OliverG's question at Yahoo! Answers: using slicing and the Trapezoidal Rule to estimate a volume

MarkFL

Staff member
Here is the question:

Calculus - Trapezoidal Rule - Cross Sections - Pools?

A circular swimming pool has diameter 28 feet. The depth of the water changes slowly from 3 feet at a point A on one side of the pool to 9 feet at a point B diametrically opposite A. Depth readings h(x) (in feet) taken along the diameter AB are given in the following table, where x is the distance (in feet) from A.

(x)==0,4,8,12,16,20,24,28
h(x)=3,3.5,4,5,6.5,8,8.5,9

Use the trapezoidal rule, with n=7, to estimate the volume of water in the pool. Approximate the number of gallons of water contained in the pool (1 gal= approx. 0.134 ft^2)

I can't understand how the trapezoidal rule can approximate a three dimensional figure.

Anyone have any ideas?
I have posted a link there to this thread so the OP can view my work.

MarkFL

Staff member
Hello OliverG,

What we are going to do here is estimate the volume of the pool by decomposing it into a series of trapezoidal prisms. For the trapezoidal faces, we need to compute circular chord lengths. For this, let's orient the origin of our coordinate axes at point $A$, and let the circular surface of the pool then be bounded by:

$$\displaystyle (x-14)^2+y^2=14^2$$

And so from this, we find the chord length $c(x)$ at $0\le x\le28$ is:

$$\displaystyle c(x)=2\sqrt{14^2-(x-14)^2}=2\sqrt{28x-x^2}$$

Now, let $h_k$ be the depth measurements given in the table. Thus, our approximating function $f$ is:

$$\displaystyle f\left(x_k \right)=2h_k\sqrt{28x_k-x_k^2}$$

where $$\displaystyle x_k=4k$$ where $$\displaystyle 0\le k\le7\,\forall\,k\in\mathbb{Z}$$

and $$\displaystyle \Delta x=x_{k+1}-x_{k}=4$$

And so we may approximate the volume of the pool with:

$$\displaystyle V\approx\sum_{k=1}^{7} \left(\frac{1}{2}\left(f\left(x_{k-1} \right)+f\left(x_{k} \right)\Delta x \right) \right)$$

Plugging in the values from our definition of $f$, we find:

$$\displaystyle V\approx$$

$$\displaystyle 4\left(3 \cdot0+2 \cdot3.5\cdot4\sqrt{6}+2 \cdot4\cdot4\sqrt{10}+2 \cdot5\cdot8\sqrt{3}+2 \cdot6.5\cdot8\sqrt{3}+2 \cdot8\cdot4\sqrt{10}+2 \cdot8.5\cdot4\sqrt{6}+9 \cdot0 \right)$$

Simplifying a bit:

$$\displaystyle V\approx16\left(7\sqrt{6}+8\sqrt{10}+20\sqrt{3}+26\sqrt{3}+16\sqrt{10}+17\sqrt{6} \right)$$

$$\displaystyle V\approx32\left(23\sqrt{3}+12\sqrt{6}+12\sqrt{10} \right)\approx3429.70807710409\text{ ft}^3$$

$$\displaystyle V\approx3429.70807710409\text{ ft}^3\frac{1\text{ gal}}{0.134\text{ ft}^3}\approx25594.8363962992\text{ gal}$$