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 29

Find the volume generated when the area bounded by $y=\text{sin}\hspace{0.17em}x,$ $0\le x\le \pi ,$ and $y=0$ is rotated about the $x\text{-axis.}$ $$\begin{array}{c} π 2 π x 1 y y=sin x \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} R ⏟ dx \end{array}\\ \text{Volume of a slice:}\hspace{0.17em}\pi R^{2}dx\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=\pi \text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=\pi }\pi R^{2}dx}& =& {\displaystyle {\int }_{x=0}^{x=\pi }\pi y^{2}dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\pi }\pi {\text{sin}}^{2}xdx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\pi }\frac{\pi }{2}({\text{sin}}^{2}x+{\text{sin}}^{2}x)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\pi }\frac{\pi }{2}({\text{sin}}^{2}x+1-{\text{cos}}^{2}x)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\pi }\frac{\pi }{2}(1-({\text{cos}}^{2}x-{\text{sin}}^{2}x))dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=\pi }\frac{\pi }{2}(1-\text{cos}\hspace{0.17em}2x)dx}\\ & =& {\displaystyle \frac{\pi }{2}(x-\frac{\text{sin}\hspace{0.17em}2x}{2}){|}_{x=0}^{x=\pi }}\\ & =& {\displaystyle \frac{\pi }{2}(\pi -\frac{\text{sin}\hspace{0.17em}2\pi }{2})-\frac{\pi }{2}(0-\frac{\text{sin}\hspace{0.17em}0}{2})}\\ & =& {\displaystyle \frac{\pi }{2}(\pi -0)}\\ & =& {\displaystyle \frac{{\pi }^2}{2}\text{.}}\end{array}$$

Find the volume generated by rotating the area bounded by the curves $y=3x-x^2$ and $y=x$ about the $x\text{-axis.}$ $$\begin{array}{c} 3 x 2 y y=x y=3x-x2 \end{array}\begin{array}{c}\text{Slice}\begin{array}{c} Ri R0 \end{array}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=2}(\pi R_0^2-\pi R_i^2)dx}& =& {\displaystyle {\int }_{x=0}^{x=2}(\pi y_{\text{top}}^2-\pi y_{\text{bottom}}^2)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}(\pi {(3x-x^2)}^2-\pi x^2)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}\pi (9x^2-6x^3+x^4-x^2)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}\pi (8x^2-6x^3+x^4)dx}\\ & =& {\displaystyle \pi (\frac{8x^3}{3}-\frac{6x^4}{4}+\frac{x^5}{5}){|}_{x=0}^{x=2}}\\ & =& {\displaystyle \pi (\frac{8·8}{3}-\frac{6·2^4}{4}+\frac{2^5}{5})-\pi (0-0+0)}\\ & =& {\displaystyle \pi 2^5(\frac{2}{3}\frac{3}{4}+\frac{1}{5})}\\ & =& {\displaystyle 32\pi (\frac{40}{60}-\frac{45}{60}+\frac{12}{60})}\\ & =& {\displaystyle \frac{32\pi ·7}{60}}\\ & =& {\displaystyle \frac{8·7\pi }{15}}\\ & =& {\displaystyle \frac{56\pi }{15}\text{.}}\end{array}$$

A barrel of height $h$ and maximum radius $R$ is constructed by rotation of the parabola $y=R-c{x}^2,$ $-\frac{h}{2}\le x\le \frac{h}{2}\text{.}$ $$\begin{array}{c} - h 2 h 2 x R - R y y=R-cx2 \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} y ⏟ dx \end{array}\text{is}\hspace{0.17em}\pi y^{2}dx\\ \text{Add up slices from}\hspace{0.17em}x=-\frac{h}{2}\hspace{0.17em}\text{to}\hspace{0.17em}\frac{h}{2}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=-\frac{h}{2}}^{x=\frac{h}{2}}\pi y^{2}dx}& =& {\displaystyle {\int }_{x=-\frac{h}{2}}^{x=\frac{h}{2}}\pi {(R-c{x}^2)}^{2}dx}\\ & =& {\displaystyle {\int }_{x=-\frac{h}{2}}^{x=\frac{h}{2}}\pi (R^2-2cR{x}^2+c^2{x}^4)dx}\\ & =& {\displaystyle \pi (R^{2}x-2cR\frac{x^3}{3}+c^2\frac{x^5}{5}){|}_{x=-\frac{h}{2}}^{x=\frac{h}{2}}}\\ & =& {\displaystyle \pi (R^2\frac{h}{2}-\frac{2cR}{3}\frac{h^3}{8}+\frac{c^2}{5}\frac{h^5}{2^5})-\pi (R^2\frac{(-h)}{2}+\frac{2cR}{3}\frac{h^3}{2^3}+\frac{(-c^2{h}^5)}{5·2^5})}\\ & =& {\displaystyle \pi (R^{2}h-\frac{2cR{h}^3}{3·8}+\frac{2c^2{h}^5}{5·2^5})}\\ & =& {\displaystyle \pi (R^{2}h-\frac{cR{h}^3}{6}+\frac{c^2{h}^5}{5·16})\text{.}}\end{array}$$

