## anonymous 5 years ago Areas between graphs of 2 functions

1. anonymous

y=x^2, x+y=2

2. anonymous

Alright, I'll lead you through the steps, how about that? :)

3. anonymous

1) Sketch a graph of both functions

4. anonymous

sounds fantastic! thank you! lol

5. anonymous

np ^_^

6. anonymous

graph was given lol (way easier)

7. anonymous

lol, okay!

8. anonymous

2) find the intersection points when y1= y2

9. anonymous

set equal to zero?

10. anonymous

yep, find the x's :)

11. anonymous

got em...-2 and 1

12. anonymous

alright, good :)

13. anonymous

now, we must find out which function is bigger :) in other words, which graph is on the top of the other?

14. anonymous

I think y= x^2 is on top of the other one, right? :)

15. anonymous

yeah i remember that from class but it sort of looks like an upside down "V" so neither is on top. Do i use the right one because its a graph of y not x?

16. anonymous

exactly! you can see the intersection in the graph right? which curve is on the top?

17. anonymous

i think for y he said right-left and for x top-bottom

18. anonymous

let me sketch the graph lol just to make sure :)

19. anonymous

yep i see the intersection lol thanks. i would upload the pic but that would take longer than a sketch

20. anonymous

hold on :)

21. anonymous

okay got it, look at your sketch, color the region bounded (where the intersection lies/ closed region lol)

22. anonymous

which line is on the top? :)

23. anonymous

in my sketch, I see y = 2-x on top :)

24. anonymous

25. anonymous

hmm, I got a different sketch , I've sovled for y for the second equation (y=2-x)

26. anonymous

try to sketch the graph using y = x^2 and y = 2-x :)

27. anonymous

A plot related to this problem is attached.

28. anonymous

yeah, that's what I got robto ^_^, birdsell are you following?

29. anonymous

its the same graph its just the given one is a snippet

30. anonymous

LOL, then draw your own to make your life easier :) , look at robto's sketch , which line is on the top ? ^_^

31. anonymous

intersection point of the 2 lines in Q1 is at 1,1

32. anonymous

are you sure? I got x = -2 and x = 1

33. anonymous

so did i :-X but the picture shows it at 1,1

34. anonymous

birdsell, dear :) , let's take it step by step 1) sketch both graphs, on your own , ignore the given :)

35. anonymous

you have y =x^2 and y = 2-x

36. anonymous

sketch them

37. anonymous

you'll have to get the same sketch as robto's ^_^

38. anonymous

im so confused :( most of them are clear that one is on top of the other

39. anonymous

calm down, did you draw the graph?

40. anonymous

and im attempting to use my calculator to graph the points but cannnt

41. anonymous

don't! you can try out values, alright for example, sketch y = x^2 first, take values for x and find y, then plot them :) same with the other equation, do that

42. anonymous

after that, you'll get a graph similar to robto's

43. anonymous

try :)

44. anonymous

The second plot version uses the same formulas except the vertical and horizontal scaling is the same.

45. anonymous

so x^2 is on top?

46. anonymous

nope, look at the sketch again :)

47. anonymous

The curve passing through (0,0) is x^2.

48. anonymous

look at the closed region,=)

49. anonymous

lmfao im dumb :)... got it

50. anonymous

no you're not, you panicked :D, alright so which line is on the top?

51. anonymous

y=2-x

52. anonymous

excellent, so this wil be you g(x) :) and the other curve is f(x) okay?

53. anonymous

got it

54. anonymous

I actually think I got it! lol wanna help with another?

55. anonymous

now to find the Area, the general formula is :$A =\int\limits_{a}^{b}[g(x) - f(x)] dx$ where g(x) is the curve that's in the top and f(x) in the bottom, replace the equations and solve : $A = \int\limits_{-2}^{1} (2-x) -(x^2) dx$ $A = \int\limits_{-2}^{1}( 2-x-x^2)dx$ $A = [2x-x^2/2 + x^3/3]$ from -2 to 1 so you'll get : $A = [2(1)-(1)^2/2-(1)^3/3] - [2(-2) -(-2)^2/2 - (-2)^3/3]$ compute it and you'll get A lol :)

56. anonymous

sure , I'll be glad to help ^_^

57. anonymous

hanks! new thread or current?

58. anonymous

thanks**

59. anonymous

lol whatever you want :), np ^_^

60. anonymous

not sure if you are familiar with slicing?

61. anonymous

yes I am , it has to do with finding the volume right =)?

