MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 34

Find the center of gravity of the arc length of one quadrant of the circle. $$\begin{array}{c} r x r y x=rcos θ y=rsin θ \end{array}$$ The center of mass will be on the line $y=x$ so its $x\text{-coordinate}$ and its $y\text{-coordinate}$ will be the same. Chop up the curve into little pieces $\begin{array}{c} dy dx ds \end{array}$ $x\text{-coordinate}$ of position of little piece is $x\text{.}$ Mass of little piece is $\left(\text{density}\right)·ds\text{.}$ Add up little pieces from $\theta =0$ to $\theta =\frac{\pi }{2}\text{.}$ $$\begin{array}{ccc}{\displaystyle x\text{-coordinate of center of mass}}& =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}x\delta ds}}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta ds}}\text{where}\delta =\text{density}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}x\delta \sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta \sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}x\delta \sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta \sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta }}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta r\text{cos}\hspace{0.17em}\theta \sqrt{{(-r\text{sin}\hspace{0.17em}\theta )}^2+{\left(r\text{cos}\hspace{0.17em}\theta \right)}^2}d\theta }}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta \sqrt{{(-r\text{sin}\hspace{0.17em}\theta )}^2+{\left(\text{cos}\hspace{0.17em}\theta \right)}^2}d\theta }}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta r\text{cos}\hspace{0.17em}\theta \sqrt{r^2{\text{sin}}^2\theta +r^2{\text{cos}}^2\theta }d\theta }}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta \sqrt{r^2{\text{sin}}^2\theta +r^2{\text{cos}}^2\theta }d\theta }}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta r\text{cos}\hspace{0.17em}\theta \hspace{0.17em}rd\theta }}{{\displaystyle {\int }_{\theta =0}^{\theta =\frac{\pi }{2}}\delta rd\theta }}}\\ & =& {\displaystyle \frac{{\displaystyle \delta r^2\text{sin}\hspace{0.17em}\theta {|}_{\theta =0}^{\theta =\frac{\pi }{2}}}}{{\displaystyle \delta r\theta {|}_{\theta =0}^{\theta =\frac{\pi }{2}}}}}\\ & =& {\displaystyle \frac{{\displaystyle \delta r^2\text{sin}\hspace{0.17em}\frac{\pi }{2}-\delta r^2·0}}{{\displaystyle \delta r\frac{\pi }{2}-\delta r·0}}}\\ & =& {\displaystyle \frac{{\displaystyle \delta r^2}}{{\displaystyle \delta r\frac{\pi }{2}}}}\\ & =& {\displaystyle \frac{2r}{\pi }\text{.}}\end{array}$$

Find the center of gravity of the area bounded by the curve $y=x-x^2$ and the line $x+y=0\text{.}$ $$\begin{array}{c} 1 x -1 1 4 y x+y=0 y=x-x2 \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} L dx \end{array}\\ \text{mass of slice: (density)}Ldx\\ y\text{-coordinate of center of mass of slice is at}\\ \text{half way between top and bottom}\\ x\text{-coordinate of center of mass of slice is at}\\ x\text{-position of slice}\\ \text{Add up slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x\hspace{0.17em}\text{at right intersection point.}\end{array}$$ When $x+y=0$ and $y=x-x^2$ intersect $y=-x=x-x^2\text{.}$ So $x^2-2x=0\text{.}$ So $x(x-2)=0\text{.}$ So $x=0$ or $x=2\text{.}$ $$\begin{array}{ccc}{\displaystyle x\text{-coordinate of center of mass of area}}& =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\left(x\text{-position of slice}\right)\delta Ldx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta Ldx}}\text{where}\hspace{0.17em}\delta \hspace{0.17em}\text{is density,}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}x\delta (y_{\text{top}}-y_{\text{bottom}})dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (y_{\text{top}}-y_{\text{bottom}})dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}x\delta ((x-x^2)-(-x))dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta ((x-x^2)-(-x))dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\delta x(2x-x^2)dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (2x-x^2)dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\delta (2x^2-x^3)dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (2x-x^2)dx}}}\\ & =& {\displaystyle \frac{{\displaystyle \delta (\frac{2x^3}{3}-\frac{x^4}{4}){|}_{x=0}^{x=2}}}{{\displaystyle \delta (\frac{2x^2}{2}-\frac{x^3}{3}){|}_{x=0}^{x=2}}}}\\ & =& {\displaystyle \frac{{\displaystyle \frac{2·2^3}{e}-\frac{2^4}{4}-(0-0)}}{{\displaystyle \frac{22^2}{2}-\frac{2^3}{3}-(0-0)}}}\\ & =& {\displaystyle \frac{{\displaystyle 2^4(\frac{1}{3}-\frac{1}{4})}}{{\displaystyle 2^3(\frac{1}{2}-\frac{1}{3})}}}\\ & =& {\displaystyle \frac{2\left(\frac{1}{12}\right)}{\left(\frac{1}{6}\right)}}\\ & =& {\displaystyle \frac{2}{2}}\\ & =& {\displaystyle 1\text{.}}\end{array}$$ $$\begin{array}{ccc}{\displaystyle y\text{-coordinate of center of mass of area}}& =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\left(y\text{-position of slice}\right)\delta Ldx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta Ldx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}(\left(\frac{y_{\text{top}}-y_{\text{bottom}}}{2}\right)+y_{\text{bottom}})\delta (y_{\text{top}}-y_{\text{bottom}})dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (y_{\text{top}}-y_{\text{bottom}})dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\delta (\frac{(x-x^2)-(-x)}{2}+(-x))({(x-x)}^2-(-x))dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta ((x-x^2)-(-x))dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\delta (-\frac{x^2}{2})(2x-x^2)dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (2x-x^2)dx}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{x=0}^{x=2}\delta (-\frac{2x^3}{2}+\frac{x^4}{2})dx}}{{\displaystyle {\int }_{x=0}^{x=2}\delta (2x-x^2)dx}}}\\ & =& {\displaystyle \frac{{\displaystyle \delta (-\frac{x^4}{4}+\frac{x^5}{10}){|}_{x=0}^{x=2}}}{{\displaystyle \delta (x^2-\frac{x^3}{3}){|}_{x=0}^{x=2}}}}\\ & =& {\displaystyle \frac{{\displaystyle -\frac{2^4}{4}+\frac{2^5}{10}}}{{\displaystyle 2^4-\frac{2^3}{3}}}}\\ & =& {\displaystyle \frac{2^4(-\frac{1}{4}+\frac{1}{5})}{2^3(2-\frac{1}{3})}}\\ & =& {\displaystyle \frac{2(-\frac{1}{20})}{\frac{5}{3}}}\\ & =& {\displaystyle \frac{-2·3}{5·20}}\\ & =& {\displaystyle \frac{-3}{5·10}}\\ & =& {\displaystyle \frac{-3}{50}\text{.}}\end{array}$$ So the center of mass is $(1,-\frac{3}{50})\text{.}$

Notes and References

These are a typed copy of Lecture 34 from a series of handwritten lecture notes for the class MATH 221 given on December 1, 2000.

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