You are given two spherical balls of wood, one of radius $r$ and a second one of radius $R\text{.}$ A circular hole is bored through each ball and the resulting napkin rings have height $h\text{.}$ Which napkin ring contains more wood? $$\begin{array}{c} h \end{array}$$ $$\begin{array}{c} -r r x -r r y x2+y2=r2 \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} } H x dx ⏟ \end{array}\text{is}\hspace{0.17em}2\pi xHdx\text{.}\\ \text{Add slices from}\hspace{0.17em}x=\sqrt{r^2-{\left(\frac{1}{2}h\right)}^2}\hspace{0.17em}\text{to}\hspace{0.17em}x=r\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=\sqrt{r^2-\frac{h^2}{4}}}^{x=r}2\pi x2ydx}& =& {\displaystyle {\int }_{x=\sqrt{r^2-\frac{h^2}{4}}}^{x=r}4\pi x\sqrt{r^2-x^2}dx}\\ & =& {\displaystyle {\int }_{x=\sqrt{r^2-\frac{h^2}{4}}}^{x=r}(-2\pi )(-2x){(r^2-x^2)}^{\frac{1}{2}}dx}\\ & =& {\displaystyle -2\pi {(r^2-x^2)}^{\frac{3}{2}}\frac{2}{3}{|}_{x=\sqrt{r^2-\frac{h^2}{4}}}^{x=r}}\\ & =& {\displaystyle -2\pi {(r^2-r^2)}^{\frac{3}{2}}\frac{2}{3}-(-2\pi ){(r^2-(r^2-\frac{h^2}{4}))}^{\frac{3}{2}}\frac{2}{3}}\\ & =& {\displaystyle 0+2\pi {\left(\frac{h^2}{4}\right)}^{\frac{3}{2}}\frac{2}{3}}\\ & =& {\displaystyle 2\pi {\left(\frac{h}{2}\right)}^3\frac{2}{3}}\\ & =& {\displaystyle \frac{4\pi }{3}\frac{h^3}{8}}\\ & =& {\displaystyle \frac{\pi h^3}{8}\text{.}}\end{array}$$ This doesn't depend on $r\text{!!}$ So both napkin rings contain the same amount of wood.

Find the volume of a tetrahedron where each side of the tetrahedron is an equilateral triangle with side length $a\text{.}$ $$\begin{array}{c} a a a a \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} s s s } dy \end{array}\\ \text{is}dy\frac{{\displaystyle \left(\text{base}\right)\left(\text{height}\right)}}{2}=\frac{{\displaystyle \left(\frac{1}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right)}}{2}dy\\ \text{Add up slices from}\hspace{0.17em}y=0\hspace{0.17em}\text{to}\hspace{0.17em}y=\text{height of tetrahedron}=H\text{.}\end{array}$$ $$\begin{array}{c} H a d \end{array}\begin{array}{c} a a a d \end{array}\frac{\frac{1}{2}a}{d}\text{cos}\hspace{0.17em}{30}^{\circ }=\frac{\sqrt{3}}{2}$$ So $$d=\frac{\frac{1}{2}a}{\frac{\sqrt{3}}{2}}=\frac{a}{\sqrt{3}}\text{.}$$ So $$H=\sqrt{a^2-{\left(\frac{a}{\sqrt{3}}\right)}^2}=\sqrt{a^2-\frac{a^2}{3}}=\sqrt{\frac{2a^2}{3}}=a\frac{\sqrt{2}}{\sqrt{3}}\text{.}$$ So we want $$2{\int }_{y=0}^{y=H}\frac{\left(\frac{1}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right)}{2}dy={\int }_{y=0}^{y=H}2·\frac{\sqrt{3}{s}^2}{8}dy\text{.}$$ $$\begin{array}{c} { { ⏟ } s y a3 s3 a23-y=H-y \end{array}$$ But $H-y=\frac{\sqrt{2}}{\sqrt{3}}s\text{.}$ So $y=H-\frac{\sqrt{2}}{\sqrt{3}}s\text{.}$ So $\frac{dy}{ds}=-\frac{\sqrt{2}}{\sqrt{3}}\text{.}$ $$\begin{array}{ccc}{\displaystyle {\int }_{y=0}^{y=H}2·\frac{\sqrt{3}}{8}{s}^{2}dy}& =& {\displaystyle {\int }_{y=0}^{y=H}2·\frac{\sqrt{3}}{8}{s}^2\frac{dy}{ds}ds}\\ & =& {\displaystyle {\int }_{y=0}^{y=H}2·\frac{\sqrt{3}}{8}{s}^2(-\frac{\sqrt{2}}{\sqrt{3}})ds}\\ & =& {\displaystyle 2·\frac{\sqrt{3}}{8}\frac{s^3}{3}(-\frac{\sqrt{2}}{\sqrt{3}}){|}_{y=0}^{y=H}}\\ & =& {\displaystyle -\frac{\sqrt{2}}{8}\frac{2}{3}{s}^3{|}_{s=a}^{s=0}}\\ & =& {\displaystyle (-\frac{\sqrt{2}·2}{8·3}{0}^3)-(-\frac{\sqrt{2}·2}{8·3}{a}^3)}\\ & =& {\displaystyle \frac{\sqrt{2}·2}{24}{a}^3}\\ & =& {\displaystyle \frac{\sqrt{2}}{12}{a}^3\text{.}}\end{array}$$ Since $H=\frac{a\sqrt{2}}{\sqrt{3}}$ we can also write this as $$\frac{\sqrt{2}}{12}{a}^3=\left(\frac{\sqrt{2}}{\sqrt{3}}a\right)\frac{\sqrt{3}}{12}{a}^2=H\frac{\sqrt{3}{a}^2}{12}=\frac{\sqrt{3}{a}^{2}H}{12}\text{.}$$

