Asked by Lol
Find the center of mass for a shape consisting of three quarters of the unit circle, removing the quarter in the third quadrant
Answers
Answered by
oobleck
You know the area of the region is A = 3π/4
Also,
x̅ = ∫x dA
y̅ = ∫y dA
The region is easily expressed in polar coordinates, making these integrals
x̅ = 1/A ∫∫x r dr dθ = 1/A ∫[0..1] ∫[-π/2..π] r^2 cosθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)
y̅ = ∫∫y r dr dθ = ∫[0..1] ∫[-π/2..π] r^2 sinθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)
Note that this is the same as for the smaller area just in QI, since the parts in QII and QIV balance each other out.
Also,
x̅ = ∫x dA
y̅ = ∫y dA
The region is easily expressed in polar coordinates, making these integrals
x̅ = 1/A ∫∫x r dr dθ = 1/A ∫[0..1] ∫[-π/2..π] r^2 cosθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)
y̅ = ∫∫y r dr dθ = ∫[0..1] ∫[-π/2..π] r^2 sinθ dθ dr
= 1/A ∫[0..1] r^2 dr = 4/(3π)
Note that this is the same as for the smaller area just in QI, since the parts in QII and QIV balance each other out.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.