Show that the centroid (x,y) for the circular region shown in Fig. 3.24 is (0, 0).

Question

Show that the centroid [latex](\overline{x},\overline{y})[/latex] for the circular region shown in Fig. 3.24 is (0, 0).

Accepted Answer

Knowing that [latex]x^{2}+y^{2}=a^{2},[/latex] or [latex]y=\sqrt{a^{2}-x^{2}},[/latex] we can compute

FIGURE 3.24 Determine the first moment of area for this circular cross section relative to two different coordinate systems.

[latex] A=\int_{-a}^{a}{\left(\int_{-\sqrt{a^{2}-x^{2}} }^{\sqrt{a^{2}-x^{2}} }{dy} ight)dx }=2\left(\frac{1}{2}a^{2} ight)\left(\frac{\pi }{2}-\frac{-\pi }{2} ight)=\pi a^{2}.[/latex]

Note from integral tables that

[latex] \int{\sqrt{a^{2}-x^{2}} }dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin ^{-1}\left(\frac{x}{\left|a ight| } ight).[/latex]

Alternatively, knowing that [latex]x=r\cos heta[/latex] and [latex]y=r\sin heta,[/latex] we can compute

[latex] A=\int_{0}^{a}{\left(\int_{0}^{2\pi }{rd heta } ight)dr }=2\pi \left(\frac{r^{2}}{2}\mid ^{a}_{0} ight)=\pi a^{2}.[/latex]

Likewise, we can compute the centroid [latex](\overline{x},\overline{y})[/latex] in either Cartesian or cylindrical coordinates. In cylindricals,

[latex]\iint{xdA}=\int_{0}^{a}{\left(\int_{0}^{2\pi }{r\cos heta rd heta } ight)dr }=\int_{0}^{a}{r^{2}dr}\int_{0}^{2\pi }{\cos heta d heta }=\left(\frac{a^{3}}{3} ight)(\sin 2\pi -\sin 0)=0[/latex]

and, similarly,

[latex]\iint{ydA}=\int_{0}^{a}{\left(\int_{0}^{2\pi }{r\sin heta rd heta } ight)dr }=\int_{0}^{a}{r^{2}dr}\int_{0}^{2\pi }{\sin heta d heta }=\left(\frac{a^{3}}{3} ight)(-\cos 2\pi -\cos 0)=0;[/latex]

therefore,

[latex] \overline{x}=0, \overline{y}=0,[/latex]

as expected. Repeat using Cartesians alone.

Question A3.2: Show that the centroid (x,y) for the circular region shown i...

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