## anonymous 5 years ago Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx. a) Find the area of R b)Find the volume of the solid generated when R is revolved about the x-axis c)Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares

1. amistre64

thats a little section beside the sin cos stuff eh

2. amistre64

since sin and cos = each other at pi/4; take your integrals from 0 to pi/4

3. amistre64

[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

4. amistre64

to revolve it around the x axis; we do a sum of areas [S] 2pi [f(x)]^2 dx

5. amistre64

take the cos first and subtract out the sin next; like cutting a hole out of a donut

6. amistre64

ack!!... pi [f(x)]^2 lol

7. amistre64

pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]

8. amistre64

makes sense?

9. anonymous

i think the integral will involve integration from x =0 to 90 degree since it's given that the problem is about to solve at the first quadrant. for question a) integration $\int\limits_{0}^{\pi/2}\int\limits_{\sin x}^{\cos x} dy dx$

10. anonymous

seems to me because the area enclosed by y-axis the interval should be from pi/4 yo pi/2...

11. amistre64

12. anonymous

nice drawing ! so, shaded area - [pi/4,pi/2]. am I wrong?

13. anonymous

amistre, is that a (cos(x))² or is it cos (x²)

14. amistre64

your wrong :) 0 to pi/4

15. amistre64

cos^2(x)

16. anonymous

ok

17. anonymous

i'm sorry, it's from o to pi/4 :)

18. amistre64

the double integral? dunno bout it ;)

19. anonymous

i thought the question want to integrate between cos x , sin x and x-axis. silly me :P you can find area under curve using double integration.

20. amistre64

cos(2t) = 2cos^2 - 1 cos^2 = (1+cos(2t))/2 might help

21. amistre64

just havent had any experience with doubles yet ;)

22. anonymous

for part A did you come up with sqrt(2) - 1?

23. amistre64

[S] cos^2(x) dx [S] 1/2 dt + [S] 2cos(2x)/4 dt ; [0,pi/4] is the first one right?

24. amistre64

whaa !?!

25. amistre64

[S] cos(t) dt -> sin(t). sin(pi/4) - sin(0) = 1/sqrt(2) - 0

26. anonymous

or .414

27. amistre64

[S] sin(t) dt -> -cos(t). -cos(pi/4) + cos(0) = -1/sqrt(2) +1

28. anonymous

i got sqrt (2) - 1 for question a

29. amistre64

1/sqrt(2) - (-1/sqrt(2) +1) 1/sqrt(2) + 1/sqrt(2) -1 (2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1 right?

30. anonymous

1/sqrt(2) + 1/sqrt(2) -1 =2/sqrt (2) - 1 =sqrt (2) - 1

31. amistre64

i see it ;)

32. amistre64

melor got it right :)

33. anonymous

okay I understand part a :)

34. amistre64

i invented a sqrt(2) in the numerator lol

35. amistre64

part b is integrateing the areas of the "donut", ie torus

36. amistre64

pi [S] cos^2 - sin^2 from 0 to pi/4 right?

37. anonymous

what's the difference between that, and having a 2 pi in front? I always get mixed up when to put pi or 2pi

38. amistre64

which is just cos(2t) thich integrates to (pi/2).sin(2t) from 0 to pi/4

39. amistre64

2pi is a surface area

40. amistre64

you want to add areas to get volume.... area of a circle = pi [f(x)]^2

41. anonymous

okay!

42. amistre64

2pi [f(x)] amounts to surface area :)

43. amistre64

adding together all the circumfrences of the circles :)

44. anonymous

so we're working with pi[S] cos²x - sin²x dx [0,pi/4] ?

45. amistre64

yes

46. anonymous

gosh, I feel silly. How do you integrate that? :/

47. amistre64

cos^2 - sin^2 = cos(2x) right? 2 cos(2x) ups to sin(2x) so we need to multiply by a comvenient form of 1; say 2/2

48. amistre64

keep the top 2 to use in the integral and slide out the bottom to under the pi

49. amistre64

we get in the end; (pi/2) sin(2x) right?

50. amistre64

go you recall your trig identitites?

51. anonymous

ehh, cos²x-sin²x= cos(2x)?

52. amistre64

the volume should be pi ------ 2sqrt(2)

53. anonymous

cos(x+y)=cos x cos y - sin x sin y substitue x into y you get, cos (2x) = cos^2 - sin^2 :)

54. amistre64

cos(a+b) = cos(a)cos(b) - sin(a)sin(b) cos(a+a); or cos(2a) = cos(a)cos(a)-sin(a)sin(a cos(2a)=cos^2(a)-sin^2(a)

55. amistre64

D(sin(2x)) = 2 cos(2x)

56. amistre64

2 cos(2x) -------- = cos(2x) right? 2

57. anonymous

I plugged pi[S] cos²x - sin²x dx [0,pi/4] into wolframalpha and it gave me pi/2?

58. amistre64

pi/2 [S] 2cos(2x) dx -> pi/2 sin(2x)

59. amistre64

at 0 = its = 0

60. amistre64

so the volume of the solid is pi/2 * sin(2(45)) = pi/2

61. amistre64

sin(2(pi/4)) = sin(pi/2) = 1

62. amistre64

so yes; the volume of the solid of rotation is pi/2 :)

63. anonymous

sweet!

64. amistre64

i gots no idea what c is saying

65. anonymous

c is asking to take the graph that we had, using the area R as the base of a solid, placing square cross sections on top of the graph to create a solid... haha I have no clue how to explain it. and I don't know how to solve it :(

66. anonymous

I guess for a square, every side is equal, so the length of each individual square placed is going to fit depending on the base R