Center of Mass and Moments
Solved Problems
Example 5.
Find the centroid of the region bounded by the cubic curve \[y = x^3,\] the vertical line \(x = 1,\) and the \(x-\)axis.
Solution.
Figure 9. Assuming that \(\rho = 1,\) we calculate the first moments \({M_x}\) and \({M_y}:\)
\[{M_x} = \frac{1}{2}\int\limits_a^b {\rho \left( x \right)\left[ {{f^2}\left( x \right) - {g^2}\left( x \right)} \right]dx} = \frac{1}{2}\int\limits_0^1 {{{\left( {{x^3}} \right)}^2}dx} = \frac{1}{2}\int\limits_0^1 {{x^6}dx} = \left. {\frac{{{x^7}}}{{14}}} \right|_0^1 = \frac{1}{{14}};\]
\[{M_y} = \int\limits_a^b {x\rho \left( x \right)\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_0^1 {\left( {x \cdot {x^3}} \right)dx} = \int\limits_0^1 {{x^4}dx} = \left. {\frac{{{x^5}}}{5}} \right|_0^1 = \frac{1}{5}.\]
The mass \(m\) of the region is given by
\[m = \int\limits_a^b {\rho \left( x \right)\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \int\limits_0^1 {{x^3}dx} = \left. {\frac{{{x^4}}}{4}} \right|_0^1 = \frac{1}{4}.\]
It follows from here that
\[\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{1}{5}}}{{\frac{1}{4}}} = \frac{4}{5},\;\; \bar y = \frac{{{M_x}}}{m} = \frac{{\frac{1}{{14}}}}{{\frac{1}{4}}} = \frac{2}{7}.\]
Example 6.
Find the centroid of the region enclosed by the curves \[y = \sqrt x,\,y = {x^2}.\]
Solution.
Figure 10. Let \(\rho\) be the density of the lamina. The total mass is equal to
\[m = \rho \int\limits_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \rho \int\limits_0^1 {\left( {\sqrt x - {x^2}} \right)dx} = \rho \left. {\left( {\frac{{2{x^{\frac{3}{2}}}}}{3} - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = \frac{\rho }{3}.\]
Compute the first moments \({M_x}\) and \({M_y}:\)
\[{M_x} = \frac{\rho }{2}\int\limits_a^b {\left[ {{f^2}\left( x \right) - {g^2}\left( x \right)} \right]dx} = \frac{\rho }{2}\int\limits_0^1 {\left[ {{{\left( {\sqrt x } \right)}^2} - {{\left( {{x^2}} \right)}^2}} \right]dx} = \frac{\rho }{2}\int\limits_0^1 {\left( {x - {x^4}} \right)dx} = \frac{\rho }{2}\left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = \frac{{3\rho }}{{20}};\]
\[{M_y} = \rho \int\limits_a^b {x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} = \rho \int\limits_0^1 {x\left( {\sqrt x - {x^2}} \right)dx} = \rho \int\limits_0^1 {\left( {{x^{\frac{3}{2}}} - {x^3}} \right)dx} = \rho \left. {\left( {\frac{{2{x^{\frac{5}{2}}}}}{5} - \frac{{{x^4}}}{4}} \right)} \right|_0^1 = \frac{{3\rho }}{{20}}.\]
Hence, the centroid of the region \(G\left( {\bar x,\bar y} \right)\) has the coordinates
\[\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{{3 \bcancel{\rho} }}{{20}}}}{{\frac{ \bcancel{\rho} }{3}}} = \frac{9}{{20}},\;\; \bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{3 \bcancel{\rho} }}{{20}}}}{{\frac{ \bcancel{\rho} }{3}}} = \frac{9}{{20}}.\]
Example 7.
Find the centroid of the region bounded by the arc of the ellipse \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\] lying in the first quadrant and the coordinate axes.
Solution.
