# Volume of Water in a swimming pool

• Jan 9th 2011, 05:48 PM
lancelot854
Volume of Water in a swimming pool
A rectangular swimming pool is 30 ft wide and 50 ft long. The table below shows the depth h(x) of the water at 5-ft intervals from one end of the pool to the other. Estimate the volume of water in the pool using the Trapezoidal rule with n=10, applied to the integral $\displaystyle \int^5_0 30(h(x))dx$.

I have to use the trapezoidal rule to solve this, and it all seems confusing, but probably isn't. Any help is appreciated, thanks. The 5 in the integral is suppose to be a 50, but I couldn't get the latex to work right -.-
• Jan 9th 2011, 05:54 PM
Prove It
Have you fitted a polynomial to your data yet to find the function $\displaystyle \displaystyle h(x)$?
• Jan 9th 2011, 05:55 PM
pickslides
Hi there lancelot854,

First thing, click on this to see $\displaystyle \displaystyle \int_0^{50} 30\times h(x)~dx$ the latex code. Hope that helps for future posts!

Now do you have a rule for $\displaystyle \displaystyle h(x)$ ? As you are integrating between 0 and 5, should you not be beaking this into intervals (n=10) $\displaystyle \displaystyle \frac{5-0}{10} \implies h = 0.5$ ?
• Jan 9th 2011, 05:57 PM
Prove It
Quote:

Originally Posted by pickslides
Hi there lancelot854,

First thing, click on this to see $\displaystyle \displaystyle \int_0^{50} 30\times h(x)~dx$ the latex code. Hope that helps for future posts!

Now do you have a rule for $\displaystyle \displaystyle h(x)$ ? As you are integrating between 0 and 5, should you not be beaking this into intervals (n=10) $\displaystyle \displaystyle \frac{5-0}{10} \implies h = 0.5$ ?

Actually, the OP is integrating from $\displaystyle \displaystyle 0$ to $\displaystyle \displaystyle 50$...
• Jan 9th 2011, 05:59 PM
pickslides
Quote:

Originally Posted by Prove It
Actually, the OP is integrating from $\displaystyle \displaystyle 0$ to $\displaystyle \displaystyle 50$...

Thanks HM, even I should be able to read this from my post!
• Jan 9th 2011, 06:14 PM
lancelot854
All the information I wrote was all that I was given, I don't understand how to proceed from this point.
• Jan 9th 2011, 06:18 PM
Prove It
I'd start by graphing the data. What type of function do you think this data models?
• Jan 9th 2011, 06:20 PM
pickslides
Hey lancelot854, you need to find a polynomial for h(x) that models your data.

Do you know how to find this?

Kind regards,
• Jan 9th 2011, 06:30 PM
lancelot854
I know graphing to would form a line, however, I forgot how to find the formula for the line.
• Jan 9th 2011, 06:31 PM
Prove It
I doubt that this function is linear anyway. Since you have 11 pieces of data, you will need to fit a polynomial of degree 11-1 = 10.
• Jan 10th 2011, 11:47 AM
LoblawsLawBlog
I don't see why you would need to fit a polynomial to your data. The whole point of the trapezoid rule is to use discrete points to estimate the integral, and you've been given your points already.

Try plotting the depth versus the distance from the end of the pool and connecting each point to the next one by a line segment. Those will be your 10 trapezoids.
• Jan 10th 2011, 08:51 PM
Prove It
Quote:

Originally Posted by LoblawsLawBlog
I don't see why you would need to fit a polynomial to your data. The whole point of the trapezoid rule is to use discrete points to estimate the integral, and you've been given your points already.

Try plotting the depth versus the distance from the end of the pool and connecting each point to the next one by a line segment. Those will be your 10 trapezoids.

Ten trapeziums will not give a very accurate answer. For a more accurate answer, you will need more trapeziums, which is why you will need a polynomial.
• Jan 11th 2011, 12:56 PM
lancelot854
I actually solved this by using the 10 trapezoids, although it gave the same answer as the one required, I'd like to know how to find the polynomial to find a more accurate answer, thanks for the help.
• Jan 11th 2011, 01:36 PM
LoblawsLawBlog
You got the same answer because that's all they wanted you to do. The problem was just testing whether you could go through the steps of the trapezoid approximation.

Fitting a polynomial to those points could give a more accurate answer if this was a "real world" problem, or it could be a disaster. It's easy to find a bunch of points that are practically collinear, but when you plot the polynomial through them it looks nothing like the real data.

But if you really want to know how to fit a polynomial to data, it just reduces to solving a system of linear equations. For example, to get a parabola through the points (1,2), (2,3), and (3, 8), you would solve
$\displaystyle a(1)^2+b(1)+c=2, a(2)^2+b(2)+c=3, a(3)^2+b(3)+c=8$ (let's see how my first attempt at latex on this forum goes). Obviously for 11 points you'll want to use a computer.

edit: Just to clarify, when I say "the" polynomial I mean the interpolating polynomial that goes through every point. A lower degree polynomial that approximates the data may work well, but that's kind of overthinking the problem.
• Jan 11th 2011, 01:59 PM
skeeter
$\displaystyle \displaystyle V = 30 \int_0^{50} h(x) \, dx \approx 30 \cdot \frac{5}{2} \left[h(0) + 2h(5) + 2h(10) + ... + 2h(40) + 2h(45) + h(50) \right]$