with the y-axis separating it in 2 while 2 cos theta is chopped in half by the x-axis to the left of sin theta. It is tangent to the origin.

11. EarthCitizen

r2 and r1 is ?

12. anonymous

The sin theta is on top.

13. anonymous

The equations are on top.

14. anonymous

But the questions is to find he area shared by the two. I tried to find the area of one circle and then the area without the area of the shared. I planned to substract them to get the shared area but one of the integrations led toa negative number or a zero. So it didn't happen.

15. anonymous

I was hoping if there were any other ways, especially with only one equation. But the other method would be great as long as I don't get negative or zero.

16. EarthCitizen

|dw:1327830143156:dw|

17. anonymous

|dw:1327830380231:dw|

18. anonymous

The top is the sin

19. EarthCitizen

$A=\int\limits_{\alpha}^{\beta}(1/2(r _{2}^{2}-r _{1}^{2})d \theta$

20. EarthCitizen

$A=\int\limits_{0}^{\pi}1/2((2\cos \theta)^{2}-(2\sin \theta)^{2})d \theta$ $=\int\limits_{0}^{\pi}1/2(4\cos ^{2}\theta-4\sin ^{2}\theta)d \theta$ =$=\int\limits_{0}^{\pi}4/2(\cos ^{2}\theta-(1-\cos ^{2}\theta)d \theta$ =$=2\int\limits_{0}^{\pi}(2\cos ^{2}\theta-1)d \theta$ $=\int\limits_{0}^{\pi}(4\cos2\theta)d \theta$

21. anonymous

That's the problem. Integrating this makes the answer 0.

22. EarthCitizen

$=\int\limits_{0}^{\pi}(2\cos2\theta)d \theta$

23. anonymous

It's 4 sin 2 theta. Plugging pi and 0 would make it zero.

24. anonymous

sin 2 pi and sin 0 are both 0.

25. EarthCitizen

$(\sin2\pi)-0 +c$

26. EarthCitizen

yh it's not symmetrical ? it's could be from the intrsc pts

27. EarthCitizen

alpha=pi/6