Chapter 5.1, Problem 24E

(a)

To determine

To find : The value of $S$ .

Answer to Problem 24E

The answer: $S=372.27873{\text{m}}^3$

Explanation of Solution

Given information:

Radius $=8m$

And depth $=4m$

Calculation:

The volume of this region can then be approximated by cylinders with a radius of $\sqrt{64-x^2}a$ and height of $\text{Δ}x$ This then gives a volume of $\pi {\left(\sqrt{64-x^2}\right)}^2\text{Δ}x=\pi \left(64-x^2\right)\text{Δ}$ for each cylinder. The total volume is then ${{\displaystyle \int }}_4^8\pi \left(64-x^2\right)dx$

Need to estimate ${{\displaystyle \int }}_4^8\pi \left(64-x^2\right)dx$ with eight cylinders type $Y_1=\pi \left(64-x^2\right)$ a left endpoint of 4, right endpoint of 8, and number of intervals of 8 into your RAM program on your graphing calculator. Since the cylinders must be circumscribed, must specifically use LRAM so the rectangles will be above the graph. This then gives $S=372.27873{\text{m}}^3$ as the volume. Since the rectangles are above the graph, this value of $S$ is an overestimate.

The required volume of these cylinders is $S=372.27873{\text{m}}^3$

(b)

To determine

To find : The error $|V-S|$ as a percentage of $V$

Answer to Problem 24E

The answer: $11\text{\%}$

Explanation of Solution

Given information:

Radius $=8m$

And depth $=4m$

Calculation:

Given that,

  $V=\frac{320\pi }{3}$ and from part $(a)$

  $S=372.27873$

This then gives as error of

  $\begin{array}{c}\frac{|V-S|}{V}\approx =\frac{\left|\frac{320\pi }{3}-372.27873\right|}{\frac{320\pi }{3}}\\ =0.11\\ =11\text{\%}\end{array}$

Therefore the required error $|V-S|$ as a percentage of $V$ is $11\text{\%}$ .