62. anonymous

yep! :):)

63. anonymous

I had a midterm on it last sunday lol!

64. anonymous

alright, which method does it want you to use? :)

65. anonymous

ughhh college... calc is killin me... up until now i have been great

66. anonymous

lol, calculus II right? practice, work hard, you'll perfect everything. Work hard now to relax in the end, believe me I sucked in this part =-= but now it's just SIMPLE :)

67. anonymous

Sorry to intterrupt, your question ask area of between x^2 , y+x=2 and x axis according to photo but the solution answers area of two function.I am confused

68. anonymous

you're finding the area of the 2 functions between the 2 curves Hakan, to do that you have to find the enclosed region :)

69. anonymous

actually calc 1. I go to a tough school and this is calc 1 carriculum

70. anonymous

serious? I'm taking this in calculus II in college lol

71. anonymous

Ill start a new thread sstarcia

72. anonymous

well, I took the area b/w the curves in calculus I but the volume in calculus II

73. anonymous

alright :)

74. anonymous

ahhh it isnt working lol

75. anonymous

how come?

76. anonymous

anyway... The solid lies between planes perpendicular to the y axis at y=0 and y=2 the cross-sections perpendicular to the y-axis are circular discs with diameters running from the y-axis to the parabola $x=\sqrt{5y^2}$

77. anonymous

But the photo of question doesn't ask that as i see.Anyway.Thanx :)

78. anonymous

lol we re-sketched it, look at robto's sketch :) np :)

79. anonymous

alright, to make your life easier bird, find y of this equation

80. anonymous

$y = x/\sqrt(5)$ I guess

81. anonymous

i got y=x/5 lol

82. anonymous

lol, take y^2 out and you'll get y, then divide by sqrt of 5 :)

83. anonymous

nvm i see now lol gotta take Sqrt of both top and bottom

84. anonymous

gotcha

85. anonymous

okay :)

86. anonymous

now: 1) sketch the given in a graph, y=0, y=2 and y=x/sqrt(5)

87. anonymous

use your calculator for y = x/sqrt(5) you'll get something like a slight curve on the right, right?

88. anonymous

it wont give me anything... when i type y=x/sqrt[5]

89. anonymous

lol, forget your graphying calculator and do it by hand. Take values for x and calculate y using the calculator :)

90. anonymous

graphing*

91. anonymous

okay

92. anonymous

got it

93. anonymous

I think you'll get something like this , I think

94. anonymous

did you get something similar to it?

95. anonymous

yep

96. anonymous

alright, now what did the question say, how does the curve rotate? about the y-axis, x-axis?

97. anonymous

find vol of disc that runs perpendicular top y axis w diameter running from x to the parabola

98. anonymous

so you'll have to use the disk method :). So that means it's rotating about the y-axis right?

99. anonymous

so cut the curve perpendiculary to the y-axis ( draw a horizontal line in the bounded region I showed you )

100. anonymous

ughh im so sorry but i have to go to class now. thank you for all of your help :)

101. anonymous

lol, sure :) good luck, and try to finish it ^_^

102. anonymous

thanks!

103. anonymous

:)

104. anonymous

:):) hiya ppl!!!!

105. anonymous

need any more help @birdsell ???

106. anonymous

lol, he's in class :)

107. anonymous

hey sstarica, couldn't type ther in grp chat ..lol...its stuck again !

108. anonymous

I don't like it either ways :) it's alright.

109. anonymous

which grade are you? ^_^

110. anonymous

im gonna be engineer too :)

111. anonymous

which field? :)

112. anonymous

software

113. anonymous

nah, I'm in the hardware field =P

114. anonymous

never mind..we can help each other:)

115. anonymous

lol, yep. I'll deal with the hardware, you deal with the software, then we'll combine them together in the end

116. anonymous

sure :):)

117. anonymous

I find it rather weird in someway. I thought software = more girls work in that field and hardware = men, but now it's the opposite LOL

118. anonymous

half of the guys I know are all wanting software and girls working in hardware.

119. anonymous

lol..girls work on both like us:)

120. anonymous

the majority of the boys in here are all hardware. It is very rare to see a girl working in the hardware field :)

121. anonymous

they say it's a guy thing, but I disagree.

122. anonymous

good...we can work on both :) right :)

123. anonymous

yes ^_^, anyways, I wish you the best of luck :)

124. anonymous

thank you !!! best of luck to u too...:) lets see wen we meet here again :)

125. anonymous

np :) and thank you ^_^, good night ~

126. anonymous

good night star..:):)