Center of mass of area described by implicit function

Mare102
Messages
1
Reaction score
0

Homework Statement



I'm trying to find the center of mass of the region (x²+y²)² =2xy in the first quadrant, but I got stuck.


The Attempt at a Solution




What I did is make the substitution x = r cos(t), y = r sin(t), which gives the equation r^{4}=2r²cos(t) sin(t), so r² = sin(2t), so r=\sqrt{sin(2t)}.
Then the integral for the area of the region becomes (as the Jacobian is r)
\int_0^{\pi/2} \int_0^{\sqrt{sin(2t)}} \! r \, dr dt

Solving this gives me 0.5.
I’m trying to find the x value of the center of mass, so I want to solve:
\int_0^{\pi/2} \int_0^{\sqrt{sin(2t)}} \! r² cos(t) \, dr dt
Which gives me
\int_0^{\pi/2} \! sin(2t)^{1.5} cos(t) \, dt

Which I’m unable to solve. Mathematica gives me a complicated integral, so does anyone know how to proceed, or maybe suggest a different approach?
 
Physics news on Phys.org
Mare102 said:

Homework Statement



I'm trying to find the center of mass of the region (x²+y²)² =2xy in the first quadrant, but I got stuck.


The Attempt at a Solution




What I did is make the substitution x = r cos(t), y = r sin(t), which gives the equation r^{4}=2r²cos(t) sin(t),
In the above, what you show is correct, but the exponent on r on the right side renders incorrectly. Use ^ for exponents inside LaTeX expressions. The same problem occurs below, in you integral for My (in your calculation for x-bar).
Mare102 said:
so r² = sin(2t), so r=\sqrt{sin(2t)}.
Then the integral for the area of the region becomes (as the Jacobian is r)
\int_0^{\pi/2} \int_0^{\sqrt{sin(2t)}} \! r \, dr dt

Solving this gives me 0.5.
I’m trying to find the x value of the center of mass, so I want to solve:
\int_0^{\pi/2} \int_0^{\sqrt{sin(2t)}} \! r² cos(t) \, dr dt
The LaTeX expression you want (and mean) is
\int_0^{\pi/2} \int_0^{\sqrt{sin(2t)}} \! r^2 cos(t) \, dr dt

Mare102 said:
Which gives me
\int_0^{\pi/2} \! sin(2t)^{1.5} cos(t) \, dt
You're missing a factor of 1/3 in the integral above. I don't have any good ideas for proceeding here, though.
Mare102 said:
Which I’m unable to solve. Mathematica gives me a complicated integral, so does anyone know how to proceed, or maybe suggest a different approach?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top