# Finding volume of swimming pool

1. Jun 30, 2011

### Charanjit

1. The problem statement, all variables and given/known data

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.

2. Relevant 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

3. 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.

2. Jun 30, 2011

### SammyS

Staff Emeritus
What's the average depth of the pool?

3. Jun 30, 2011

### DaveC426913

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.

4. Jul 1, 2011

### hunt_mat

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.

5. Jul 1, 2011

### Redbelly98

Staff Emeritus
We have two winning entries:

6. Jul 1, 2011

### DaveC426913

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?

7. Jul 1, 2011

### SammyS

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

8. Jul 2, 2011

### hunt_mat

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....

9. Jul 2, 2011

### Redbelly98

Staff Emeritus
z is not a function of just r.

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

10. Jul 2, 2011

### ArcanaNoir

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook