Question

Find center of mass

Answer 1

Of the upper half (\[y \ge 0\] of the disk bounded by the circle x^2+y^2=4 with p(x,y)=1+y/2

Answer 2

i got the mass of 8/3 +2pi is that right?

Answer 3

\[\int_{0}^{2} 1+\frac12y~dy\]

2 + 2/4 is what i see along the y axis

Answer 4

the center of mass is a point at which the total mass can be depicted

Answer 5

or in this case, maybe a line for the y cg and another for the xcg; where they cross would be the center of mass

Answer 6

what i was taught is first to calculate the mass and then calculate integrals for moments with respect to the y-axis and x-axis and then divide My/M (my moment by mass ) to get x

Answer 7

hmmm, that does ring a bell

Answer 8

http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

yeah, i might need to read this for a refresher :)

Answer 9

this is your density function? p(x,y)=1+y/2

Answer 10

i get the concept but i have trouble setting up the integral

Answer 11

i found the mass \int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{ 2 }dydx = \frac{ 8 }{ 3 }+2\pi

Answer 12

\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{2 } dydx =\frac{ 8 }{ 3 }+2\pi\]

Answer 13

wrap that with a:
```
\[ \]
```
yeah, there ya go

Answer 14

would the integral for my be the same just with extra y? like Mx\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\]

Answer 15

would that be Mx?

Answer 16

http://people.usd.edu/~jflores/MultiCalc02/WebBook/Chapter_16/Graphics16/MI_5/Html16_5/16.5%20Applications%20of%20Double%20Integrals.htm

that does look appriopriate to me

Answer 17

then i would get 0, ooo it should be zero since circle is centered at 0

Answer 18

\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\]
\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y+\frac12 y^2~dy~dx\]
\[\int\limits_{-2}^{2}\frac12(\sqrt{4-x^2})^2+\frac16 (\sqrt{4-x^2})^3~dx\]
\[k\int\limits_{-2}^{2}~dx\]
\[k(2-(-2))=4k\]
right?

Answer 19

lol .... i totally looked over the x parts in the second go around :/

Answer 20

whats k?

Answer 21

i thought id clean it up by using a constant k, because i overran the xs and simply forgot to integrate them
\[\frac12\int\limits_{-2}^{2}(\sqrt{4-x^2})^2~dx+\frac16\int\limits_{-2}^{2} (\sqrt{4-x^2})^3~dx\]
\[\frac12\int\limits_{-2}^{2}4-x^2~dx+\frac16\int\limits_{-2}^{2} (4-x^2)\sqrt{4-x^2}~dx\]
\[8-\frac{8}3+6\pi\]

Answer 22

http://www.wolframalpha.com/input/?i=integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284-x%5E2%29

http://www.wolframalpha.com/input/?i=integrate+%28integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284-x%5E2%29%29+from+x%3D-2+to+2

i seem to have mismathed it someplace .. ah well. i have to get going so ill try to take another look at this tomorrow; good luck

Answer 23

thanks i think i got it, just the integration is painful

Answer 24

I get \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\]

Answer 25

it is easy to integrate if you switch to polar coordinates

Answer 26

oo I didn't think of that, and yea it seems right

Answer 27

thanks a lot

Answer 28

\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\]

\[\int\limits_{0}^{\pi}\int\limits_{0}^{2}r\sin(\theta)\left(1+\frac{ r\sin(\theta) }{ 2 }\right)rdrd\theta=\pi+\frac{16}{3}\]