MHB OliverG's Calculus Q: Estimating Pool Volume w/ Trap Rule

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The discussion focuses on estimating the volume of a circular swimming pool using the trapezoidal rule. The pool has a diameter of 28 feet and varying depths from 3 to 9 feet. The volume is approximated by decomposing the pool into trapezoidal prisms, with calculations based on depth measurements and circular chord lengths. The final estimated volume is approximately 3429.71 cubic feet, which converts to about 25,594.84 gallons. The method effectively demonstrates how the trapezoidal rule can be applied to a three-dimensional figure.
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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.
 
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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:

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

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

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

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

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

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

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

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

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

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

$$V\approx$$

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

Simplifying a bit:

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

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

In gallons this is about:

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