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 26

Computing Areas and Volumes

(1)	Carefully draw the region.
(2)	Slice it up; draw a typical slice.
(3)	Find the volume of a slice.
(4)	Add up the volumes of the slices with an integral.

Typical slices might look like: $$\begin{array}{c} ⏞ L { ⏟ { L } Area } Area } Area } Area } height distance around \end{array}$$

Calculate the area of the region bounded by the parabolas $y=x^2$ and $y^2=x\text{.}$ $$\begin{array}{c} -1 1 x -1 1 y \end{array}\begin{array}{c} 1 x 1 y L y=x2 y2=x \end{array}$$ Slice: $\begin{array}{c} L dx \end{array}$
Area of Slice: $Ldx$
Add slices from $x=0$ to $x=1\text{.}$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=1}Ldx}& =& {\displaystyle {\int }_{x=0}^{x=1}(y_{\text{upper}}-y_{\text{lower}})dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=1}(\sqrt{x}-x^2)dx}\\ & =& {\displaystyle \frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^3}{3}{|}_{x=0}^{x=1}}\\ & =& {\displaystyle (\frac{2}{3}{1}^{\frac{3}{2}}-\frac{1}{3})-(\frac{2}{3}{0}^{\frac{3}{2}}-\frac{0^3}{3})}\\ & =& {\displaystyle \frac{2}{3}-\frac{1}{3}=\frac{1}{3}\text{.}}\end{array}$$

Find the area of the region bounded by $y=-1,$ $y=2,$ $x=y^3$ and $x=0\text{.}$ $$\begin{array}{c} -1 1 x -1 1 2 y y=2 y=-1 x=y3 \end{array}$$ $$\begin{array}{ccc}\begin{array}{c}\text{Type 1 slice:}\begin{array}{c} L1 dy \end{array}\\ \text{Area of slice:}{L}_{1}dy\\ \text{Add slices from}\hspace{0.17em}y=0\hspace{0.17em}\text{to}\hspace{0.17em}y=2\text{.}\end{array}& & \begin{array}{c}\text{Type 2 slice:}\begin{array}{c} L1 dy \end{array}\\ \text{Area of slice:}{L}_{2}dy\\ \text{Add slices from}\hspace{0.17em}y=-1\hspace{0.17em}\text{to}\hspace{0.17em}y=0\text{.}\end{array}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{y=0}^{y=2}{L}_{1}dy+{\int }_{y=-1}^{y=0}{L}_{2}dy}& =& {\displaystyle {\int }_{y=0}^{y=2}xdy+{\int }_{y=-1}^{y=0}(-x)dy}\\ & =& {\displaystyle {\int }_{y=0}^{y=2}{y}^{3}dy+{\int }_{y=-1}^{y=0}-y^{3}dy}\\ & =& {\displaystyle \frac{y^4}{4}{|}_{y=0}^{y=2}+\frac{-y^4}{4}{|}_{y=-1}^{y=0}}\\ & =& {\displaystyle (\frac{2^4}{4}-\frac{0}{4})+(-\frac{0^4}{4}-\left(\frac{-{(-1)}^4}{4}\right))}\\ & =& {\displaystyle 2^2+\frac{1}{4}}\\ & =& {\displaystyle 4\frac{1}{4}\text{.}}\end{array}$$

Find the volume of a sphere of radius $r\text{.}$ $$\begin{array}{c} - r r R x - r r y x2+y2=r2 \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} } dy R \end{array}\\ \text{Volume of slice:}\pi R^{2}dy\\ \text{Add slices from}\hspace{0.17em}y=-r\hspace{0.17em}\text{to}\hspace{0.17em}y=r\end{array}$$ $$\begin{array}{ccc}{\displaystyle \text{Volume of sphere}}& =& {\displaystyle {\int }_{y=-r}^{y=3}\pi R^{2}dy}\\ & =& {\displaystyle {\int }_{y=-r}^{y=r}\pi x^{2}dy}\\ & =& {\displaystyle {\int }_{y=-r}^{y=r}\pi (r^2-y^2)dy}\\ & =& {\displaystyle \pi (r^{2}y-\frac{y^3}{3}){|}_{y=-r}^{y=r}}\\ & =& {\displaystyle \pi (r^2·r-\frac{r^3}{3})-\pi (r^2(-r)-\frac{{(-r)}^3}{3})}\\ & =& {\displaystyle \pi \frac{2}{3}{r}^3+\pi r^3-\frac{\pi r^3}{3}}\\ & =& {\displaystyle \frac{2}{3}\pi r^3+\frac{2}{3}\pi r^3}\\ & =& {\displaystyle \frac{4}{3}\pi r^3\text{.}}\end{array}$$

