Question

Find the centroid of the plane region bounded by the curves \(y=0\) and \(y=x^2-4\). Assume that the density is \(\delta =1\)

Answer 1

so I know that this is symmetric and therefor \(\bar{x}=0\) and since the density is zero \[m=\int^2_{-2}x^2-4\,dx=-\frac{3}{32}\]

Answer 2

so then I did \[\bar{y}=-\frac{3}{32}\int^2_{-2} \int^0_{x^2-4}y\,dy\,dx\]

Answer 3

Then did something go wrong?

Answer 4

is that correct so far?

Answer 5

well except for when you said "density is zero"

Answer 6

oops I meant one

Answer 7

I must have done the integration wrong then, thanks!

Answer 8

well not really
If you're finding M then you pretty much have everything right but the -1

Answer 9

Your actual area should be
\[y_1 - y_2 = 0-(x^2-4)\]
Which comes out to
3/32, if I trust your integration.

Answer 10

See, it's not the function that you put in to find M. You were doing
\[\int\limits_{x_1}^{x_2}y_2dx\]
Where it should be
\[\int\limits_{x_1}^{x_2}\int\limits_{y_1}^{y_2}\delta(x, y)dydx = \int\limits_{x_1}^{x_2}\int\limits_{y_1}^{y_2}1dydx =\int\limits_{x_1}^{x_2}y_2-y_1dx \] (I messed up the subscripts in my above answer) The point is, mass should be positive.

Answer 11

And, now that I check your integration, the answer is actually M = 32/3

Answer 12

But you were right regarding the symmetry argument (which only holds true since \[\delta(x, y)\]
is symmetric, too). Thus we know xbar = 0

Answer 13

@richyw i found two minor mistakes .
first the formula you wrote is for X bar nor for Y bar
second integral should be multiplied by -32/3 not -3/32

Answer 14

lol sami late for the party?

Answer 15

yes m late..lol

Answer 16

alright thank you. I have the answer now!

Answer 17

Anyways, your formula for ybar is right, with my additions:
\[y_{centroid} = \int\limits_{x_1}^{x_2}\int\limits_{y_1}^{y_2}y \times \delta(x, y)dydx/M\]
(Note that we've got ourselves a type 2 here, so we integrate with y on the inside)
Substitute density for one:
\[ \int\limits\limits_{x_1}^{x_2}\int\limits\limits_{y_1}^{y_2}y \times dydx/(32/3) = (3/32)\int\limits\limits_{x_1}^{x_2}\int\limits\limits_{y_1}^{y_2}y \times dydx\]
FToC:\[(3/64)\int\limits\limits\limits_{x_1}^{x_2}y^2|_{x^2-4}^{0} dx = (-3/64)\int\limits\limits\limits_{x_1}^{x_2}(x^2-4)^2 dx \]

Answer 18

And then it's just first semester calc from there!