Figure 11. It is convenient to write the equation of the arc 0f the ellipse in explicit form:
\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\;\; \Rightarrow \frac{{{y^2}}}{{{b^2}}} = 1 - \frac{{{x^2}}}{{{a^2}}},\;\; \Rightarrow {y^2} = \frac{{{b^2}}}{{{a^2}}}\sqrt {{a^2} - {x^2}} ,\;\; \Rightarrow y = \frac{b}{a}\sqrt {{a^2} - {x^2}} .\]
Determine the mass of the region assuming that \(\rho = 1.\)
\[m = \int\limits_0^a {f\left( x \right)dx} = \frac{b}{a}\int\limits_0^a {\sqrt {{a^2} - {x^2}} dx} .\]
This integral can be evaluated using the substitution
\[x = a\sin t,\;\; \Rightarrow \sqrt {{a^2} - {x^2}} = a\cos t,\;\; dx = a\cos tdt.\]
When \(x = 0,\) \(t = 0,\) and when \(x = a,\) \(t = \frac{\pi}{2}.\) So the mass of the region is given by
\[m = \frac{b}{a}\int\limits_0^{\frac{\pi }{2}} {a\cos t \cdot a\cos tdt} = ab\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} = \frac{{ab}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} = \frac{{ab}}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}} = \frac{{ab}}{2} \cdot \frac{\pi }{2} = \frac{{\pi ab}}{4}.\]
Calculate the first moments \({M_x}\) and \({M_y}.\)
\[{M_x} = \frac{1}{2}\int\limits_0^a {{f^2}\left( x \right)dx} = \frac{{{b^2}}}{{2{a^2}}}\int\limits_0^a {\left( {{a^2} - {x^2}} \right)dx} = \frac{{{b^2}}}{{2{a^2}}}\left. {\left( {{a^2}x - \frac{{{x^3}}}{3}} \right)} \right|_0^a = \frac{{{b^2}}}{{2{a^2}}}\left( {{a^3} - \frac{{{a^3}}}{3}} \right) = \frac{{{b^2}a}}{3}.\]
\[{M_y} = \int\limits_0^a {xf\left( x \right)dx} = \frac{b}{a}\int\limits_0^a {x\sqrt {{a^2} - {x^2}} dx} .\]
Make the substitution:
\[{a^2} - {x^2} = {z^2},\;\; \Rightarrow - 2xdx = 2zdz,\;\; \Rightarrow xdx = - zdz.\]
When \(x = 0,\) \(z = a,\) and when \(x = a,\) \(z = 0.\) Hence,
\[{M_y} = \frac{b}{a}\int\limits_a^0 {\left( { - {z^2}} \right)dz} = \frac{b}{a}\int\limits_0^a {{z^2}dz} = \left. {\frac{b}{a}\frac{{{z^3}}}{3}} \right|_0^a = \frac{{b{a^2}}}{3}.\]
The centroid of the region \(G\left( {\bar x,\bar y} \right)\) has the following coordinates:
\[\bar x = \frac{{{M_y}}}{m} = \frac{{\frac{{b{a^2}}}{3}}}{{\frac{{\pi ab}}{4}}} = \frac{{4a}}{{3\pi }},\;\; \bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{{b^2}a}}{3}}}{{\frac{{\pi ab}}{4}}} = \frac{{4b}}{{3\pi }}.\]
Example 8.
An isosceles triangle with vertices \(A( - 1,0),\) \(B(1,0),\) \(C(0,2)\) has the density distribution according to the law \[\rho \left( y \right) = 1 + {y^2}.\] Find the center of mass of the triangle.
Solution.
Figure 12. By symmetry, the center of mass \(G\left( {\bar x,\bar y} \right)\) of the triangle must lie on the \(y-\)axis, so we need to determine only the \(\bar y-\)coordinate.
