Finding Center of Mass of Thin Region in 1st Quadrant

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SUMMARY

The discussion focuses on calculating the center of mass for a thin infinite region in the first quadrant, bounded by the coordinate axes and the curve y=e-2x, with a density function of ρ(x,y) = xy. The mass M was computed using the double integral M = ∫∫ xy dydx, resulting in M = 1/32 after proper integration. The participants emphasized the necessity of correctly setting up the integrals for Mx and My, suggesting the addition of an extra variable in the integrand to facilitate the calculations.

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  • Understanding of double integrals in calculus
  • Familiarity with the concept of center of mass
  • Knowledge of exponential functions and their properties
  • Experience with integration techniques, particularly in polar coordinates
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  • Study the derivation of the center of mass for different density functions
  • Learn advanced integration techniques for handling complex integrals
  • Explore the application of polar coordinates in double integrals
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Homework Statement


Find the center of mass of a thin infinite refion in the first quadrant bounded by the coordinate axes and the curve y=e-2x, if ρ(density)= xy


Homework Equations





The Attempt at a Solution


I set M= ∫∫ xy dydx where y goes from 0 to y=e-2x, and x goes from 0 to infinity. I solved this and got 1/8

but when i try to calculate Mx and My the integration are impossible.
 
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You're doing the integral wrong, mate! What have you got so far?
I get \int_0^{e^{-2x}} y dy = \left[y^2/2\right]_0^{e^{-2x}} = e^{-4x}/2
M = \int_0^\infty dx \int_0^{e^{-2x}} dy xy = \int_0^\infty dx x \int_0^{e^{-2x}} dy y = \int_0^\infty x e^{-4x}/2 = 1 / 32
Now put one additional x or y, respectively into your integrand and do the calculation for Mx and My.
 

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