Calculus

Double Integrals

Physical Applications of Double Integrals

Page 1
Problem 1
Page 2
Problems 2-3

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 on problem description to see 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
\[
{m = \iint\limits_R {dA} }
= {\int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]dx} }
= {\int\limits_0^1 {\left[ {\left. y \right|_{{x^2}}^{\sqrt x }} \right]dx} }
= {\int\limits_0^1 {\left( {\sqrt x – {x^2}} \right)dx} }
= {\int\limits_0^1 {\left( {{x^{\large\frac{1}{2}\normalsize}} – {x^2}} \right)dx} }
= {\left. {\left( {\frac{{2{x^{\large\frac{3}{2}\normalsize}}}}{3} – \frac{{{x^3}}}{3}} \right)} \right|_0^1 }
= {\frac{2}{3} – \frac{1}{3} }={ \frac{1}{3}.}
\]

A lamina formed by two parabolas

Figure 1.

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
\[
{{\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}},\;\;}}\kern-0.3pt
{{\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} }={ \frac{9}{{20}}.}}
\]

Page 1
Problem 1
Page 2
Problems 2-3