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

[MarkFL]

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]

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:

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