# Thread: integration... I think

1. ## integration... I think

I need help with the problem below (I'm not sure where to start...this was in the same section as rotating a region and finding volumes and integrations.....I'm not sure where the fits in) ....Thanks!
Here is the problem:

The circumference of a tree at different heights above the ground is given in the table below. Assume that all horizontal cross-sections of the tree are circles.

Height (inches) 0 30 60 90 120
Circumference (inches) 30 26 20 12 4

(a) What is an upper estimate of the volume of the tree from the ground to a height of 30 inches?

(b) What is an upper estimate of the total volume of the tree? Use N=4.

2. ## Riemann Sum

I guess we are allowed to assume that the diameter of the tree always gets thinner (or stays the same) the higher one measures on the tree. In other words between two of the given heights the tree isn't bulging outwards .

Think about flipping the tree on its side and think of radius as a function of height r(h) given by the table below. Then you will have a series of points:
$\displaystyle r(h): (0,\frac{30}{2\pi}), (30,\frac{26}{2\pi}), (60, \frac{20}{2\pi}), (90,\frac{12}{2\pi}), (120,\frac{4}{2\pi})$

An upper estimate of the volume of the whole tree can be found by using the Left Reimann sum where we find the volume of the 5 cylinders with radii:
$\displaystyle \frac{30}{2\pi}, \frac{26}{2\pi}, \frac{20}{2\pi}, \frac{12}{2\pi}, \frac{4}{2\pi}$

So the first cylinder (base of the tree) has a height of 30 (from h=0 to h=30) and a radius of $\displaystyle r(0) = \frac{30}{2\pi}$

The second cylinder has a height of 30 (from h=30 to h=60) and a radius of $\displaystyle r(30) = \frac{26}{2\pi}$

There might be better approximations of the actual volume of the tree. For example instead of defining rectangles to approximate the shape of the tree we could use trapezoids by connecting consecutive points using straight lines. You can then use the techniques of rotating a region to find the volume of each section. (Averaging the left and right Reimann sums will give the same result).