anonymous
  • anonymous
Areas between graphs of 2 functions
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
y=x^2, x+y=2
anonymous
  • anonymous
Alright, I'll lead you through the steps, how about that? :)
anonymous
  • anonymous
1) Sketch a graph of both functions

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

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