### Mass and Static Moments of a Lamina

Suppose we have a lamina which occupies a region \(R\) in the \(xy\)-plane and is made of non-homogeneous material. Its density at a point \(\left( {x,y} \right)\) in the region \(R\) is \(\rho \left( {x,y} \right).\) The total mass of the lamina is expressed through the double integral as follows:

\[m = \iint\limits_R {\rho \left( {x,y} \right)dA} .\]

The static moment of the lamina about the \(x\)-axis is given by the formula

\[{M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA} .\]

Similarly, the static moment of the lamina about the \(y\)-axis is

\[{M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA} .\]

The coordinates of the center of mass of a lamina occupying the region \(R\) in the \(xy\)-plane with density function \(\rho \left( {x,y} \right)\) are described by the formulas

\[

{\bar x = \frac{{{M_y}}}{m} }

= {\frac{1}{m}\iint\limits_R {x\rho \left( {x,y} \right)dA} }

= {\frac{{\iint\limits_R {x\rho \left( {x,y} \right)dA} }}{{\iint\limits_R {\rho \left( {x,y} \right)dA} }},}

\]

\[

{\bar y = \frac{{{M_x}}}{m} }

= {\frac{1}{m}\iint\limits_R {y\rho \left( {x,y} \right)dA} }

= {\frac{{\iint\limits_R {y\rho \left( {x,y} \right)dA} }}{{\iint\limits_R {\rho \left( {x,y} \right)dA} }}.}

\]

When the mass density of the lamina is \(\rho \left( {x,y} \right) = 1\) for all \(\left( {x,y} \right)\) in the region \(R,\) the center of mass is defined only by the shape of the region and is called the centroid of \(R.\)

### Moments of Inertia of a Lamina

The moment of inertia of a lamina about the \(x\)-axis is defined by the formula

\[{{I_x} }={ \iint\limits_R {{y^2}\rho \left( {x,y} \right)dA} .}\]

Similarly, the moment of inertia of a lamina about the \(y\)-axis is given by

\[{{I_y} }={ \iint\limits_R {{x^2}\rho \left( {x,y} \right)dA} .}\]

The polar moment of inertia is

\[{{I_0} }={ \iint\limits_R {\left( {{x^2} + {y^2}} \right)\rho \left( {x,y} \right)dA} .}\]

### Charge of a Plate

Suppose electrical charge is distributed over a region which has area \(R\) in the \(xy\)-plane and its charge density is defined by the function \({\sigma \left( {x,y} \right)}.\) Then the total charge \(Q\) of the plate is defined by the expression

\[Q = \iint\limits_R {\sigma \left( {x,y} \right)dA} .\]

### Average of a Function

We give here the formula for calculation of the average value of a distributed function. Let \({f \left( {x,y} \right)}\) be a continuous function over a closed region \(R\) in the \(xy\)-plane. The average value \(\mu\) of the function \({f \left( {x,y} \right)}\) in the region \(R\) is given by the formula

\[\mu = \frac{1}{S}\iint\limits_R {f\left( {x,y} \right)dA} ,\]

where \(S = \iint\limits_R {dA} \) is the area of the region of integration \(R.\)

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the centroid of the lamina cut by the parabolas \({y^2} = x\) and \(y = {x^2}.\)### Example 2

Calculate the moments of inertia of the triangle bounded by the straight lines \(x + y = 1,\) \(x = 0,\) \(y = 0\) (Figure \(2\)) and having density \(\rho \left( {x,y} \right) = xy.\)### Example 3

Electric charge is distributed over the disk \({x^2} + {y^2} = 1\) so that its charge density is \(\sigma \left( {x,y} \right) =\) \( 1 + {x^2} + {y^2}\) \(\left( {\text{Kl/m}^2} \right).\) Calculate the total charge of the disk.### Example 1.

Find the centroid of the lamina cut by the parabolas \({y^2} = x\) and \(y = {x^2}.\)Solution.

The lamina has the form shown in Figure \(1.\)

Since it is homogeneous, we suppose that the density \(\rho \left( {x,y} \right) = 1.\) The mass of the lamina is

\[

{m = \iint\limits_R {dA} }

= {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]dx} }

= {\int\limits_0^1 {\left[ {\left. y \right|_{{x^2}}^{\sqrt x }} \right]dx} }

= {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} }

= {\int\limits_0^1 {\left( {{x^{\large\frac{1}{2}\normalsize}} – {x^2}} \right)dx} }

= {\left. {\left( {\frac{{2{x^{\large\frac{3}{2}\normalsize}}}}{3} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 }

= {\frac{2}{3} – \frac{1}{3} }={ \frac{1}{3}.}

\]

Now we find the moment of the lamina about the \(x\)-axis and \(y\)-axis.

\[

{{M_x} = \iint\limits_R {ydA} }

= {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {ydy} } \right]dx} }

= {\int\limits_0^1 {\left[ {\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_{{x^2}}^{\sqrt x }} \right]dx} }

= {\frac{1}{2}\int\limits_0^1 {\left( {x – {x^4}} \right)dx} }

= {\frac{1}{2}\left. {\left( {\frac{{{x^2}}}{2} – \frac{{{x^5}}}{5}} \right)} \right|_0^1 }

= {\frac{1}{2}\left( {\frac{1}{2} – \frac{1}{5}} \right) }={ \frac{3}{{20}},}

\]

\[

{{M_y} = \iint\limits_R {xdA} }

= {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]xdx} }

= {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)xdx} }

= {\int\limits_0^1 {\left( {{x^{\large\frac{3}{2}\normalsize}} – {x^3}} \right)dx} }

= {\left. {\left( {\frac{{2{x^{\large\frac{5}{2}\normalsize}}}}{5} – \frac{{{x^4}}}{4}} \right)} \right|_0^1 }

= {\frac{2}{5} – \frac{1}{4} }={ \frac{3}{{20}}.}

\]

Thus, the coordinates of the center of mass are

\[

{{\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}},\;\;}}\kern-0.3pt

{{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}}.}}

\]