Compute ${\int }_{-a}^a\sqrt{a^2-x^2}dx\text{.}$

If $x=a\text{sin}\hspace{0.17em}\theta ,$ then $$\begin{array}{ccc}{\displaystyle {\int }_{-a}^a\sqrt{a^2-x^2}dx}& =& {\displaystyle {\int }_{-a}^a\sqrt{a^2-a^2{\text{sin}}^2\theta }dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\sqrt{a^2{\text{cos}}^2\theta }dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}a\text{cos}\hspace{0.17em}\theta dx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}a\text{cos}\hspace{0.17em}\theta \frac{dx}{d\theta }d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}a\text{cos}\hspace{0.17em}\theta \hspace{0.17em}a\text{cos}\hspace{0.17em}\theta \hspace{0.17em}d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}{a}^2{\text{cos}}^2\theta d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\frac{1}{2}{a}^2({\text{cos}}^2\theta +{\text{cos}}^2\theta )d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\frac{1}{2}{a}^2({\text{cos}}^2\theta +1-{\text{sin}}^2\theta )d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\frac{1}{2}{a}^2({\text{cos}}^2\theta -{\text{sin}}^2\theta +1)d\theta }\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\frac{1}{2}{a}^2(\text{cos}2\theta +1)d\theta }\\ & =& {\displaystyle \frac{1}{2}{a}^2(\frac{\text{sin}\hspace{0.17em}2\theta }{2}+\theta ){|}_{x=-a}^{x=a}}\\ & =& {\displaystyle \frac{1}{2}{a}^2(\frac{\text{sin}\hspace{0.17em}2\theta }{2}+\theta ){|}_{\text{sin}\hspace{0.17em}\theta =-1}^{\text{sin}\hspace{0.17em}\theta =1}}\\ & =& {\displaystyle \frac{1}{2}{a}^2(\frac{\text{sin}\hspace{0.17em}2\theta }{2}+\theta ){|}_{\theta =-\frac{\pi }{2}}^{\theta =\frac{\pi }{2}}}\\ & =& {\displaystyle \frac{1}{2}{a}^2(\frac{\text{sin}\hspace{0.17em}\pi }{2}+\frac{\pi }{2})-\frac{1}{2}{a}^2(\frac{\text{sin}(-\pi )}{2}-\frac{\pi }{2})}\\ & =& {\displaystyle \frac{1}{2}{a}^2\frac{\pi }{2}-\frac{1}{2}{a}^2(-\frac{\pi }{2})}\\ & =& {\displaystyle \frac{\pi a^2}{4}\text{.}}\end{array}$$

Compute ${\int }_{x=-a}^{x=a}\sqrt{a^2-x^2}dx\text{.}$ $$\begin{array}{c} - a a x a y y=a2-x2 \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} L dx \end{array}\\ \text{Area of slice:}\hspace{0.17em}Ldx\\ \text{Area slices from}\hspace{0.17em}x=-a\hspace{0.17em}\text{to}\hspace{0.17em}x=a\text{.}\end{array}$$ $$\begin{array}{ccc}{\displaystyle \frac{\pi a^2}{2}}& =& {\displaystyle \text{Area of semicircle}}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}Ldx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}ydx}\\ & =& {\displaystyle {\int }_{x=-a}^{x=a}\sqrt{a^2-x^2}dx\text{.}}\end{array}$$ So $$\frac{\pi a^2}{2}={\int }_{x=-a}^{x=a}\sqrt{a^2-x^2}dx\text{.}$$

Find the volume of a right circular cone of height $h$ and radius $r\text{.}$ $$\begin{array}{c} - r r x h y y=-hrx+h \end{array}\begin{array}{c}\text{Slice:}\begin{array}{c} } H R dx ⏟ \end{array}\\ \text{Volume of slice:}\hspace{0.17em}2\pi RHdx\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=r\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=0}^{x=r}2\pi RHdx}& =& {\displaystyle {\int }_{x=0}^{x=r}2\pi xydx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}2\pi x(-\frac{h}{r}x+h)dx}\\ & =& {\displaystyle {\int }_{x=0}^{x=r}(-\frac{2\pi h}{r}{x}^2+2\pi hx)dx}\\ & =& {\displaystyle -\frac{2\pi h}{r}\frac{x^3}{3}+\pi h{x}^2{|}_{x=0}^{x=r}}\\ & =& {\displaystyle (-\frac{2\pi h}{r}\frac{r^3}{3}+\pi h{r}^2)-(-0+0)}\\ & =& {\displaystyle -\frac{2}{3}\pi r^{2}h+\pi r^{2}h}\\ & =& {\displaystyle \frac{1}{3}\pi r^{2}h\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 26 from a series of handwritten lecture notes for the class MATH 221 given on November 8, 2000.

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