Question

The region bounded by the parabola y=x^2 and above by the line y=x-2 Density = 8x^2. Find the center of mass of the thin plate covering the given region

Answer 1

any ideas?

Answer 2

LOL what?

Answer 3

you asked a question, im asking what ideas you have come up with for a solution process.

Answer 4

OOO ok

Answer 5

I have that M= \[\int\limits_{-1}^{2} (8x^2)[(x+2)^2-(x^2)^2]\]

Answer 6

hmm, i was considering a multiple integation setup, something like:\[\int_D\rho(x,y)~dA\]

Answer 7

limits of y = x^2 to x-2
x = intersectiions

Answer 8

does that sound like the same process?

Answer 9

youre curves dont intersect, are they written correctly?

Answer 10

o its y=x+2 sorry

Answer 11

\[\int \limits^{x=2}_{x=-1}~\int\limits^{y=x+2}_{y=x^2}8x^2~dy~dx\]

Answer 12

ok

Answer 13

I havent gotten to double intergals yet

Answer 14

well, its just working inside out

8(x^3)/3 from x^2 to x+2

8((x+2)^3)/3 - 8(x^6)/3

that takes care of dy, then its just the usual dx stuff

Answer 15

pfft, been awhile, pretty sure i messed up

Answer 16

there are no ys to integrate, so its just y'=1 which integrates to y

so 8x^2[(x+2) - (x^2)] dx

Answer 17

Ok

Answer 18

which is just a product of density times the difference in our functions

Answer 19

OOOOOOOO

Answer 20

so out total mass seems to be

\[8\int_{-1}^{2}x^3+2x^2-x^4dx=63/20\]for the total mass

Answer 21

126/5 i forgot the 8 out front lol

Answer 22

ooooooooo ok

Answer 23

thats where I was going with my intergration

Answer 24

i was wondering why you had squared your functions is all

Answer 25

do you agree that the center of mass will balance out to an average x and y point?

if so, then its really equating:

\[8\int_{-1}^{b}x^3+2x^2-x^4dx=\frac{126}{2(5)}\]

Answer 26

where b is the balance point for x

Answer 27

O ok

Answer 28

-b^5/5+b^4/4+2b^3/3 - (-(-1)^5/5+(-1)^4/4+2(-1)^3/3) = 126/80

-b^5/5+b^4/4+2b^3/3 - (1/5+1/4-2/3) = 126/80

which works out to about 1.27118 if my idea is good, unless you can think of a formula for it

Answer 29

thanks so much!

Answer 30

youre welcome, make sure you chk with your course material to be sure that my idea is sound. all i got to go on is memory at the moment

Answer 31

finding the average y part of the point would be something similar but trickier since my idea would have to try to find inverses to play with