Question

Can i get help with this problem dealing with finding the area between curves:

x=y^2-y and x=2y^2-2y-6

Answer 1

I hope youre taking calculus. Take the integrals of both, then subtract the line thats underneath from the one thats over it.

Answer 2

i dont quite get you. What line?

Answer 3

what line do i subtract?

Answer 4

Youre given two equations which can be graphed. One line will be above the other on the graph. Once you take the integrals of both, you now will have the areas of both curves. to find the areas between them, subtract the smaller one from the larger one.

Answer 5

http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx

Go to example three

Answer 6

okay am i integrating with respects to y?

Answer 7

it might help if you solve for y first too.

Answer 8

oh crap. you might have to do implicit differentiation first. have you done that?

Answer 9

oh so solve for y in both equations

Answer 10

yes

Answer 11

solve for y, then integrate.

Answer 12

i still dont understand how to get started

Answer 13

yeah we can do this one quickly

Answer 14

first of all i hate this x = stuff so lets make it
\[y=x^2-x\] and
\[y=2x^2-2x-6\]

Answer 15

we are going to get a number out of this, so the variable is irrelevant

Answer 16

we need only the limits of integration, which is where these two parabolas meet

Answer 17

okay

Answer 18

set them equal and solve

Answer 19

and thats it, nothing special?

Answer 20

like resolve for y or somehting like that?

Answer 21

i get x = 0 and x = 2

Answer 22

then you have to integrate the bigger one minus the smaller one over that interval

Answer 23

now, which one would be the smaller one and bigger one?

Answer 24

damn damn damn i messed up

Answer 25

lets go back to
\[x=y^2-y\]
\[ x=2y^2-2y-6\]

Answer 26

lets match these up

Answer 27

how?

Answer 28

set them equal and solve

Answer 29

\[2y^2-2y-6=y^2-y\]
\[y^2-y-6=0\]
\[(y-3)(y+2)=0\]
\[y=-2,y=3\]

Answer 30

so those are our limits of integration from -2 to 3

Answer 31

we are going to integrate along the y axis, that is wrt y. no different than wrt x, just dy instead of dx

Answer 32

but how come, the curves are top and bottome, i thought that when integrating with respects to y there should be a right and left curve

Answer 33

when i graphed them i saw that they were like two parabolas one on top and one on bottome

Answer 34

there is

Answer 35

we go from y = -2 to y = 3

Answer 36

lets graph

Answer 37

which is left curve will be f and which will be g

Answer 38

it will be
\[y^2-y-(2y^2-2y-6)\]
\[-y^2+y+6\] from -2 to 3

Answer 39

\[\int_{-2}^3(-y^2+y+6)dy\]

Answer 40

ooooooooooooh i have an idea. lets see if it works!

Answer 41

lets cheat and see if we can get it instantly. i know it will not explain, but we have it now in any case

Answer 42

hold on

Answer 43

all you have to do is ask!
http://www.wolframalpha.com/input/?i=area+between+the+curves+x%3Dy^2-y+and+x%3D2y^2-2y-6

Answer 44

with the method you have shown me, do i have to compensate in some way for the portion below the x-axis

Answer 45

yeah i screwed that up so forget about it. it was just plain wrong

Answer 46

oh i mean the switching x and y was wrong. the method above is correct

Answer 47

oh man , i need to do this for HW

Answer 48

oh, so then the other way is still good

Answer 49

you don't have to worry about "above" and "below" because you are taking one function minus another

Answer 50

second way is good. switching was stupid

Answer 51

so then if i evluate the integral of -y^2+y+6 at -2 and 3 i should get my answer right

Answer 52

here look at this
here is your integral not as an area between two curves. you see that everything is positive
http://www.wolframalpha.com/input/?i=integrate+-2+to+3++%28-y^2%2By%2B6%29+dy

Answer 53

okay so then just integrat that and then evaluate right

Answer 54

yes, find the anti-derivative, plug in 3, plug in -2, subtract, get
\[\frac{125}{3}\] and be done

Answer 55

yup you got it!

Answer 56

thanks man, i appreaciate it

Answer 57

yw