Find the volume generated by rotating the area bounded by $y=3x-x^2$ and $y=x$ about the $y\text{-axis}$ $$\begin{array}{c} x y \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} } H ⏟ dx R \end{array}\text{is}\hspace{0.17em}2\pi RHdx\text{.}\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=2\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=2}2\pi RHdx}& =& {\displaystyle {\int }_{x=0}^{x=2}2\pi x(y_{\text{upper}}-y_{\text{lower}})dx}\\ & =& {\displaystyle {\&Intersect;}_{x=0}^{x=2}2\pi x(3x-x^2-x)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}2\pi x(2x-x^2)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=2}2\pi (2x^2-x^3)dx}\\ & =& {\displaystyle 2\pi (\frac{2x^3}{3}-\frac{x^4}{4}){|}_{x=0}^{x=2}}\\ & =& {\displaystyle 2\pi (\frac{2}{3}·8-\frac{16}{4})-2\pi (0-0)}\\ & =& {\displaystyle 2\pi (\frac{16}{3}-\frac{16}{4})}\\ & =& {\displaystyle 2\pi ·16(\frac{1}{3}-\frac{1}{4})}\\ & =& {\displaystyle \frac{32\pi }{12}}\\ & =& {\displaystyle \frac{16\pi }{6}}\\ & =& {\displaystyle \frac{8\pi }{3}\text{.}}\end{array}$$

Find the volume generated by revolving the triangle with vertices $(1,1),$ $(1,2)$ and $(2,2)$ about the $y\text{-axis.}$ $$\begin{array}{c} x y y=x \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} } H R dx ⏟ \end{array}\text{is}\hspace{0.17em}2\pi RHdx\text{.}\\ \text{Add slices from}\hspace{0.17em}x=1\hspace{0.17em}\text{to}\hspace{0.17em}x=2\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=1}^{x=2}2\pi RHdx}& =& {\displaystyle {\int }_{x=1}^{x=2}2\pi x(2-x)dx}\\ & =& {\displaystyle {\int }_1^{2}2\pi (2x-x^2)dx}\\ & =& {\displaystyle 2\pi (x^2-\frac{x^3}{3}){|}_{x=1}^{x=2}}\\ & =& {\displaystyle 2\pi (4-\frac{8}{3})-2\pi (1-\frac{1}{3})}\\ & =& {\displaystyle 2\pi \left(\frac{4}{3}\right)-2\pi \left(\frac{2}{3}\right)}\\ & =& {\displaystyle \frac{4\pi }{3}\text{.}}\end{array}$$

Find the volume of a slice obtained by chopping off the end of a sphere of radius $r,$ if the slice has thickness $h$ (at its thickest point). $$\begin{array}{c} r r x r r h y x2+y2=r2 \end{array}\begin{array}{c}\text{Volume of a slice}\begin{array}{c} } dy R \end{array}\text{is}\hspace{0.17em}\pi R^{2}dy\text{.}\\ \text{Add slices from}\hspace{0.17em}y=h\hspace{0.17em}\text{to}\hspace{0.17em}y=r\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{y=h}^{y=r}\pi R^{2}dy}& =& {\displaystyle {\int }_{y=h}^{y=r}\pi x^{2}dy}\\ & =& {\displaystyle {\int }_{y=h}^{y=r}\pi (r^2-y^2)dy}\\ & =& {\displaystyle \pi (r^{2}y-\frac{y^3}{3}){|}_{y=h}^{y=r}}\\ & =& {\displaystyle \pi (r^3-\frac{r^3}{3})-\pi (r^{2}h-\frac{h^3}{3})}\\ & =& {\displaystyle \pi \frac{2}{3}{r}^3-\pi r^{2}h+\frac{\pi h^3}{3}}\\ & =& {\displaystyle \frac{\pi }{3}(2r^3-3r^{2}h+h^3)\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 29 from a series of handwritten lecture notes for the class MATH 221 given on November 15, 2000.

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