Areas between graphs of 2 functions

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Areas between graphs of 2 functions

Mathematics
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y=x^2, x+y=2
Alright, I'll lead you through the steps, how about that? :)
1) Sketch a graph of both functions

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sounds fantastic! thank you! lol
np ^_^
graph was given lol (way easier)
lol, okay!
2) find the intersection points when y1= y2
set equal to zero?
yep, find the x's :)
got em...-2 and 1
alright, good :)
now, we must find out which function is bigger :) in other words, which graph is on the top of the other?
I think y= x^2 is on top of the other one, right? :)
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?
exactly! you can see the intersection in the graph right? which curve is on the top?
i think for y he said right-left and for x top-bottom
let me sketch the graph lol just to make sure :)
yep i see the intersection lol thanks. i would upload the pic but that would take longer than a sketch
hold on :)
okay got it, look at your sketch, color the region bounded (where the intersection lies/ closed region lol)
which line is on the top? :)
in my sketch, I see y = 2-x on top :)
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hmm, I got a different sketch , I've sovled for y for the second equation (y=2-x)
try to sketch the graph using y = x^2 and y = 2-x :)
A plot related to this problem is attached.
1 Attachment
yeah, that's what I got robto ^_^, birdsell are you following?
its the same graph its just the given one is a snippet
LOL, then draw your own to make your life easier :) , look at robto's sketch , which line is on the top ? ^_^
intersection point of the 2 lines in Q1 is at 1,1
are you sure? I got x = -2 and x = 1
so did i :-X but the picture shows it at 1,1
birdsell, dear :) , let's take it step by step 1) sketch both graphs, on your own , ignore the given :)
you have y =x^2 and y = 2-x
sketch them
you'll have to get the same sketch as robto's ^_^
im so confused :( most of them are clear that one is on top of the other
calm down, did you draw the graph?
and im attempting to use my calculator to graph the points but cannnt
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
after that, you'll get a graph similar to robto's
try :)
The second plot version uses the same formulas except the vertical and horizontal scaling is the same.
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so x^2 is on top?
nope, look at the sketch again :)
The curve passing through (0,0) is x^2.
look at the closed region,=)
lmfao im dumb :)... got it
no you're not, you panicked :D, alright so which line is on the top?
y=2-x
excellent, so this wil be you g(x) :) and the other curve is f(x) okay?
got it
I actually think I got it! lol wanna help with another?
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 :)
sure , I'll be glad to help ^_^
hanks! new thread or current?
thanks**
lol whatever you want :), np ^_^
not sure if you are familiar with slicing?
yes I am , it has to do with finding the volume right =)?
yep! :):)
I had a midterm on it last sunday lol!
alright, which method does it want you to use? :)
ughhh college... calc is killin me... up until now i have been great
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 :)
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
you're finding the area of the 2 functions between the 2 curves Hakan, to do that you have to find the enclosed region :)
actually calc 1. I go to a tough school and this is calc 1 carriculum
serious? I'm taking this in calculus II in college lol
Ill start a new thread sstarcia
well, I took the area b/w the curves in calculus I but the volume in calculus II
alright :)
ahhh it isnt working lol
how come?
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 \[x=\sqrt{5y^2}\]
But the photo of question doesn't ask that as i see.Anyway.Thanx :)
lol we re-sketched it, look at robto's sketch :) np :)
alright, to make your life easier bird, find y of this equation
\[y = x/\sqrt(5)\] I guess
i got y=x/5 lol
lol, take y^2 out and you'll get y, then divide by sqrt of 5 :)
nvm i see now lol gotta take Sqrt of both top and bottom
gotcha
okay :)
now: 1) sketch the given in a graph, y=0, y=2 and y=x/sqrt(5)
use your calculator for y = x/sqrt(5) you'll get something like a slight curve on the right, right?
it wont give me anything... when i type y=x/sqrt[5]
lol, forget your graphying calculator and do it by hand. Take values for x and calculate y using the calculator :)
graphing*
okay
got it
I think you'll get something like this , I think
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did you get something similar to it?
yep
alright, now what did the question say, how does the curve rotate? about the y-axis, x-axis?
find vol of disc that runs perpendicular top y axis w diameter running from x to the parabola
so you'll have to use the disk method :). So that means it's rotating about the y-axis right?
so cut the curve perpendiculary to the y-axis ( draw a horizontal line in the bounded region I showed you )
ughh im so sorry but i have to go to class now. thank you for all of your help :)
lol, sure :) good luck, and try to finish it ^_^
thanks!
:)
:):) hiya ppl!!!!
need any more help @birdsell ???
lol, he's in class :)
hey sstarica, couldn't type ther in grp chat ..lol...its stuck again !
I don't like it either ways :) it's alright.
which grade are you? ^_^
im gonna be engineer too :)
which field? :)
software
nah, I'm in the hardware field =P
never mind..we can help each other:)
lol, yep. I'll deal with the hardware, you deal with the software, then we'll combine them together in the end
sure :):)
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
half of the guys I know are all wanting software and girls working in hardware.
lol..girls work on both like us:)
the majority of the boys in here are all hardware. It is very rare to see a girl working in the hardware field :)
they say it's a guy thing, but I disagree.
good...we can work on both :) right :)
yes ^_^, anyways, I wish you the best of luck :)
thank you !!! best of luck to u too...:) lets see wen we meet here again :)
np :) and thank you ^_^, good night ~
good night star..:):)

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