# Finding volume of swimming pool

• Charanjit
In summary: However, once I get used to them, they're actually pretty easy. I think the key to success is breaking the problem down into simpler pieces and then tackling those one at a time. That's what I did with this one. Start with the height and work your way down. I'm sorry if that wasn't clear, but hopefully it was helpful.
Charanjit

## Homework Statement

A swimming pool is circular with a 40-ft. diameter. The depth is constant along east-west lines and increases linearly from 2 ft.at the south end to 7 ft. at the north end. Find the volume of water in the pool.

## Homework Equations

Dont know how to enter integrals here, but we need to use the polar coordinates here.

Refer to this: http://tutorial.math.lamar.edu/Classes/CalcIII/DIPolarCoords.aspx

## The Attempt at a Solution

I know how to integrate and find the volume. I am having trouble decomposing the problem into a picture and finding the limits.

What's the average depth of the pool?

Bah.

Dump a known quantity of dye or marker chemical in the pool. Run the pumps till it's evenly distributed. Take a water sample and test the concentration of the dye. Poof. You have the volume.

Works on closed bodies of water of any size & shape.

I think that there is an easier way. Take the shape and flip another one and place it underneath it so the two shapes form a perfect cylinder. As yourself, what will be the height of the cylinder? Once you know this, you can easily calculate the volume of the cylinder which will be twice the volume of the original pool in question.

We have two winning entries:

SammyS said:
What's the average depth of the pool?

hunt_mat said:
I think that there is an easier way. Take the shape and flip another one and place it underneath it so the two shapes form a perfect cylinder. As yourself, what will be the height of the cylinder? Once you know this, you can easily calculate the volume of the cylinder which will be twice the volume of the original pool in question.

Or simply duplicate the sloped section, then halve it.

So, one cylinder that's 2 ft high, plus one cylinder that's 7-2=5 ft high, which will be halved.

Question: if if this a calculus problem, are these solutions cheating?

DaveC426913 said:
...
Question: if if this a calculus problem, are these solutions cheating?

Probably, but OP has not returned to the scene of the crime!

Okay, here is my idea, the has to be split up into two parts, that which is just a cylinder and that with the sloped part. I think that the key is to come up with the equation for the slope, and this will be $z=z(r)$, so if we sit our cylinder so the centre of it is is sat at $r=0$ then the equation of the slope will be:
$$\frac{5}{40}=\frac{z-0}{r-20}$$
So this gives us a $z=z(r)$, which we can invert to give $r=r(z)$, and the integral becomes:
$$V=\int_{0}^{5}\int_{0}^{2\pi}\int_{0}^{r(z)}rdrd \theta dz$$

Or something like that...

z is not a function of just r.

Please, let's let the OP reply before we offer more help.

I know it was said not to reply anymore until we hear from OP, and I feel bad about posting, but I must have the same calculus book because I remember this problem from just a few weeks ago. I think I can give a valuable response since I know the format being asked for in this specific case, and I can answer using only what Charanjit should know.

First, think about the pool. It is a cylinder. The height doesn't matter. Taking your cue from the section title, "double integrals in polar coordinates", let the polar element take care of the circle, that is, the base of the pool. What is r, what is theta? (we're talking about a circle here, it's clear what they should be) Now, forget all about the 3 dimensional aspect of the pool and see only a flat surface, the pool from the side. Really force the dimension out of your head. On a regular 2D graph (x,y), translate the height of the water as if you had a FLAT wall of water starting at a depth of 2 feet, and rising up to 7. Did you start at the origin? If you overlay a circle (the base of the pool) you may find that it is better to CENTER your water depth line. Be sure you start at X=-20 and end at 20. Once you do, you can figure out your cartesian equation of the line, and the do the polar integral translation stuff (change x's to rcos(theta) and y's (you shouldn't have any here) to rsin(theta) and don't forget to multiply the whole thing by another r. Integrate the way you should, dr dtheta, and use the upper and lower bounds that are natural for simple polar area problems using a circle with radius 20 centered at the origin that you chose way back before this whole picture got confusing. If your line equation is correct, and you made no other mistakes, you will arrive at the correct answer. I checked in the back of the book, I got it.

I find polar integrals to be visually counterintuitive. My integral never looks like it will give me the answer I'm aiming for, but if you don't let that bother you, and always let round stuff be polar and cornered stuff be Cartesian, it all works out in the end.

## 1. How do I calculate the volume of a swimming pool?

To calculate the volume of a swimming pool, you will need to measure the length, width, and average depth of the pool. Then, multiply these measurements together and multiply the result by 7.5 to convert the volume to gallons. The formula is: Volume = Length x Width x Average Depth x 7.5.

## 2. Can I use any unit of measurement to calculate the volume of a swimming pool?

Yes, you can use any unit of measurement as long as all the measurements are in the same unit. For example, if you measure the length in feet, the width in meters, and the depth in inches, you will need to convert all the measurements to the same unit before calculating the volume.

## 3. What is the average depth of a swimming pool?

The average depth of a swimming pool is the depth at the deepest point of the pool and the shallowest point of the pool added together and divided by 2. This will give you the average depth of the pool, which is needed to calculate the volume.

## 4. Do I need to account for the shape of the swimming pool when calculating the volume?

Yes, you will need to account for the shape of the swimming pool when calculating the volume. The formula for calculating the volume is based on the assumption that the pool has a rectangular shape. If the pool has a different shape, you will need to use a different formula or break the pool down into smaller sections with similar shapes and calculate the volume for each section separately.

## 5. How accurate are volume calculations for swimming pools?

The accuracy of volume calculations for swimming pools depends on the accuracy of the measurements taken. If the measurements are precise, the volume calculation should also be precise. However, other factors such as irregularities in the pool's shape, water displacement, and the accuracy of the formula used can also affect the accuracy of the volume calculation.

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