# 14.6: Moments and Centers of Mass - Mathematics

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Using a single integral we were able to compute the center of mass for a one-dimensional object with variable density, and a two dimensional object with constant density. With a double integral we can handle two dimensions and variable density.

Just as before, the coordinates of the center of mass are

[ar x={M_yover M} qquad ar y={M_xover M},]

where (M) is the total mass, (M_y) is the moment around the (y)-axis, and (M_x) is the moment around the (x)-axis. (You may want to review the concepts in Section 9.6.)

The key to the computation, just as before, is the approximation of mass. In the two-dimensional case, we treat density (sigma) as mass per square area, so when density is constant, mass is ((hbox{density})(hbox{area})). If we have a two-dimensional region with varying density given by (sigma(x,y)), and we divide the region into small subregions with area (Delta A), then the mass of one subregion is approximately (sigma(x_i,y_j)Delta A), the total mass is approximately the sum of many of these, and as usual the sum turns into an integral in the limit:

[M=int_{x_0}^{x_1}int_{y_0}^{y_1} sigma(x,y),dy,dx,]

and similarly for computations in cylindrical coordinates. Then as before

[eqalign{
M_x &= int_{x_0}^{x_1}int_{y_0}^{y_1} ysigma(x,y),dy,dxcr
M_y &= int_{x_0}^{x_1}int_{y_0}^{y_1} xsigma(x,y),dy,dx.cr
}]

Example (PageIndex{1})

Find the center of mass of a thin, uniform plate whose shape is the region between (y=cos x) and the (x)-axis between (x=-pi/2) and (x=pi/2). Since the density is constant, we may take (sigma(x,y)=1).

It is clear that (ar x=0), but for practice let's compute it anyway. First we compute the mass:

[egin{align*} M&=int_{-pi/2}^{pi/2} int_0^{cos x} 1,dy,dx [4pt]&=int_{-pi/2}^{pi/2} cos x,dx [4pt]&=left.sin x ight|_{-pi/2}^{pi/2}=2.end{align*}]

Next,

[egin{align*} M_x&=int_{-pi/2}^{pi/2} int_0^{cos x} y,dy,dx [4pt]&=int_{-pi/2}^{pi/2} {1over2}cos^2 x,dx[4pt]&={piover4}.end{align*}]

Finally,

[egin{align*} M_y&=int_{-pi/2}^{pi/2} int_0^{cos x} x,dy,dx [4pt]&=int_{-pi/2}^{pi/2} xcos x,dx[4pt]&=0.end{align*}]

So (ar x=0) as expected, and (ar y=pi/4/2=pi/8). This is the same problem as in example 9.6.4; it may be helpful to compare the two solutions.

Example (PageIndex{2})

Find the center of mass of a two-dimensional plate that occupies the quarter circle (x^2+y^2le1) in the first quadrant and has density (k(x^2+y^2)). It seems clear that because of the symmetry of both the region and the density function (both are important!), (ar x=ar y). We'll do both to check our work.

Jumping right in:

[egin{align*} M&=int_0^1 int_0^{sqrt{1-x^2}} k(x^2+y^2),dy,dx [4pt]&=kint_0^1 x^2sqrt{1-x^2}+{(1-x^2)^{3/2}over3},dx. end{align*}]

This integral is something we can do, but it's a bit unpleasant. Since everything in sight is related to a circle, let's back up and try polar coordinates. Then (x^2+y^2=r^2) and

[egin{align*} M&=int_0^{pi/2} int_0^{1} k(r^2),r,dr,d heta [4pt]&=kint_0^{pi/2}left.{r^4over4} ight|_0^1,d heta [4pt]&=kint_0^{pi/2} {1over4},d heta [4pt]&=k{piover8}.end{align*}]

Much better. Next, since (y=rsin heta),

[egin{align*} M_x&=kint_0^{pi/2} int_0^{1} r^4sin heta,dr,d heta
[4pt]&=kint_0^{pi/2} {1over5}sin heta,d heta
[4pt]&=kleft.-{1over5}cos heta ight|_0^{pi/2}={kover5}.end{align*}]

Similarly,

[egin{align*} M_y&=kint_0^{pi/2} int_0^{1} r^4cos heta,dr,d heta
[4pt]&=kint_0^{pi/2} {1over5}cos heta,d heta
[4pt]&=kleft.{1over5}sin heta ight|_0^{pi/2}={kover5}.end{align*}]

Finally, (ar x = ar y = {8over5pi}).

## Watch the video: Massenmittelpunkt 1 (June 2022).

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