Question

how do you find the area enclosed by a line, a parabola, and the x-axis?

Answer 1

integrate :)

Answer 2

i know but integrate what

Answer 3

we need to know what parabola and what line

Answer 4

if you have x^2 from 0 to 1 its int (0 to 1) of x^2 dx

Answer 5

as an example

Answer 6

Find the area enclosed by \[y=\sqrt{2x}, y=6-2x, and the xaxis\]

Answer 7

first find where they intersect

Answer 8

but they intersect at three different places

Answer 9

those equation only intersect in one place :)

Answer 10

rearrange for x in the equations, your parabola will then be the lower limit, the straight line will be the upper limit and your y limits will be from 0 to the y component of the intersection point, then it is a trivial double integral

Answer 11

i will scan and show you my drawing

Answer 12

\[\int\limits_{0}^{y component of intersection}\int\limits_{(y^2)/2}^{1/2y+3}..and so on\]

Answer 13

i always have to draw picture first

Answer 14

me too by visual inspection you can simplify the problem a lot

Answer 15

dana, have you looked at the pic?

Answer 16

i also put what i integrated

Answer 17

oops that second integral should e from 2 to 3 i messed up my line a bit

Answer 18

\[\int\limits_{0}^{2}\int\limits_{.5y^2}^{-0.5y-3}dxdy\]

Answer 19

yeaa that makes sence now, thank you!

Answer 20

there was not suppose to be an e in that sentence lol

Answer 21

because the x intercept of that y=6-2x is 3 not 4

Answer 22

yea his way if fine too, the double integral is just another way you can check your answer

Answer 23

she*

Answer 24

her*

Answer 25

oh sorry

Answer 26

wasn't paying attention to the name there sorry

Answer 27

but on the other hand that double integral is pretty epic? can someone be my fan i think i deserve it for going all out on this question

Answer 28

i will be ur fan

Answer 29

thanks! i'm your fan too.

Answer 30

helped people a few times but it seems a lot of people just make an account to ask a question and then leave without saying thank you

Answer 31

your welcome dana lol

Answer 32

he didnt become our fan :(

Answer 33

dana, if you come back you could have done int(3-y/2-y^2/2,y=0..2) and got the same thing you could have look at the functions as if y-axis was x-axis

Answer 34

ohh thats true, to switch the problem around. and i did say thank you, for the record. you all were a lot of help! :)

Answer 35

lol im sorry i didn't notice im an idiot

Answer 36

are aunder a curve, use integration & subtract lower curve from upper