## anonymous 5 years ago Would the radius be from 1 to 2? Thanks!

1. anonymous

2. anonymous

it is 1

3. anonymous

So, wat would the interval be? if I were to integrate it.

4. anonymous

5. anonymous

Yea,

6. anonymous

x is the point of intersection of both the circles, y is from 0 to 1

7. anonymous

If you're talking radius, the radius is from the first circle to the second circle.

8. anonymous

Sigh, sorry, I'm being vague again -.- I mean I'm using polar system, so how would the iterated integral look like?

9. anonymous

10. anonymous

What are you having problem with?

11. anonymous

radius from x^2+y^2-1 to x^2+y^2-2y angle is you have drawn the angles in the v absolute value shape from right to left. whatever angle you have there

12. anonymous

the boundaries for the radius should be between 1< r< 2y where y= r sin (theta) 1<r< rsin (thetat) this make sense because the boundary for the r the farthest is not constant so there should be a variable. Furthermore I think you should double check your boundaries for theta.

13. anonymous

I like doing double and triple integrals so if you have more questions feel free to ask them :)

14. anonymous

Ok, thx! I will lol. I'll examine the radius , but I think my theta is correct b/c I set the x^2+y^2=1 and x^2+(y-1)^2 = 1 to each other and solved to find the point of interception. Then, I used x = rcostheta, where x is one of the point of interception, r =1, and solved for theta. Right approach?

15. anonymous

oh, i mean, when i solved for the point of inteception, in terms of y, i got 1/2, which I then plugged into the equation to get x = +/- sqrtof3/2 , which I plugged into x=rcostheta

16. anonymous

well it says the first quadrant so I assume its (angle of intersection)< Theta < Pi / 2

17. anonymous

im not challenging your point of intersection angle but aren't you going beyond the first quadrant into your second?

18. anonymous

Oh, pellet, I totally forgot that its in the 1st quandratn LOLLOL

19. anonymous

Thanks , omg, faillll

20. anonymous

Hey, wouldn't the radius be sqrt(2y)? sinec its x^2 + y^2 = (sqrt(2y))^2 but I also solved for y, which is y =1/2, so if i plug that in, i get 1, weird.

21. anonymous

when going to polar coordinates you have x^2 + y^2 = r^2 r^2 = 2* y $r= \sqrt{2y}$ but we also know y = r sin theta so I got it wrong sorry it should be $\sqrt{2 * r \sin (\theta)}$

22. anonymous

$1<r< \sqrt{2* r \sin(\theta)}$

23. anonymous

lol, ok, thanks a lot.

24. anonymous

i got 1 because x^2 + y^2 = 1 so r^2 = 1 r=1 its really easy thats why I didnt show that one

25. anonymous

but if i get 1<r<2∗rsin(θ)√

26. anonymous

when i integrate

27. anonymous

using polar system

28. anonymous

i won't get rid of a variable, b/c there's two in the interval

29. anonymous

yeah I just saw that

30. anonymous

ok i said r^2 = 2 y where y = r sin (theta) r^2 = 2 * r sin (theta) divide r to both side and you have r= 2 sin (theta) idk thats the only way I can see this happening. try it out and see if it works.

31. anonymous

lol, ok. Thanks. Looks right. i'll get back to you when i'm done.