The density \(\rho\) of the triangular lamina varies along the \(y-\)axis. Therefore, to calculate the first moment \({M_x},\) we use the formula
\[{M_x} = \int\limits_c^d {y\rho \left( y \right)\left[ {f\left( y \right) – g\left( y \right)} \right]dy}.\]
The legs of the isosceles triangle have the following equations:
\[AC:\;x = g\left( y \right) = \frac{y}{2} - 1;\]
\[BC:\;x = f\left( y \right) = -\frac{y}{2} + 1.\]
Since
\[f\left( y \right) - g\left( y \right) = \left( { - \frac{y}{2} + 1} \right) - \left( {\frac{y}{2} - 1} \right) = 2 - y,\]
the first moment \({M_x}\) is equal to
\[{M_x} = \int\limits_0^2 {y\left( {1 + {y^2}} \right)\left( {2 - y} \right)dy} = \int\limits_0^2 {\left( {2y - {y^2} + 2{y^3} - {y^4}} \right)dy} = \left. {\left( {{y^2} - \frac{{{y^3}}}{3} + \frac{{{y^4}}}{2} - \frac{{{y^5}}}{5}} \right)} \right|_0^2 = 4 - \frac{8}{3} + 8 - \frac{{32}}{5} = \frac{{44}}{{15}}.\]
Compute the mass of the lamina:
\[m = \int\limits_c^d {\rho \left( y \right)\left[ {f\left( y \right) - g\left( y \right)} \right]dy} = \int\limits_0^2 {\left( {1 + {y^2}} \right)\left( {2 - y} \right)dy} = \int\limits_0^2 {\left( {2 - y + 2{y^2} - {y^3}} \right)dy} = \left. {\left( {2y - \frac{{{y^2}}}{2} + \frac{{2{y^3}}}{3} - \frac{{{y^4}}}{4}} \right)} \right|_0^2 = \cancel{4} - 2 + \frac{{16}}{3} - \cancel{4} = \frac{{10}}{3}.\]
Hence,
\[\bar y = \frac{{{M_x}}}{m} = \frac{{\frac{{44}}{{15}}}}{{\frac{{10}}{3}}} = \frac{{22}}{{25}}.\]
Thus, the center of mass of the triangular lamina is at the point
\[G\left( {\bar x,\bar y} \right) = G\left( {0,\frac{{22}}{{25}}} \right).\]
Example 9.
Find the centroid of a circular sector with radius \(R\) and angle \(\alpha.\)
Solution.
Figure 13. Let \(G\left( {\bar x,\bar y} \right)\) be the centroid of the circular sector. It is convenient to consider the circle as a polar curve given by the equation \(r\left( \theta \right) = R,\) where the angle \(\theta\) varies from \(0\) to \(\alpha.\)
We can compute the coordinates \(\bar x,\) \(\bar y\) by the formulas
\[\bar x = \frac{2}{3}\frac{{\int\limits_0^\alpha {{r^3}\left( \theta \right)\cos \theta d\theta } }}{{\int\limits_0^\alpha {{r^2}\left( \theta \right)d\theta } }},\;\; \bar y = \frac{2}{3}\frac{{\int\limits_0^\alpha {{r^3}\left( \theta \right)\sin \theta d\theta } }}{{\int\limits_0^\alpha {{r^2}\left( \theta \right)d\theta } }}.\]
This yields:
\[\bar x = \frac{2}{3}\frac{{\int\limits_0^\alpha {{R^3}\cos \theta d\theta } }}{{\int\limits_0^\alpha {{R^2}d\theta } }} = \frac{{2R}}{3}\frac{{\int\limits_0^\alpha {\cos \theta d\theta } }}{{\int\limits_0^\alpha {d\theta } }} = \frac{{2R}}{3}\frac{{\left. {\left( {\sin \theta } \right)} \right|_0^\alpha }}{\alpha } = \frac{{2R\sin \alpha }}{{3\alpha }};\]
\[\bar y = \frac{2}{3}\frac{{\int\limits_0^\alpha {{R^3}\sin \theta d\theta } }}{{\int\limits_0^\alpha {{R^2}d\theta } }} = \frac{{2R}}{3}\frac{{\int\limits_0^\alpha {\sin \theta d\theta } }}{{\int\limits_0^\alpha {d\theta } }} = \frac{{2R}}{3}\frac{{\left. {\left( { - \cos \theta } \right)} \right|_0^\alpha }}{\alpha } = \frac{{2R\left( {1 - \cos \alpha } \right)}}{{3\alpha }}.\]
Thus, the centroid of the circular sector is located at the point
\[G\left( {\bar x,\bar y} \right) = G\left( {\frac{{2R\sin \alpha }}{{3\alpha }},\frac{{2R\left( {1 - \cos \alpha } \right)}}{{3\alpha }}} \right).\]
Example 10.
