Science:Math Exam Resources/Courses/MATH103/April 2009/Question 06 (b)

MATH103 April 2009

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[hide] Question 06 (b)
Seeds are dispersed from a tree in all directions uniformly. The density of seeds that lands on the ground at a distance r away is approximately described by the function $s(r)=50\sin \left({\frac {\pi r}{10}}\right){\text{ seeds/m}}^{2},\quad 0\leq r\leq 10{\text{ m}}$ What is the total number of seeds on the ground within a 10 m radius of the tree?

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[show] Hint 1
We have the density $\ s$ of seeds per $\ m^{2}$ given and are searching for the number of seeds. So, an integral must be involved in the calculation.

[show] Hint 2
The seeds lay radially symmetric around the tree. So, it is a circular area with radius $\ 10m$ , where the seeds are distributed. The volume of the seeds are a solid of revolution with the function $\ s(r)$ rotated around the tree (=y-axis).

[show] Hint 3
For the solid of revolution use the shell method $\ \int 2\pi xh(x)dx$ . For the integration use integration by parts.

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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. The total number of seeds is the sum of all $\ m^{2}$ around the tree times the passend concentration of seeds. Because the area around the tree is a ring, the field of seeds looks like a the function rotated around the tree (y-axis). To calculate this integral, we use the shell method $\displaystyle \int 2\pi xh(x)dx$ For integration we use integration by parts. ${\begin{aligned}\int _{0}^{10}2\pi \underbrace {r} _{=u}50\underbrace {\sin \left({\frac {\pi r}{10}}\right)dr} _{dv}&=100\pi \left[\left.-r\cos \left({\frac {\pi r}{10}}\right){\frac {10}{\pi }}\right\|_{0}^{10}+{\frac {10}{\pi }}\int _{0}^{10}\cos \left({\frac {\pi r}{10}}\right)dr\right]\\&=100\left[-10\cdot 10\cos(\pi )+0+10\left.\sin \left({\frac {\pi r}{10}}\right){\frac {10}{\pi }}\right\|_{0}^{10}\right]\\&=10000.\end{aligned}}$ There are 10000 seeds around the tree.