Chapter 5: Applications of Integration

Section 5.7: Centroids

Example 5.7.6

$$ Determine the centroid of $C$, the upper half of the unit circle whose center is at the origin. $$ Solution Mathematical Solution $$ The curve $C$ can be defined by the equation $y\=\sqrt{1-x^2}$, $x\in \left[-1\,1\right]$, so $S\=\mathrm{\π}$,    $\mathrm{ds}\=\sqrt{1\+y{\prime }^2}\mathrm{dx}\=\sqrt{1\+{\left(-\frac{x}{\sqrt{1- x^2}}\right)}^2}\mathrm{dx}\=\frac{\mathrm{dx}}{\sqrt{1-x^2}}$  and $\stackrel{\&conjugate0;}{x}$ $\=\frac{1}{\mathrm{\pi }}{\int }_{-1}^1\frac{x}{\sqrt{1-x^2}} \mathrm{dx}\=0$ $\stackrel{\&conjugate0;}{y}$ $\=\frac{1}{\mathrm{\π}}{\int }_{-1}^1\sqrt{1-x^2} \frac{\mathrm{dx}}{\sqrt{1- x^2}}$ $$ $\=\frac{1}{\mathrm{\π}}{\int }_{-1}^{1}1 \mathrm{dx}$ $$ $\=\frac{2}{\mathrm{\π}}$ Clearly, since $y\left(0\right)\=1\ne 2\/\mathrm{\π}$, the centroid does not fall on the curve itself.   Maple Solution $$ Define the function $y$  •  Context Panel: Assign Function $y\left(x\right)\=\sqrt{1-x^2}$$\stackrel{\text{assign as function}}{\to }$$y$ Calculate the coordinates of the centroid •  Write the integral for $\stackrel{\&conjugate0;}{x}$. •  Context Panel: Evaluate and Display Inline $\frac{1}{\mathrm{\π}}{\int }_{-1}^{1}x\sqrt{1\+y\prime {\left(x\right)}^2} \mathit{\ⅆ}x$ = $0$$$ •  Write the integral for $\stackrel{\&conjugate0;}{y}$. •  Context Panel: Evaluate and Display Inline $\frac{1}{\mathrm{\π}}{\int }_{-1}^{1}y\left(x\right)\sqrt{1\+y\prime {\left(x\right)}^2} \mathit{\ⅆ}x$ = $\frac{2}{\mathrm{\π}}$$$ $$

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