Find the centroid of the region enclosed by the cardioid \[r\left( \theta \right) = 1 + \cos \theta .\]
Solution.
Figure 14. Let the centroid be at the point \(G\left( {\bar x,\bar y} \right).\) By symmetry, the \(\bar y\) coordinate of the centroid is equal to zero. We calculate the \(\bar x\) coordinate by the formula
\[\bar x = \frac{2}{3}\frac{{\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\cos \theta d\theta } }}{{\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }}.\]
The integral in the denominator \({\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } }\) is equal to \(2A,\) where \(A\) is the area of the polar region. We can easily evaluate it:
\[\int\limits_\alpha ^\beta {{r^2}\left( \theta \right)d\theta } = 2\int\limits_0^\pi {{{\left( {1 + \cos \theta } \right)}^2}d\theta } = 2\int\limits_0^\pi {\left( {1 + 2\cos \theta + {{\cos }^2}\theta } \right)d\theta } = 2\int\limits_0^\pi {\left( {\frac{3}{2} + 2\cos \theta + \frac{{\cos 2\theta }}{2}} \right)d\theta } = 2\left. {\left( {\frac{3}{2}\theta + 2\sin \theta + \frac{{\sin 2\theta }}{4}} \right)} \right|_0^\pi = 3\pi .\]
Consider now the integral in the numerator:
\[\int\limits_\alpha ^\beta {{r^3}\left( \theta \right)\cos \theta d\theta } = \int\limits_\alpha ^\beta {{{\left( {1 + \cos \theta } \right)}^3}\cos \theta d\theta } .\]
The integrand is written in the form:
\[{\left( {1 + \cos \theta } \right)^3}\cos \theta = \cos \theta + 3{\cos ^2}\theta + 3{\cos ^3}\theta + {\cos ^4}\theta .\]
Using the trig identities
\[{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2},\]
\[{\cos ^3}\theta = \frac{{3\cos \theta + \cos 3\theta }}{4},\]
\[{\cos ^4}\theta = \frac{{\cos 4\theta + 4\cos 2\theta + 3}}{8},\]
we can rewrite the integrand as follows:
\[{\left( {1 + \cos \theta } \right)^3}\cos \theta = \frac{{15}}{8} + \frac{{13\cos \theta }}{4} + 2\cos 2\theta + \frac{{3\cos 3\theta }}{4} + \frac{{\cos 4\theta }}{8}.\]
Integrating from \(\theta = 0\) to \(\theta = 2\pi\) gives the following answer:
\[\int\limits_0^{2\pi } {{{\left( {1 + \cos \theta } \right)}^3}\cos \theta d\theta } = \frac{{15}}{8} \cdot 2\pi = \frac{{15\pi }}{4}\]
Hence, the \(\bar x\) coordinate of the centroid is equal to
\[\bar x = \frac{2}{3}\frac{{\frac{{15\pi }}{4}}}{{3\pi }} = \frac{5}{6},\]
so the answer is \(G\left( {\bar x,\bar y} \right) = \left( {\frac{5}{6},0} \right).\)
