anonymous
  • anonymous
Find center of mass
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
i got the mass of 8/3 +2pi is that right?
amistre64
  • amistre64
\[\int_{0}^{2} 1+\frac12y~dy\] 2 + 2/4 is what i see along the y axis

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
the center of mass is a point at which the total mass can be depicted
amistre64
  • amistre64
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
anonymous
  • anonymous
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
amistre64
  • amistre64
hmmm, that does ring a bell
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx yeah, i might need to read this for a refresher :)
amistre64
  • amistre64
this is your density function? p(x,y)=1+y/2
anonymous
  • anonymous
i get the concept but i have trouble setting up the integral
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}1+\frac{ y }{2 } dydx =\frac{ 8 }{ 3 }+2\pi\]
amistre64
  • amistre64
wrap that with a: ``` \[ \] ``` yeah, there ya go
anonymous
  • anonymous
would the integral for my be the same just with extra y? like Mx\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\]
anonymous
  • anonymous
would that be Mx?
amistre64
  • amistre64
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
anonymous
  • anonymous
then i would get 0, ooo it should be zero since circle is centered at 0
amistre64
  • amistre64
\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y(1+\frac{ y }{ 2 })dydx\] \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y+\frac12 y^2~dy~dx\] \[\int\limits_{-2}^{2}\frac12(\sqrt{4-x^2})^2+\frac16 (\sqrt{4-x^2})^3~dx\] \[k\int\limits_{-2}^{2}~dx\] \[k(2-(-2))=4k\] right?
amistre64
  • amistre64
lol .... i totally looked over the x parts in the second go around :/
anonymous
  • anonymous
whats k?
amistre64
  • amistre64
i thought id clean it up by using a constant k, because i overran the xs and simply forgot to integrate them \[\frac12\int\limits_{-2}^{2}(\sqrt{4-x^2})^2~dx+\frac16\int\limits_{-2}^{2} (\sqrt{4-x^2})^3~dx\] \[\frac12\int\limits_{-2}^{2}4-x^2~dx+\frac16\int\limits_{-2}^{2} (4-x^2)\sqrt{4-x^2}~dx\] \[8-\frac{8}3+6\pi\]
amistre64
  • amistre64
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
anonymous
  • anonymous
thanks i think i got it, just the integration is painful
Zarkon
  • Zarkon
I get \[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\]
Zarkon
  • Zarkon
it is easy to integrate if you switch to polar coordinates
anonymous
  • anonymous
oo I didn't think of that, and yea it seems right
anonymous
  • anonymous
thanks a lot
Zarkon
  • Zarkon
\[\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\] \[\int\limits_{0}^{\pi}\int\limits_{0}^{2}r\sin(\theta)\left(1+\frac{ r\sin(\theta) }{ 2 }\right)rdrd\theta=\pi+\frac{16}{3}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.