## math_proof 3 years ago Find center of mass

1. math_proof

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

2. math_proof

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

3. amistre64

$\int_{0}^{2} 1+\frac12y~dy$ 2 + 2/4 is what i see along the y axis

4. amistre64

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

5. 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

6. math_proof

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

7. amistre64

hmmm, that does ring a bell

8. amistre64

http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx yeah, i might need to read this for a refresher :)

9. amistre64

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

10. math_proof

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

11. math_proof

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

12. math_proof

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

13. amistre64

wrap that with a:    yeah, there ya go

14. math_proof

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

15. math_proof

would that be Mx?

16. amistre64
17. math_proof

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

18. 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?

19. amistre64

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

20. math_proof

whats k?

21. amistre64

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

22. 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

23. math_proof

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

24. Zarkon

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

25. Zarkon

it is easy to integrate if you switch to polar coordinates

26. math_proof

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

27. math_proof

thanks a lot

28. Zarkon

$\int\limits_{-2}^{2}\int\limits_{0}^{\sqrt{4-x^2}}y\left(1+\frac{ y }{ 2 }\right)dydx=\pi+\frac{16}{3}$ $\int\limits_{0}^{\pi}\int\limits_{0}^{2}r\sin(\theta)\left(1+\frac{ r\sin(\theta) }{ 2 }\right)rdrd\theta=\pi+\frac{16}{3}$