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 33

Find the area of the surface obtained by rotating the curve determined by $x=a{\text{cos}}^3\theta ,$ $y=a{\text{sin}}^3\theta$ about the $x\text{-axis.}$

To graph this: $${\text{cos}}^3\theta =\frac{x}{a}\text{and}{\text{sin}}^3\theta =\frac{y}{a}\text{.}$$ So $$\text{cos}\hspace{0.17em}\theta ={\left(\frac{x}{a}\right)}^{\frac{1}{3}}\text{and}\text{sin}\hspace{0.17em}\theta ={\left(\frac{y}{a}\right)}^{\frac{1}{3}}\text{.}$$ Since $${\text{cos}}^2\theta +{\text{sin}}^2\theta =1,{\left(\frac{x}{a}\right)}^{\frac{2}{3}}+{\left(\frac{y}{a}\right)}^{\frac{2}{3}}=1\text{.}$$ So we have to graph $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\text{.}$ $$\begin{array}{c} - a a x - a a y x2+y2=a2 \end{array}\begin{array}{c} x y x1+y1=a1 \end{array}\begin{array}{c} - a a x - a a y x23+y23=a23 \end{array}$$ So when this is rotated about the $x\text{-axis}$ $$\begin{array}{c} - a a x - a a y \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} R ⏟ ds \end{array}\\ \text{Surface area of a slice:}\hspace{0.17em}\pi R^{2}ds\text{.}\\ \text{Add up slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=a\hspace{0.17em}\text{and then multiply by}\hspace{0.17em}2\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle \text{Surface area}}& =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi R^{2}ds}\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi y^2\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi y^2\frac{\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2}}{d\theta }d\theta }\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi y^2\sqrt{{\left(dx\right)}^2+{\left(dy\right)}^2{\left(d\theta \right)}^2}d\theta }\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi y^2\sqrt{{\left(\frac{dx}{d\theta }\right)}^2+{\left(\frac{dy}{d\theta }\right)}^2}d\theta \text{.}}\end{array}$$ Since $x=a{\text{cos}}^3\theta$ and $y=a{\text{sin}}^3\theta$ $$\frac{dx}{d\theta }=-3a{\text{cos}}^2\theta \text{sin}\hspace{0.17em}\theta \text{and}\frac{dy}{d\theta }=3a{\text{sin}}^2\theta \hspace{0.17em}\text{cos}\hspace{0.17em}\theta \text{.}$$ So $$\begin{array}{ccc}{\displaystyle \text{Surface area}}& =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi y^2\sqrt{{(-3a{\text{cos}}^2\text{sin}\hspace{0.17em}\text{sin}\hspace{0.17em}\theta )}^2+{\left(3a{\text{sin}}^2\theta \hspace{0.17em}\text{cos}\hspace{0.17em}\theta \right)}^2}d\theta }\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi a^2{\text{sin}}^6\theta \sqrt{9a^2{\text{cos}}^4\theta \hspace{0.17em}{\text{sin}}^2\theta +9a^2{\text{sin}}^4\hspace{0.17em}\theta \hspace{0.17em}{\text{cos}}^2\theta }d{t}_h}\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi a^2{\text{sin}}^6\theta \sqrt{9a^2{\text{sin}}^2\theta {\text{cos}}^2\theta ({\text{cos}}^2\theta +{\text{sin}}^2\theta )}d\theta }\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}\pi a^2{\text{sin}}^6\theta \hspace{0.17em}3a\text{sin}\theta \hspace{0.17em}\text{cos}\hspace{0.17em}\theta \hspace{0.17em}d\theta }\\ & =& {\displaystyle 2{\int }_{x=0}^{x=a}3\pi a^3{\text{sin}}^7\theta \text{cos}\hspace{0.17em}\theta \hspace{0.17em}d\theta }\\ & =& {\displaystyle 6\pi a^3\frac{{\text{sin}}^8\theta }{8}{|}_{x=0}^{x=a}}\\ & =& {\displaystyle \frac{6\pi a^3}{8}{\text{sin}}^8\theta {|}_{a{\text{cos}}^3\theta =0}^{a{\text{cos}}^3\theta =a}}\\ & =& {\displaystyle \frac{3\pi a^3}{4}{\text{sin}}^8\theta {|}_{\text{cos}\hspace{0.17em}\theta =0}^{\text{cos}\hspace{0.17em}\theta =1}}\\ & =& {\displaystyle \frac{3a^3\pi }{4}{\text{sin}}^8\theta {|}_{\theta =\frac{\pi }{2}}^{\theta =0}}\\ & =& {\displaystyle \frac{3\pi a^3}{4}{\text{sin}}^{8}0-\frac{3\pi a^3}{4}{\text{sin}}^8\frac{\pi }{2}}\\ & =& {\displaystyle -\frac{3\pi a^3}{4}·1^8}\\ & =& {\displaystyle -\frac{3\pi a^3}{4}\text{.}}\end{array}$$ So the surface area is $\frac{3\pi a^3}{4}\text{.}$

Moments and center of mass

A moment and a center of mass are the same thing.

