A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find center of mass
anonymous
 4 years ago
Find center of mass

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Of 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
 4 years ago
Best ResponseYou've already chosen the best response.0i got the mass of 8/3 +2pi is that right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int_{0}^{2} 1+\frac12y~dy\] 2 + 2/4 is what i see along the y axis

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the center of mass is a point at which the total mass can be depicted

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or 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
 4 years ago
Best ResponseYou've already chosen the best response.0what i was taught is first to calculate the mass and then calculate integrals for moments with respect to the yaxis and xaxis and then divide My/M (my moment by mass ) to get x

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1hmmm, that does ring a bell

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx yeah, i might need to read this for a refresher :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1this is your density function? p(x,y)=1+y/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get the concept but i have trouble setting up the integral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i found the mass \int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}1+\frac{ y }{ 2 }dydx = \frac{ 8 }{ 3 }+2\pi

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}1+\frac{ y }{2 } dydx =\frac{ 8 }{ 3 }+2\pi\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1wrap that with a: ``` \[ \] ``` yeah, there ya go

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would the integral for my be the same just with extra y? like Mx\[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}y(1+\frac{ y }{ 2 })dydx\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://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
 4 years ago
Best ResponseYou've already chosen the best response.0then i would get 0, ooo it should be zero since circle is centered at 0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}y(1+\frac{ y }{ 2 })dydx\] \[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}y+\frac12 y^2~dy~dx\] \[\int\limits_{2}^{2}\frac12(\sqrt{4x^2})^2+\frac16 (\sqrt{4x^2})^3~dx\] \[k\int\limits_{2}^{2}~dx\] \[k(2(2))=4k\] right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lol .... i totally looked over the x parts in the second go around :/

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i thought id clean it up by using a constant k, because i overran the xs and simply forgot to integrate them \[\frac12\int\limits_{2}^{2}(\sqrt{4x^2})^2~dx+\frac16\int\limits_{2}^{2} (\sqrt{4x^2})^3~dx\] \[\frac12\int\limits_{2}^{2}4x^2~dx+\frac16\int\limits_{2}^{2} (4x^2)\sqrt{4x^2}~dx\] \[8\frac{8}3+6\pi\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284x%5E2%29 http://www.wolframalpha.com/input/?i=integrate+%28integrate+y%281%2By%2F2%29+from+y%3D0..sqrt%284x%5E2%29%29+from+x%3D2+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
 4 years ago
Best ResponseYou've already chosen the best response.0thanks i think i got it, just the integration is painful

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0I get \[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\]

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0it is easy to integrate if you switch to polar coordinates

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oo I didn't think of that, and yea it seems right

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{2}^{2}\int\limits_{0}^{\sqrt{4x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}\] \[\int\limits_{0}^{\pi}\int\limits_{0}^{2}r\sin(\theta)\left(1+\frac{ r\sin(\theta) }{ 2 }\right)rdrd\theta=\pi+\frac{16}{3}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.