Question

Areas between graphs of 2 functions

Answer 1

y=x^2, x+y=2

Answer 2

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

Answer 3

1) Sketch a graph of both functions

Answer 4

sounds fantastic! thank you! lol

Answer 5

np ^_^

Answer 6

graph was given lol (way easier)

Answer 7

lol, okay!

Answer 8

2) find the intersection points when y1= y2

Answer 9

set equal to zero?

Answer 10

yep, find the x's :)

Answer 11

got em...-2 and 1

Answer 12

alright, good :)

Answer 13

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

Answer 14

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

Answer 15

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?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

hold on :)

Answer 21

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

Answer 22

which line is on the top? :)

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

A plot related to this problem is attached.

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

birdsell, dear :) , let's take it step by step

1) sketch both graphs, on your own , ignore the given :)

Answer 34

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

Answer 35

sketch them

Answer 36

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

Answer 37

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

Answer 38

calm down, did you draw the graph?

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

try :)

Answer 43

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

Answer 44

so x^2 is on top?

Answer 45

nope, look at the sketch again :)

Answer 46

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

Answer 47

look at the closed region,=)

Answer 48

lmfao im dumb :)... got it

Answer 49

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

Answer 50

y=2-x

Answer 51

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

Answer 52

got it

Answer 53

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

Answer 54

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 :)

Answer 55

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

Answer 56

hanks! new thread or current?

Answer 57

thanks**

Answer 58

lol whatever you want :), np ^_^

Answer 59

not sure if you are familiar with slicing?

Answer 60

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

Answer 61

yep! :):)

Answer 62

I had a midterm on it last sunday lol!

Answer 63

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

Answer 64

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

Answer 65

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 :)

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

Ill start a new thread sstarcia

Answer 71

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

Answer 72

alright :)

Answer 73

ahhh it isnt working lol

Answer 74

how come?

Answer 75

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 \[x=\sqrt{5y^2}\]

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

\[y = x/\sqrt(5)\] I guess

Answer 80

i got y=x/5 lol

Answer 81

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

Answer 82

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

Answer 83

gotcha

Answer 84

okay :)

Answer 85

now:

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

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

graphing*

Answer 90

okay

Answer 91

got it

Answer 92

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

Answer 93

did you get something similar to it?

Answer 94

yep

Answer 95

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