## anonymous 4 years ago Find the area of the region that lies inside both curves r = A*sin(theta) and r = B*sin(theta), A > 0, B > 0?

1. nikvist

$r_1=A\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{A}{2}\right)^2=\left(\frac{A}{2}\right)^2$$r_2=B\sin\theta\quad\Rightarrow\quad x^2+\left(y-\frac{B}{2}\right)^2=\left(\frac{B}{2}\right)^2$$S=\frac{\pi}{4}\cdot\left(\min{\left\{A,B\right\}}\right)^2$

2. perl

this does not work

3. perl

try A = 2 , B = 1

4. perl

should be 3pi/2 , not sure how you got your answer

5. perl

ok looks like you did a polar conversion r = A sin theta r^2 = r* A sin theta x^2 + y^2 = A r* sin theta x^2 + y^2 = Ay x^2 + y^2 - Ay = 0 complete square x^2 + (y -A/2)^2 = (A/2)^2 similiarly x^2 + (y-B/2)^2 = (B/2)^2 directly integrating polar is much easier, btw so now you have two cartesian equations , and you did ... ?

6. perl

im not sure how to to integrate it using cartesian equations, hmmm

7. perl

ok i guess it can be done, but it is tricky

8. perl

wlog A > B > 0 so

9. nikvist

These are circles. Why do you integrate?

10. perl

|dw:1328800285296:dw|

11. perl

you want the area inside bounded by the two circles, you changed to cartesian so i integrated cartesian. the polar integration is much easier

12. perl

|dw:1328800478524:dw|

13. perl

i am not sure why you changed to cartesian, but either way works

14. nikvist

Perl,can you draw these circles?

15. perl

|dw:1328800689137:dw|

16. nikvist

Which does region lie inside both circles?

17. perl

lol

18. perl

I guess i solved a related question

19. perl

yours is still wrong though

20. perl

|dw:1328801098170:dw|

21. perl

|dw:1328801200075:dw|

22. nikvist

limits of integration are 0 and pi, not 0 and 2pi.

23. perl

ok

24. perl

ok i agree now

25. perl

so i calculated area twice?

26. nikvist

Between pi and 2pi circles do not exist.

27. perl

ok so you did integrate using polar, good , so you were just using the cartesian as an aid to graph it

28. perl

i set A sin theta = B sin theta

29. perl

solutions are pi*k

30. perl

why dont circles exist between pi and 2pi?

31. nikvist

because r cannot be negative

32. perl

yes it can in polar coordinates

33. perl

errr (r, theta ) = (-r, theta + pi)

34. perl

it just keeps going round and round in circles :)

35. perl

but clearly i went twice around

36. perl

that is why I got pi instead of pi/2

37. nikvist

$r=\sqrt{x^2+y^2}\ge0$

38. perl

in polar we allow negative r

39. perl

well if you have a TI 84 or graphing calculator, you can graph r = sin theta , and watch what happens when theta > pi . i think

40. perl

anyways, we define (-r, theta +pi) = (r,theta) , so its not a problem.

41. perl

for instance, suppose theta = 3pi/2, and given r(theta) = sin theta . we have ( r, theta ) = ( sin 3pi/2, 3pi/2) = (-1, 3pi/2) = (1 , 3pi/2 + - pi) = ( 1, pi/2) , so we can make any negative radius positive, using that trick

42. nikvist

For this problem, your solution is very complicated and unnecessary.

43. perl

which one? i was solving a different problem

44. perl

finding the region bounded by the two semicircles, i think its called the 'lune'

45. nikvist

these are not semicircles

46. perl

right, i solved a different problem , the area bounded by the arcs . the lune

47. perl

might not be the lune either not sure what you would call it, the area bounded by the inner and outer circles?

48. perl

and i got | a^2 - b^2) |* pi/4 as the area of the region bounded by the inner and outer circles

49. nikvist

Find the area of the region that lies inside both curve

50. perl

yes i misread it

51. perl

i have the answer now, thanks

52. perl

but i answered a different question , which is interesting

53. perl

i see , you didnt do any integration. you just used the area formula, pi * radius ^2 . nice

54. perl

i Have a difficult question, if you can help me...

55. perl

its elementary though

56. perl

Can i post it here?

57. perl

solve for all real and non real the system of equations. xy - z/3 = xyz +1 yz - x/3 = xyz -1 xz - y/3 = xyz - 9

58. perl

if you have a solution, please email riemanngalois@gmail.com I already have the integer solution (3,0,-3) and i know that you can plug it into wolfram. I want to know how did wolfram solve this. thanks

59. perl

as in , i would like to know the other five solutions, i believe 4 real solutions and 2 non real

60. nikvist

brute force solution

61. perl

wow, i wish i could do that. did you use a computer algebra system for help, like maple or matlab?

62. perl

yes so we have to test the solutions, to see if any are extraneous. looks good so you did equation 1 - equation 2 , then you solved equation 3 . the logic of this gets me a bit confused. you got the quadratic of z, treating y as a constant. then z = f(y). so then x = g(y) back substituting. so shouldnt it be ( g(y), y, f(y) ) , infinite solutions>?

63. perl

ok so you did (1) - (2), (3) -> you got z = f(y) , x = g(y) . you can then plug these into any of the three equations? does it matter?