The center of mass is the average position of the mass in an object. $$\begin{array}{ccc}{\displaystyle \text{Center of mass}}& =& {\displaystyle \frac{\text{position of mass}·\text{mass}}{\text{mass}}}\\ & =& {\displaystyle \frac{\int \left(\text{position of a slice}\right)·\left(\text{mass of a slice}\right)}{\int \left(\text{mass of a slice}\right)}\text{.}}\end{array}$$ Note: $$\text{mass of a slice}=\left(\text{volume of slice}\right)·\left(\text{density of the slice}\right)\text{.}$$ Center of mass and center of gravity are the same thing.

Find the center of mass of a solid hemisphere of radius $r$ if its density at a point $P$ is proportional to the distance between $P$ and the base of the hemisphere. $$\begin{array}{c} - r r x r y \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} R dy } \end{array}\\ \text{Volume of a slice:}\hspace{0.17em}\pi R^{2}dy\\ \text{Density of a slice}=\text{Height of a slice}\\ \text{Mass of a slice}=\pi R^{2}dy\left(\text{height of slice}\right)\\ \text{Add slices from}\hspace{0.17em}y=0\hspace{0.17em}\text{to}\hspace{0.17em}y=r\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle \text{Center of mass}}& =& {\displaystyle \frac{{\displaystyle \int \left(\text{center of mass of slice}\right)\left(\text{mass of slice}\right)}}{{\displaystyle \int \left(\text{mass of slice}\right)}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{y=0}^{y=r}y\pi R^{2}dy\left(\text{height of slice}\right)}}{{\displaystyle {\int }_{y=0}^{y=r}\pi R^{2}dy\left(\text{height of slice}\right)}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{y=0}^{y=r}y\pi x^{2}dy·y}}{{\displaystyle {\int }_{y=0}^{y=r}\pi x^{2}dy·y}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{y=0}^{y=r}\pi x^2{y}^{2}dy}}{{\displaystyle {\int }_{y=0}^{y=r}\pi x^{2}ydy}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{y=0}^{y=r}\pi (r^2-y^2)y^{2}dy}}{{\displaystyle {\int }_{y=0}^{y=r}\pi (r^2-y^2)ydy}}}\\ & =& {\displaystyle \frac{{\displaystyle {\int }_{y=0}^{y=r}(\pi r^2{y}^2-\pi y^4)dy}}{{\displaystyle {\int }_{y=0}^{y=r}(\pi r^{2}y-\pi y^3)dy}}}\\ & =& {\displaystyle \frac{{\displaystyle \frac{\pi r^2{y}^3}{3}-\frac{\pi y^5}{5}{|}_{y=0}^{y=r}}}{{\displaystyle \frac{\pi r^2{y}^2}{2}-\frac{\pi y^4}{4}{|}_{y=0}^{y=r}}}}\\ & =& {\displaystyle \frac{{\displaystyle \frac{\pi r^5}{3}-\frac{\pi r^5}{5}-(0-0)}}{{\displaystyle \frac{\pi r^4}{2}-\frac{\pi r^4}{4}-(0-0)}}}\\ & =& {\displaystyle \frac{{\displaystyle \pi r^5(\frac{1}{3}-\frac{1}{5})}}{{\displaystyle \pi r^4(\frac{1}{2}-\frac{1}{4})}}}\\ & =& {\displaystyle r\frac{\left(\frac{2}{15}\right)}{\left(\frac{1}{4}\right)}}\\ & =& {\displaystyle r\frac{2}{15}·4}\\ & =& {\displaystyle \frac{8r}{15}\text{.}}\end{array}$$ So the center of mass is at $(0,\frac{8r}{15})\text{.}$

Notes and References

These are a typed copy of Lecture 33 from a series of handwritten lecture notes for the class MATH 221 given on November 29, 2000.

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