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

MATH103 April 2009

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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?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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.

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).

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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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.

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