## anonymous 4 years ago Need help in integral area involve 3d objects

1. anonymous

2. anonymous

3. anonymous

ah I see those are multiple choices...

4. anonymous

did you get the 'sketch the region' one?

5. anonymous

basically just asking you to pick which of those areas is bounded by y=5, x=4 and the function...

6. anonymous

you there?

7. anonymous

the first choice is bounded by y=10, x=4 and the function... the second choice is bounded by y=10, x=0, x=4 and the function...

8. anonymous

yes i am here..

9. anonymous

10. anonymous

For the first picture, i think it is the last one for the 2nd picture, i am thinking either 1st or 4th one , what do u think ?

11. anonymous

naw, the last picture is bounded by x=0 y=0 and the function...

12. anonymous

(and x=4)

13. anonymous

so for the first picture, u also think is the last one?

14. anonymous

for my function, x is not 0 u mean?

15. anonymous

just trace around each of those red regions and see what functions or lines your tracing...

16. anonymous

what lines do you trace for the last choice?

17. anonymous

|dw:1352777609189:dw|

18. anonymous

yea, that's what i have too. dont really know how i got that..but yeah.. Do you possibly know how to volume the volume of this?

19. anonymous

|dw:1352777630991:dw|

20. anonymous

|dw:1352777716676:dw|

21. anonymous

dont u need to integrate the area function to volume , and then something to do with the height..?

22. anonymous

so... is that the region described in your problem statement??

23. anonymous

yes

24. anonymous

really? your problem statement makes no mention of x=0 or y=0 ...

25. anonymous

it does mention y=5 however...

26. anonymous

well, but it says rotate about y= 10, and y= 5

27. anonymous

naw..

28. anonymous

then u can kinda know there is a hole in between

29. anonymous

umm..wait..so which picture do you think it is for the first set of picture? i thought we have the same answer

30. anonymous

the region is bounded by x=4 y=5 and y=5e^-x

31. anonymous

like I said trace the boundary of the regions you have been given to find which one works with that description..

32. anonymous

there's no point in guessing here... even though this is multiple guess

33. anonymous

i still think is the last one..

34. anonymous

that's why i want to know which one u think it is

35. anonymous

why is it the last one..

36. anonymous

are you talking about the first part or the second part?

37. anonymous

first part ..

38. anonymous

because if you do the first part correctly, the answer to the second part should be obvious

39. anonymous

ok..

40. anonymous

so do you agree that it is either the bottom left or bottom right picture is correct due to the fact that given y=5

41. anonymous

then why are you saying it's the last region?

42. anonymous

why is it not the last picture ..lol |dw:1352778179844:dw|

43. anonymous

did you see the sketches where I traced the bounding functions??

44. anonymous

what are they??

45. anonymous

the curve is y= 5e^-x , y =5 , x=4

46. anonymous

what functions bound the region in the last picture?

47. anonymous

hint: I already did it for you.

48. anonymous

x=0, y=0..which i dont really get u ..o_o

49. anonymous

trace the boundary

50. anonymous

|dw:1352778560923:dw|

51. anonymous

what's that portion?

52. anonymous

say "y=0"

53. anonymous

|dw:1352778627593:dw|

54. anonymous

what's that portion?

55. anonymous

are you trying to say that radius doesnt bound to x=0?

56. anonymous

which it does indeed make sense when i look in the 2nd part

57. anonymous

I'd focus on the first part, if I were you..

58. anonymous

picking which region is bounded by y=5, x=4 and y= 5e^-x

59. anonymous

i'm not sure what u mean by bounded..do you mean by starting from there？

60. anonymous

trace the boundary of each region

61. anonymous

like I did... for the last region pictured in your choices...

62. anonymous

if i do what u did, then the answer will be the bottom left

63. anonymous

yes

64. anonymous

but my question with that picture is.. don't u need to do 5- 5e^-t in order to get tht area?

65. anonymous

not necessarily..

66. anonymous

you'll be using washers probably, since you haven't learned about shells yet (I assume)...

67. anonymous

so the inner radius is 5 and the outer radius is 10-5e^-x

68. anonymous

yea i didnt yet.. i didnt even learn these stuff much when my assignment is due 2 days after..

69. anonymous

ya, i do get that part

70. anonymous

|dw:1352779152955:dw|

71. anonymous

is that region not what the last picture look like?

72. anonymous

for the second part... yep.

73. anonymous

can i ask what does "region " mean..

74. anonymous

the red areas in the first part.

75. anonymous

yea..what does that red area mean

76. anonymous

is it like the path of rotation or something?

77. anonymous

no.

78. anonymous

it's the thing you're sweeping around the specified axis in order to generate a solid

79. anonymous

|dw:1352779636199:dw|

80. anonymous

take that region as an example... it's a rectangle...

81. anonymous

rotate about the x axis and you sweep out a cylinder..

82. anonymous

so it's kinda you are sweeping that region, and that sweeping motion generate the solid figure?

83. anonymous

|dw:1352779702320:dw|

84. anonymous

yes, exactly.

85. anonymous

okay thank you so so much!!!!!!

86. anonymous

Do you still have time to guide me on filding the volume?

87. anonymous

the second multiple choice question? or a new question?

88. anonymous

it's the same question. but it asks to find the volume

89. anonymous

cuz finding the volume is actually the first part..but the graphing part seems easier so i skipped to that first ..

90. anonymous

sure... we are using the washers in the sketch (and in the last choice to the second question)

91. anonymous

92. anonymous

area of a washer will be pi( (10 - 5e^-x)^2 - 5^2 )

93. anonymous

yea..that's all i have too.. not sure how to integrate that..

94. anonymous

$\pi \int\limits_{ }^{} (10-5e^{-x})^2 - 5^2 dx$

95. anonymous

$\pi \int\limits_{0}^{4} 75 -100e^{-x} +25e^{-2x} dx$

96. anonymous

that should be pretty easy to integrate...

97. anonymous

oh ok..i didnt combine them.. ok thanks, i will try that out :)

98. anonymous

@Algebraic! umm.. how can i leave it as exact numbers..?

99. anonymous

i got pi ((300+10 e^-4 - 25/2 e^-8)- (10-25/2))