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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

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 :)

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

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 :)

hold on :)

which line is on the top? :)

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

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.

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

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

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

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

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.

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?

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

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?

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]

graphing*

okay

got it

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

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?

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!!!!

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:)

sure :):)

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

lol..girls work on both like us:)

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..:):)