Center of mass of area described by implicit function

Click For Summary
SUMMARY

The discussion focuses on calculating the center of mass for the region defined by the implicit function (x²+y²)² = 2xy in the first quadrant. The user successfully transformed the coordinates using the substitution x = r cos(t) and y = r sin(t), leading to the area integral ∫₀^{π/2} ∫₀^{√{sin(2t)}} r \, dr \, dt, which evaluates to 0.5. However, they encountered difficulties in finding the x-coordinate of the center of mass, represented by the integral ∫₀^{π/2} ∫₀^{√{sin(2t)}} r² cos(t) \, dr \, dt, which simplifies to ∫₀^{π/2} sin(2t)^{1.5} cos(t) \, dt. The user seeks alternative methods or insights to solve this integral.

PREREQUISITES
  • Understanding of polar coordinates and transformations
  • Familiarity with double integrals and area calculations
  • Knowledge of Jacobians in coordinate transformations
  • Experience with integral calculus, particularly trigonometric integrals
NEXT STEPS
  • Research techniques for solving trigonometric integrals, particularly ∫ sin(2t)^{1.5} cos(t) \, dt
  • Explore the use of numerical integration methods in Mathematica for complex integrals
  • Study the application of symmetry in calculating centers of mass for implicit functions
  • Investigate alternative coordinate systems that may simplify the integration process
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and physics, as well as anyone involved in solving complex integrals related to center of mass calculations.

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?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K