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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, I'll lead you through the steps, how about that? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01) Sketch a graph of both functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sounds fantastic! thank you! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0graph was given lol (way easier)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) find the intersection points when y1= y2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, we must find out which function is bigger :) in other words, which graph is on the top of the other?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think y= x^2 is on top of the other one, right? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah 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
 5 years ago
Best ResponseYou've already chosen the best response.0exactly! you can see the intersection in the graph right? which curve is on the top?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think for y he said rightleft and for x topbottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me sketch the graph lol just to make sure :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep i see the intersection lol thanks. i would upload the pic but that would take longer than a sketch

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay got it, look at your sketch, color the region bounded (where the intersection lies/ closed region lol)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which line is on the top? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in my sketch, I see y = 2x on top :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, I got a different sketch , I've sovled for y for the second equation (y=2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try to sketch the graph using y = x^2 and y = 2x :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A plot related to this problem is attached.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that's what I got robto ^_^, birdsell are you following?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the same graph its just the given one is a snippet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL, then draw your own to make your life easier :) , look at robto's sketch , which line is on the top ? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0intersection point of the 2 lines in Q1 is at 1,1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure? I got x = 2 and x = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so did i :X but the picture shows it at 1,1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0birdsell, dear :) , let's take it step by step 1) sketch both graphs, on your own , ignore the given :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have y =x^2 and y = 2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you'll have to get the same sketch as robto's ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im so confused :( most of them are clear that one is on top of the other

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0calm down, did you draw the graph?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and im attempting to use my calculator to graph the points but cannnt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don'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
 5 years ago
Best ResponseYou've already chosen the best response.0after that, you'll get a graph similar to robto's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The second plot version uses the same formulas except the vertical and horizontal scaling is the same.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope, look at the sketch again :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The curve passing through (0,0) is x^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0look at the closed region,=)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lmfao im dumb :)... got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you're not, you panicked :D, alright so which line is on the top?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0excellent, so this wil be you g(x) :) and the other curve is f(x) okay?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I actually think I got it! lol wanna help with another?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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 : \[A = \int\limits_{2}^{1} (2x) (x^2) dx\] \[A = \int\limits_{2}^{1}( 2xx^2)dx\] \[A = [2xx^2/2 + x^3/3]\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0sure , I'll be glad to help ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hanks! new thread or current?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol whatever you want :), np ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure if you are familiar with slicing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes I am , it has to do with finding the volume right =)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I had a midterm on it last sunday lol!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, which method does it want you to use? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ughhh college... calc is killin me... up until now i have been great

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry 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
 5 years ago
Best ResponseYou've already chosen the best response.0you're finding the area of the 2 functions between the 2 curves Hakan, to do that you have to find the enclosed region :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually calc 1. I go to a tough school and this is calc 1 carriculum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0serious? I'm taking this in calculus II in college lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ill start a new thread sstarcia

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, I took the area b/w the curves in calculus I but the volume in calculus II

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh it isnt working lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyway... The solid lies between planes perpendicular to the y axis at y=0 and y=2 the crosssections perpendicular to the yaxis are circular discs with diameters running from the yaxis to the parabola \[x=\sqrt{5y^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the photo of question doesn't ask that as i see.Anyway.Thanx :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol we resketched it, look at robto's sketch :) np :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, to make your life easier bird, find y of this equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y = x/\sqrt(5)\] I guess

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, take y^2 out and you'll get y, then divide by sqrt of 5 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm i see now lol gotta take Sqrt of both top and bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now: 1) sketch the given in a graph, y=0, y=2 and y=x/sqrt(5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use your calculator for y = x/sqrt(5) you'll get something like a slight curve on the right, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it wont give me anything... when i type y=x/sqrt[5]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, forget your graphying calculator and do it by hand. Take values for x and calculate y using the calculator :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you'll get something like this , I think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get something similar to it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, now what did the question say, how does the curve rotate? about the yaxis, xaxis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find vol of disc that runs perpendicular top y axis w diameter running from x to the parabola

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you'll have to use the disk method :). So that means it's rotating about the yaxis right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so cut the curve perpendiculary to the yaxis ( draw a horizontal line in the bounded region I showed you )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ughh im so sorry but i have to go to class now. thank you for all of your help :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, sure :) good luck, and try to finish it ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0need any more help @birdsell ???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, he's in class :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey sstarica, couldn't type ther in grp chat ..lol...its stuck again !

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't like it either ways :) it's alright.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which grade are you? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im gonna be engineer too :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nah, I'm in the hardware field =P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0never mind..we can help each other:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, yep. I'll deal with the hardware, you deal with the software, then we'll combine them together in the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0half of the guys I know are all wanting software and girls working in hardware.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol..girls work on both like us:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the majority of the boys in here are all hardware. It is very rare to see a girl working in the hardware field :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they say it's a guy thing, but I disagree.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good...we can work on both :) right :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes ^_^, anyways, I wish you the best of luck :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you !!! best of luck to u too...:) lets see wen we meet here again :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np :) and thank you ^_^, good night ~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good night star..:):)
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