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Section 8.3 : Center Of Mass
1. Find the center of mass for the region bounded by \(y = 4 - {x^2}\) that is in the first quadrant.
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Start SolutionLet’s start out with a quick sketch of the region, with the center of mass indicated by the dot (the coordinates of this dot are of course to be determined in the final step…..).
We’ll also need the area of this region so let’s find that first.
\[A = \int_{0}^{2}{{4 - {x^2}\,dx}} = \left. {\left( {4x - \frac{1}{3}{x^3}} \right)} \right|_0^2 = \frac{{16}}{3}\] Show Step 2Next, we need to compute the two moments. We didn’t include the density in the computations below because it will only cancel out in the final step.
\[\begin{align*}{M_x} & = \int_{0}^{2}{{\frac{1}{2}{{\left( {4 - {x^2}} \right)}^2}\,dx}} = \int_{0}^{2}{{\frac{1}{2}\left( {16 - 8{x^2} + {x^4}} \right)\,dx}} = \left. {\frac{1}{2}\left( {16x - \frac{8}{3}{x^3} + \frac{1}{5}{x^5}} \right)} \right|_0^2 = \frac{{128}}{{15}}\\ {M_y} & = \int_{0}^{2}{{x\left( {4 - {x^2}} \right)\,dx}} = \int_{0}^{2}{{4x - {x^3}\,dx}} = \left. {\left( {2{x^2} - \frac{1}{4}{x^4}} \right)} \right|_0^2 = 4\end{align*}\] Show Step 3Finally, the coordinates of the center of mass is,
\[\overline{x} = \frac{{{M_y}}}{M} = \frac{{\rho \left( 4 \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{3}{4}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\overline{y} = \frac{{{M_x}}}{M} = \frac{{\rho \left( {\frac{{128}}{{15}}} \right)}}{{\rho \left( {\frac{{16}}{3}} \right)}} = \frac{8}{5}\]The center of mass is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\frac{3}{4},\frac{8}{5}} \right)}}\).