# Homework Help: 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.