Chapter 15.3, Problem 27E

To determine

Find the amount of material is required for patch.

Answer to Problem 27E

  $10Squarefeet$

Explanation of Solution

Given information:

A patch is to be sewn onto a flag, if each unit in the coordinate system represents one foot.

  

Calculation:

The amount of material is required for patch, find the area of the patched region shown in given figure,

  $y=\frac{1}{2}{x}^{3}andthex-axisfromx=1to3$

By integration find the area,

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{3}{\int }}\frac{1}{2}{x}^{3}dx}\\ =\frac{1}{2}{\displaystyle \underset{0}{\overset{3}{\int }}{x}^{3}dx}-\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\end{array}$

  $firstconsider,{\displaystyle \underset{0}{\overset{3}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(x_i\right)}^3\Delta x}$

  $\Delta x=\frac{3-0}{n}=\frac{3}{n}and{x}_i=a+i\Delta x=0+i\frac{3}{n}$

  $putallin{\displaystyle \underset{0}{\overset{3}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(x_i\right)}^3\Delta x}$

  ${\displaystyle \underset{0}{\overset{3}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{3i}{n}\right)}^3}\frac{3}{n}$

  $\begin{array}{l}=81\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}\\ =81\underset{n\to \infty }{\lim }\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\\ =81\underset{n\to \infty }{\lim }\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{1}^3}+2^3+\mathrm{..}+n^3\end{array}$

  $=81\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(1^3+2^3+\mathrm{..}+n^3\right)$

  $=81\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right)$

  $=81\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(\frac{n^2{\left(1+\frac{1}{n}\right)}^2}{4n^2}\right)$

  $=\frac{81}{4}$

Now consider,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(x_i\right)}^3\Delta x}\\ \Delta x=\frac{1-0}{n}=\frac{1}{n}and{x}_i=a+i\Delta x=0+i\frac{1}{n}\\ putallin{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(x_i\right)}^3\Delta x}\\ {\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n{\left(\frac{i}{n}\right)}^3}\frac{1}{n}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(1^3+2^3+\mathrm{..}+n^3\right)\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right)\\ =\underset{n\to \infty }{\lim }\frac{1}{n^4}\left(\frac{n^2{\left(1+\frac{1}{n}\right)}^2}{4n^2}\right)\\ =\frac{1}{4}\end{array}$

Now,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2}{x}^{3}dx}=\frac{1}{2}{\displaystyle \underset{0}{\overset{3}{\int }}{x}^{3}dx}-\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ =\frac{1}{2}\left(\frac{81}{4}\right)-\frac{1}{2}\left(\frac{1}{4}\right)\\ =\frac{80}{8}=10\end{array}$

The area of patched region is, $10Squarefeet$ ,

Hence, material is required to sew the patch on flag is $10Squarefeet$ .