Calculus

Double Integrals

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Physical Applications of Double Integrals

  • 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 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 \(1.\)

    A lamina formed by two parabolas
    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

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

    Page 1
    Problem 1
    Page 2
    Problems 2-3