## anonymous one year ago hello, advance definite integral area bound by 4 curves. https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-xpl1/v/t34.0-12/11939020_821736254611990_2090835384_n.jpg?oh=3cd85831dc162d90184347e50a890cc8&oe=55DD3400&__gda__=1440572593_c4c4257c8f45c3517dc933a3312f81db

1. anonymous

@King.Void.

2. Astrophysics

First thing you want to do is sketch it out

3. anonymous

yes u want to or should i?

4. Astrophysics

You should

5. anonymous

|dw:1440417292230:dw|

6. anonymous

srry i dont know how to do it properly its a bit rough

7. Astrophysics

|dw:1440417409578:dw|

8. anonymous

yes

9. anonymous

when it says label co-ordinates of 4 vertices does that mean what u did on the graph

10. anonymous

|dw:1440417708934:dw|

11. Astrophysics

So what is the area you should be integrating

12. anonymous

is it where the all curves are

13. anonymous

|dw:1440418416259:dw|

14. Astrophysics

I guess it wants it above y=8

15. Astrophysics

So we probably need a better drawing

16. phi

it wants this region

17. Astrophysics

Yeah exactly

18. phi

the top point is at 9,9^3 or (9,729)

19. Astrophysics

|dw:1440418815806:dw|

20. Astrophysics

Just putting your pic up in draw phi :P

21. anonymous

$\int\limits_{9}^{2}x ^{3} dx +\int\limits_{9}^{2} 8 dx -\int\limits_{9}^{8} x dx$

22. anonymous

is it something like this

23. phi

part (b) wants the integral with respect to y

24. anonymous

yes

25. Astrophysics

You should have dy

26. Astrophysics

So you're doing horizontal strips not vertical, that's part c

27. anonymous

how?

28. phi

|dw:1440419331010:dw|

29. phi

the first integral has a right bound formed by y=x the right bound is the curve y= x^3 each rectangle will have width x1-x2 and height dy we need x1 and x2 in terms of y x1 (right bound) is x=y so y x2 (left bound is x^3= y so x= y^(1/3) the integral is $\int_8^9 y-y^\frac{1}{3} \ dy$ then we need a second integral for y=9 to 9^3

30. phi

follow?

31. anonymous

yes

32. phi

can you post the 2nd integral?

33. anonymous

$\int\limits_{9}^{3} y$

34. anonymous

$\int\limits_{3}^{9} y dy$

35. anonymous

srry the first one was a mistake

36. phi

then we need a second integral for y=9 to 9^3

37. anonymous

$\int\limits_{3}^{9} 8$

38. phi

First, identify the dimensions of the "thin rectangle" |dw:1440420324066:dw| we want the area of that "box" we want a positive number, so we would do Xright - Xleft that gives the width then multiply by the height dy (Xr - Xleft)* dy that is the first step

39. anonymous

ok

40. phi

next step: we don't want to use an x variable. but we know Xright is a constant = 9 so we have (8 - Xleft) dy now we need to get rid of Xleft (i.e. put it in terms of y) if we are given a y value, can you find Xleft?

41. phi

**(9 - Xleft) dy

42. anonymous

(9-x) dy

43. phi

ok, but that x you have in (9-x) is some number at a specific y say the "box" is at y=8. what is the value of that x? (remember that x is the x of the point on the curve y=x^3 when y is 8)

44. anonymous

$\sqrt[3]{y}$

45. phi

yes, in general, (and specifically, when y=8, x is 2) but we want the general answer, because as we change y we want the "correct x" x= y^(1/3) (I would use exponents, easier to integrate)

46. phi

so fix up this: (9-x) dy in other words, replace that x with a function of y

47. anonymous

(9-y^(1/3) ) dy

48. anonymous

$\int\limits_{3}^{9} [ 9- y ^{\frac{ 1 }{ 3 }} ]$

49. anonymous

dy

50. phi

that is the area of a single thin box we know "add them up" , which is to say, integrate starting from the lowest y (which is point B in the graph) up to the highest y (which is off the graph)

51. anonymous

ok

52. anonymous

$\int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}} dy$ +$\int\limits_{9}^{3} [9-y ^{\frac{ 1 }{ 3}}] dy$

53. phi

how are you getting 3 for the upper limit?

54. anonymous

srry mistake

55. anonymous

\int\limits_{3}^{9} [9-y ^{\frac{ 1 }{ 3}}] dy

56. phi

no, the upper limit is where x^3 meets x=9

57. phi

the left side curve is y= x^3 and the right boundary is x=9 the left side gets closer to x=9 as y gets bigger they intersect when y= 729

58. anonymous

how did u get Y= 729

59. anonymous

didnt u just use the two integrals and then integrate

60. phi

the very first part of the problem is to find the 4 points where the various curves meet the shape is like this: |dw:1440422110612:dw|

61. phi

we find where y=x^3 meets x=9 obviously they meet when x=9 in the first equation, and the y value is y=9^3 = 729

62. phi

notice we use two integrals. the first one has a different "right boundary" we use that one from y=8 to y=9 then we switch to the new right boundary (of x=9), and use that all the way up to y=729

63. anonymous

yea

64. phi

so the problem is $\int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}}\ dy + \int\limits_{9}^{729} 9-y ^{\frac{ 1 }{ 3}} \ dy$

65. anonymous

oh wow

66. anonymous

6.45+1573.29 =1579.74

67. anonymous

is this right?

68. phi

yes

69. anonymous

thank u so much u have alot of patience

70. anonymous

part c?

71. anonymous

i show u the solution i have done

72. phi

part c is more complicated. the first step is draw in a thin vertical rectangle, and figure out its height and width

73. anonymous

|dw:1440423107863:dw|

74. anonymous

so do we get the values respect to x

75. anonymous

$\int\limits_{2}^{9} x ^{3} dx - \int\limits_{2}^{9} 8 dx - \int\limits_{8}^{9}$

76. anonymous

$\int\limits_{8}^{9} x dx *$

77. phi

there are a few ideas we have to keep clear. I would begin with a picture and draw in the rectangle.

78. phi

|dw:1440423583733:dw|