Physical Applications of Double Integrals
Solved Problems
Example 1.
Find the centroid of the lamina cut by the parabolas \[{y^2} = x, y = {x^2}.\]
Solution.
The lamina has the form shown in Figure \(1.\)
Figure 1. Since it is homogeneous, we suppose that the density \(\rho \left( {x,y} \right) = 1.\) The mass of the lamina is
\[m = \iint\limits_R {dA} = \int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]dx} = \int\limits_0^1 {\left[ {\left. y \right|_{{x^2}}^{\sqrt x }} \right]dx} = \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} = \int\limits_0^1 {\left( {{x^{\frac{1}{2}}} - {x^2}} \right)dx} = \left. {\left( {\frac{{2{x^{\frac{3}{2}}}}}{3} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}.\]
Now we find the moment of the lamina about the \(x\)-axis and \(y\)-axis.
\[{M_x} = \iint\limits_R {ydA} = \int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {ydy} } \right]dx} = \int\limits_0^1 {\left[ {\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_{{x^2}}^{\sqrt x }} \right]dx} = \frac{1}{2}\int\limits_0^1 {\left( {x - {x^4}} \right)dx} = \frac{1}{2}\left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{5}} \right) = \frac{3}{{20}},\]
\[{M_y} = \iint\limits_R {xdA} = \int\limits_0^1 {\left[ {\int\limits_{{x^2}}^{\sqrt x } {dy} } \right]xdx} = \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)xdx} = \int\limits_0^1 {\left( {{x^{\frac{3}{2}}} - {x^3}} \right)dx} = \left. {\left( {\frac{{2{x^{\frac{5}{2}}}}}{5} - \frac{{{x^4}}}{4}} \right)} \right|_0^1 = \frac{2}{5} - \frac{1}{4} = \frac{3}{{20}}.\]
Thus, the coordinates of the center of mass are
\[\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} = \frac{9}{{20}},\;\; \bar y = \frac{{{M_x}}}{m} = \frac{{\frac{3}{{20}}}}{{\frac{1}{3}}} = \frac{9}{{20}}.\]
Example 2.
Calculate the moments of inertia of the triangle bounded by the straight lines \[x + y = 1, x = 0, y = 0\] and having density \[\rho \left( {x,y} \right) = xy.\]
Solution.
Figure 2. The moment of inertia about the \(x\)-axis is
\[ {I_x} = \iint\limits_R {{y^2}\rho \left( {x,y} \right)dxdy} = \int\limits_0^1 {\left[ {\int\limits_0^{1 - x} {{y^2}xydy} } \right]dx} = \int\limits_0^1 {\left[ {\int\limits_0^{1 - x} {{y^3}dy} } \right]xdx} = \int\limits_0^1 {\left[ {\left. {\left( {\frac{{{y^4}}}{4}} \right)} \right|_0^{1 - x}} \right]xdx} = \frac{1}{4}\int\limits_0^1 {{{\left( {1 - x} \right)}^4}xdx} = \frac{1}{4}\int\limits_0^1 {\left( {1 - 4x + 6{x^2} - 4{x^3} + {x^4}} \right)xdx} = \frac{1}{4}\int\limits_0^1 {\left( {x - 4{x^2} + 6{x^3} - 4{x^4} + {x^5}} \right)dx} = \frac{1}{4}\left. {\left( {\frac{{{x^2}}}{2} - \frac{{4{x^3}}}{3} + \frac{{6{x^4}}}{4} - \frac{{4{x^5}}}{5} + \frac{{{x^6}}}{6}} \right)} \right|_0^1 = \frac{1}{4}\left( {\frac{1}{2} - \frac{4}{3} + \frac{3}{2} - \frac{4}{5} + \frac{1}{6}} \right) = \frac{{1}}{{120}}. \]
Similarly, we can find the moment of inertia about the \(y\)-axis:
\[{I_y} = \iint\limits_R {{x^2}\rho \left( {x,y} \right)dxdy} = \int\limits_0^1 {\left[ {\int\limits_0^{1 - x} {{x^2}xydy} } \right]dx} = \int\limits_0^1 {\left[ {\int\limits_0^{1 - x} {ydy} } \right]{x^3}dx} = \int\limits_0^1 {\left[ {\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_0^{1 - x}} \right]{x^3}dx} = \frac{1}{2}\int\limits_0^1 {{{\left( {1 - x} \right)}^2}{x^3}dx} = \frac{1}{2}\int\limits_0^1 {\left( {1 - 2x + {x^2}} \right){x^3}dx} = \frac{1}{2}\int\limits_0^1 {\left( {{x^3} - 2{x^4} + {x^5}} \right)dx} = \frac{1}{2}\left. {\left( {\frac{{{x^4}}}{4} - \frac{{2{x^5}}}{5} + \frac{{{x^6}}}{6}} \right)} \right|_0^1 = \frac{1}{2}\left( {\frac{1}{4} - \frac{2}{5} + \frac{1}{6}} \right) = \frac{1}{{120}}.\]
Example 3.
Electric charge is distributed over the disk \[{x^2} + {y^2} = 1\] so that its charge density is \[\sigma \left( {x,y} \right) = 1 + {x^2} + {y^2} \left( {\text{C/m}^2} \right).\] Calculate the total charge of the disk.
Solution.
In polar coordinates, the region occupied by the disk is defined by the set
\[\left\{ {\left( {r,\theta } \right)|\;0 \le r \le 1,\; 0 \le \theta \le 2\pi } \right\}.\]
The total charge is
\[Q = \iint\limits_R {\sigma \left( {x,y} \right)dxdy} = \int\limits_0^{2\pi } {d\theta } \int\limits_0^1 {\left( {1 + {r^2}} \right)rdr} = 2\pi \int\limits_0^1 {\left( {r + {r^3}} \right)dr} = 2\pi \left. {\left( {\frac{{{r^2}}}{2} + \frac{{{r^4}}}{4}} \right)} \right|_0^1 = 2\pi \left( {\frac{1}{2} + \frac{1}{4}} \right) = \frac{{3\pi }}{2}\;\left( {\text{C}} \right).\]
