Science:Math Exam Resources/Courses/MATH101/April 2005/Question 06 (a)

MATH101 April 2005

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[hide] Question 06 (a)
A tree trunk is 60 feet long and has circular cross sections. The diameters in feet measured at 10-foot intervals starting at the bottom of the tree are 10, 8, 7, 6, 5, 4, and 3. The volume of the tree, in cubic feet, is computed using the formula $V=\int _{0}^{60}A(y)\,dy,$ where $A(y)$ is the cross-sectional area of the trunk, in square feet, $y$ feet above the bottom of the tree. (a) Using a Simpson's Rule approximation for $V$ , give an estimate for the volume of wood in the tree trunk. (Do not simplify your answer.)

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[show] Hint 1
To get the volume of the tree you will need to write a Simpson's Rule sum involving the cross sectional area of the trunk. Look closely at what you are given when evaluating the sum.

[show] Hint 2
The formula for the Simpson's rule is given by $S_{n}:={\frac {\Delta x}{3}}\left(f(x_{0})+4f(x_{1})+2f(x_{2})+...+4f(x_{n-1})+f(x_{n})\right)$ where as always $n$ is an even integer.

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This is a straightforward application of Simpson's Rule to the integral $V=\int _{0}^{60}A(y)\,dy.$ Notice that ${\frac {\Delta x}{3}}={\frac {\tfrac {60-0}{6}}{3}}={\frac {10}{3}}$ By Simpson's Rule, the integral can be approximated by the sum ${\begin{aligned}V\approx {\frac {10}{3}}(A(0)+4A(10)+2A(20)+4A(30)+2A(40)+4A(50)+A(60))\end{aligned}}$ Remembering the given information about the diameter $D(y)$ of the tree trunk at various heights, $y$ , we make the appropriate substitutions to determine the cross sectional area at each height: $A(y)=\pi \left({\frac {D(y)}{2}}\right)^{2}={\frac {\pi }{4}}D(y)^{2}$ Plugging into our sum we get the approximate volume for the tree trunk. ${\begin{aligned}V&\approx {\frac {10\pi }{12}}(10^{2}+4\times 8^{2}+2\times 7^{2}+4\times 6^{2}+2\times 5^{2}+4\times 4^{2}+3^{2})\end{aligned}}$