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Reposting from stackoverflow :) - Told to go here :)

So I've got a issue with a math assignment, and I were hoping someone could at least point me to what I should be doing, because currently I'm just wasting my time, thats for sure. - Please don't solve it all, thats up for me to do so :)

The assignment sounds like this:

A swimming pool is circular with a 10-metre diameter. The depth is constant along east-west lines and increases linearly from 1meter at the south end to 2meters at the north end. - Find the volume of water in the pool.

I figured out that its about using double integrals on polar coordinates, however from there I kinda stalled, first of all; (1) if the pool is circular and has a 10-metre diameter, am I certain that the diameter is 10metres everywhere on the pool? - or just on a single spot? - That is could the pool be oval? (2) How far goes a 'side' alike west, east, north and south, on such a circular pool? - if its circular when will I know when I'm in the east line, and what about the north-east? (3) Would it seem likely that I'm going to describe the pool as a polar function first, before even considering the use of Integrals?

I know this might not be the correct fora, for math questions, however stackoverflow has always seemed like a nice helpfull community :)


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So pretty much just make the cylinder with a diameter of 10, and then a plane which truncates it, at z=-1 at the southern end, and z=-2 at the northern end, and then just find the volume within the cylinder, from that plane to the xy-plane? – Emil Madsen Nov 17 '10 at 11:22
Using an integral here seems like overkill. – Raskolnikov Nov 17 '10 at 11:26
up vote 8 down vote accepted

Imaging one more inverted pool of same size below our original pool. They two pools will exactly fit in and will together form a cylinder of diameter 10 metres and height 3 metres.

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Clever. :) You know the volume of a cylinder, you can cut the result in half afterwards. – J. M. Nov 17 '10 at 13:01

To address the question that stalled you: a circular pool isn't oval. If the problem meant for you to consider an oval pool, it would have said "an elliptical pool with (say) horizontal axis $a$ and vertical axis $b$," or something like that.

You don't need polar coordinates for this, nor do you need a double integral. Single-variable calculus will do the trick just fine. For starters, orient the pool so that the circle which defines it is centered at the origin, and look down at the pool from overhead.

Think of the water in it as if it was a big block of jello. Now imagine slicing that block of jello horizontally infinitely many times, so that you have a stack of jello slices, each of which is infinitely thin, and each of which has its own width and depth, and each of which is located at its own $y$ coordinate (that is to say, for each slice, the points that make up the slice all have the same $y$-coordinate). Each slice is essentially a rectangular box with volume width times depth times "thinness". In the figure below, the blue lines represent the individual jello slices, with the thick one being the slice located at $y = -3$. figure of pool

Using the equation of the circle (which, since it's centered at the origin, is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle), you can figure out the width of each jello slice. Knowing that the depth increases linearly, you can also figure out the depth of each slice as a function of its $y$ coordinate.

[Actually, it's not important that the depth increase linearly, per se, just that it changes north-south (i.e. as a function of $y$) and not east-west (i.e. as a function of $x$). If it changed east-west instead, you'd slice the jello vertically instead of horizontally. If it changed in all directions, then you'd need a double integral.].

Anyway, once you know the width and the depth, you can compute the volume of each slice: $$V_{\mbox{slice}} = \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$ where $dy$ represents the infinite "thinness" of the slice. Now it's just a question of adding them up, which is exactly the point of a definite integral.

So the total volume is: $$V_{\mbox{total}} = \int_{-r}^{-r} V_{\mbox{slice}} = \int_{-r}^{r} \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$

I'll leave you to figure out the appropriate expressions (in terms of $y$) for the width and the depth of each slice.

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