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 27

Use integration to find the area of the triangle with vertices $(-1,1),$ $(0,5)$ and $(3,2)\text{.}$ $$\begin{array}{c} -1 3 x 5 y y-1=4(x+1) y-5=-33(x-0) y-1=14(x+1) L1 L2 \end{array}$$ $$\begin{array}{ccc}\begin{array}{c}\text{Type 1 slice:}\begin{array}{c} L1 dx \end{array}\\ \text{Area of type 1 slice:}\hspace{0.17em}{L}_{1}dx\\ \text{Add slices from}\hspace{0.17em}x=-1\hspace{0.17em}\text{to}\hspace{0.17em}x=0\end{array}& & \begin{array}{c}\text{Type 2 slice:}\begin{array}{c} L2 dx \end{array}\\ \text{Area of type 2 slice:}\hspace{0.17em}{L}_{2}dx\\ \text{Add slices from}\hspace{0.17em}x=0\hspace{0.17em}\text{to}\hspace{0.17em}x=3\end{array}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\int }_{x=-1}^{x=0}{L}_{1}dx+{\int }_{x=0}^{x=3}{L}_{2}dx}& =& {\displaystyle {\int }_{x=-1}^{x=0}(y_{\text{top 1}}-y_{\text{bottom}})dx+{\int }_{x=0}^{x=3}(y_{\text{top 2}}-y_{\text{bottom}})dx}\\ & =& {\displaystyle {\int }_{x=-1}^{x=0}((4(x+1)+1)-(\frac{1}{4}(x+1)+1))dx+{\int }_{x=0}^{x=3}((-x+5)-(\frac{1}{4}(x+1)+1))dx}\\ & =& {\displaystyle {\int }_{x=-1}^{x=0}(4x+5-\frac{1}{4}x-\frac{1}{4}-1)dx+{\int }_{x=0}^{x=3}(-x+5-\frac{1}{4}x-\frac{1}{4}-1)dx}\\ & =& {\displaystyle {\int }_{x=-1}^{x=0}(\frac{15}{4}x+\frac{15}{4})dx+{\int }_{x=0}^{x=3}(-\frac{5}{4}x+\frac{15}{4})dx}\\ & =& {\displaystyle (\frac{15}{4}\frac{x^2}{2}+\frac{15}{4}x){|}_{x=-1}^{x=0}+(-\frac{5}{4}\frac{x^2}{2}+\frac{15}{4}x){|}_{x=0}^{x=3}}\\ & =& {\displaystyle (\frac{15}{4}·0+\frac{15}{4}·0)-(\frac{15}{4}\frac{{(-1)}^2}{2}+\frac{15}{4}(-1))+(-\frac{5}{4}·\frac{3^2}{2}+\frac{15}{4}·3)-(0+0)}\\ & =& {\displaystyle 0+0-\frac{15}{8}+\frac{15}{4}-\frac{45}{8}+\frac{45}{4}=\frac{15}{8}+\frac{45}{8}=\frac{60}{8}=7\frac{1}{2}\text{.}}\end{array}$$

Find the curved surface area of a cone of radius $r$ and height $h$ (a right circular cone). $$\begin{array}{c} s r h \end{array}$$ Cut the cone open and lay it out to get $$\begin{array}{c} s C θ \end{array}$$ The region $C$ is a portion of a circle of radius $s,$ where $s$ is the slant height of the cone. The area of $C$ is $\frac{1}{2}\theta s^2\text{.}$ The arc length along the border of $C$ is $\theta s\text{.}$ This arc length is also the length around the circle at the base of the cone, which is $2\pi r\text{.}$ So $$\theta s=2\pi r\text{.}$$ so $$\begin{array}{ccc}{\displaystyle \text{curved surface area}}& =& {\displaystyle \frac{1}{2}\theta s^2}\\ & =& {\displaystyle \frac{1}{2}\left(\theta s\right)s}\\ & =& {\displaystyle \frac{1}{2}\left(2\pi r\right)s}\\ & =& {\displaystyle \pi rs}\\ & =& {\displaystyle \pi r\sqrt{h^2+r^2}\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 27 from a series of handwritten lecture notes for the class MATH 221 given on November 10, 2000.

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