We can partition into subrectangles with of them in the direction and in the direction Suppose each subrectangle has width and height Then a subrectangle containing the point x has approximate mass x and the mass of is approximately 1 1 y wher ID: 22773 Download Pdf

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We can partition into subrectangles with of them in the direction and in the direction Suppose each subrectangle has width and height Then a subrectangle containing the point x has approximate mass x and the mass of is approximately 1 1 y wher

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Mass, Centers of Mass, and Double Integrals Suppose a 2-D region has density x, y at each point x, y . We can partition into subrectangles, with of them in the -direction, and in the -direction. Suppose each subrectangle has width and height . Then a subrectangle containing the point ( x, has approximate mass ( x, ) and the mass of is approximately =1 =1 , y ) where , y is a point in the i, j -th subrectangle. Letting and go to inﬁnity, we have mass of ZZ x, y dA. Similary, the moment with respect to the axis can be calculated as ZZ y x, y dA and the moment with respect

to the axis can be calculated as ZZ x x, y dA. The we may calculate the center of mass of via center of mass of = ( x, ) = Example 1 Let be the unit square, x, y ) : 0 . Suppose the density of is given by the function x, y ) = + 1 so that is denser near the -axis. As a result, we would expect the center of mass to be bel ow the geometric center, (1 2) . However, since the density does not depend on , we do expect = 1

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We have: ZZ + 1 dA + 1 dy dx ln( + 1) dx ln 2 dx = ln 2 = 0 693147 .... ZZ + 1 dA + 1 dy dx ln( + 1)) dx (1 ln 2) dx = 1 ln 2 = 0 306852819 .... ZZ + 1 dA + 1 dy

dx ln 2 dx ln 2 ln 2 = 0 346573590 .... Thus the center of mass is ( x, ) = ln 2 ln 2 ln 2 ln 2 442095 ... Example 2 (Polar) Let . Let be the polar region r, ) : Suppose has constant density . Then: ZZ ρ dA ZZ dA area of πz ZZ ρy dA π/ sin θ dθ dr cos π/ dr dr (1 ZZ ρxdA π/ cos θ dθ dr sin π/ dr dr (1 Thus, the center of mass is ( x, ) = (1 (1 (1 (1

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An interesting feature of this region is that if is sufﬁciently large, the center of mass will be outside the region. This happens when the distance from the center

of mass to (0 0) is less than . That is, when (1 (1 < z. By factoring, we see that this is equivalent to (1 + (1 + < z. The critical value is the positive solution to 0 = which is about 82337397 ... . Thus, if z > 82337397 ... , the region is very thin, and the center of mass lies outside of the region.

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